id
stringlengths 1
260
| contents
stringlengths 1
234k
|
|---|---|
21973
|
\section{Subset Relation is Antisymmetric}
Tags: Subsets
\begin{theorem}
The relation "is a subset of" is antisymmetric:
:$\paren {R \subseteq S} \land \paren {S \subseteq R} \iff R = S$
\end{theorem}
\begin{proof}
This is a direct statement of the definition of set equality:
:$R = S := \paren {R \subseteq S} \land \paren {S \subseteq R}$
{{qed}}
\end{proof}
|
21974
|
\section{Subset Relation is Compatible with Subset Product}
Tags: Compatible Relations, Subset Products
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\powerset S$ be the power set of $S$.
Let $\circ_\PP$ be the operation induced on $\powerset S$ by $\circ$.
Then the subset relation $\subseteq$ is compatible with $\circ_\PP$.
\end{theorem}
\begin{proof}
Let $A, B, C \in \powerset S$.
Let $A \subseteq B$.
Let $x \in A \circ_\PP C$.
Then for some $a \in A$ and some $c \in C$:
:$x = a \circ c$
Since $A \subseteq B$, $a \in B$.
Thus $x \in B \circ_\PP C$.
Since this holds for all $x \in A \circ_\PP C$:
:$A \circ_\PP C \subseteq B \circ_\PP C$
The same argument shows that:
:$C \circ_\PP A \subseteq C \circ_\PP B$
{{qed}}
Category:Compatible Relations
Category:Subset Products
\end{proof}
|
21975
|
\section{Subset Relation is Compatible with Subset Product/Corollary 1}
Tags: Compatible Relations, Subset Products
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\powerset S$ be the power set of $S$.
Let $\circ_\PP$ be the operation induced on $\powerset S$ by $\circ$.
Let $A, B, C, D \in \powerset S$.
Let $A \subseteq B$ and $C \subseteq D$.
Then:
:$A \circ_\PP C \subseteq B \circ_\PP D$
\end{theorem}
\begin{proof}
By Subset Relation is Compatible with Subset Product, $\subseteq$ is compatible with $\circ_\PP$.
By Subset Relation is Transitive, $\subseteq$ is transitive.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
{{qed}}
Category:Subset Products
Category:Compatible Relations
\end{proof}
|
21976
|
\section{Subset Relation is Compatible with Subset Product/Corollary 2}
Tags: Subset Products, Subset Product
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $A, B \in \powerset S$, the power set of $S$.
{{improve|No need to use power set. $A \subseteq B \subseteq S$ sufficient.}}
Let $A \subseteq B$.
Let $x \in S$.
Then:
:$x \circ A \subseteq x \circ B$
:$A \circ x \subseteq B \circ x$
\end{theorem}
\begin{proof}
This follows from Subset Relation is Compatible with Subset Product and the definition of the subset product with a singleton.
{{qed}}
Category:Subset Products
\end{proof}
|
21977
|
\section{Subset Relation is Ordering}
Tags: Order Theory, Orderings, Subsets, Subset
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$ be any subset of $\powerset S$, that is, an arbitrary set of subsets of $S$.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set.
\end{theorem}
\begin{proof}
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
\end{proof}
|
21978
|
\section{Subset Relation is Ordering/General Result}
Tags: Order Theory, Subsets, Subset
\begin{theorem}
Let $\mathbb S$ be a set of sets or class.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set.
\end{theorem}
\begin{proof}
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering:
\end{proof}
|
21979
|
\section{Subset Relation on Power Set is Partial Ordering}
Tags: Power Set, Orderings, Partial Orderings, Subsets, Order Theory, Subset
\begin{theorem}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\struct {\powerset S, \subseteq}$ be the relational structure defined on $\powerset S$ by the relation $\subseteq$.
Then $\struct {\powerset S, \subseteq}$ is an ordered set.
The ordering $\subseteq$ is partial {{iff}} $S$ is neither empty nor a singleton; otherwise it is total.
\end{theorem}
\begin{proof}
From Subset Relation is Ordering, we have that $\subseteq$ is an ordering on any set of subsets of a given set.
Suppose $S$ is neither a singleton nor the empty set.
Then $\exists a, b \in S$ such that $a \ne b$.
Then $\set a \in \powerset S$ and $\set b \in \powerset S$.
However, $\set a \nsubseteq \set b$ and $\set b \nsubseteq \set a$.
So by definition, $\subseteq$ is a partial ordering.
Now suppose $S = \O$.
Then $\powerset S = \set \O$ and, by Empty Set is Subset of All Sets, $\O \subseteq \O$.
Hence, trivially, $\subseteq$ is a total ordering on $\powerset S$.
Now suppose $S$ is a singleton: let $S = \set a$.
Then $\powerset S = \set {\O, \set a}$.
So there are only two elements of $\powerset S$, and we see that $\O \subseteq \set a$ from Empty Set is Subset of All Sets.
So, trivially again, $\subseteq$ is a total ordering on $\powerset S$.
{{qed}}
\end{proof}
|
21980
|
\section{Subset and Image Admit Infima and Mapping is Increasing implies Infimum of Image Succeeds Mapping at Infimum}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a increasing mapping.
Let $D \subseteq S$ such that
:$D$ admits a infimum in $S$ and $f \sqbrk D$ admits a infimum in $T$.
Then $\map f {\inf D} \precsim \map \inf {f \sqbrk D}$
\end{theorem}
\begin{proof}
By definition of infimum:
:$\inf D$ is lower bound for $D$.
By Increasing Mapping Preserves Lower Bounds:
:$\map f {\inf D}$ is a lower bound for $f \sqbrk D$.
Thus by definition of infimum:
:$\map f {\inf D} \precsim \map \inf {f \sqbrk D}$
{{qed}}
\end{proof}
|
21981
|
\section{Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a increasing mapping.
Let $D \subseteq S$ such that
:$D$ admits a supremum in $S$ and $f \sqbrk D$ admits a supremum in $T$.
Then:
:$\map \sup {f \sqbrk D} \precsim \map f {\sup D}$
\end{theorem}
\begin{proof}
By definition of supremum:
:$\sup D$ is upper bound for $D$.
By Increasing Mapping Preserves Upper Bounds:
:$\map f {\sup D}$ is upper bound for $f \sqbrk D$.
Thus by definition of supremum:
:$\map \sup {f \sqbrk D} \precsim \map f {\sup D}$
{{qed}}
\end{proof}
|
21982
|
\section{Subset equals Image of Preimage implies Surjection}
Tags: Subset equals Image of Preimage implies Surjection, Surjections
\begin{theorem}
Let $f: S \to T$ be a mapping.
Let:
:$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$
where $f \sqbrk B$ denotes the image of $B$ under $f$.
Then $f$ is a surjection.
\end{theorem}
\begin{proof}
Let $g$ be such that:
:$\forall B \in \mathcal P \left({T}\right): B = \left({f_g \circ f_{g^{-1}}}\right) \left({B}\right)$
In particular, it holds for $T$ itself.
Hence:
{{begin-eqn}}
{{eqn | l = T
| r = \left({f_g \circ f_{g^{-1} } }\right) \left({T}\right)
| c =
}}
{{eqn | l = T
| r = f_g \left({f_{g^{-1} } \left({T}\right)}\right)
| c = Definition of Composition of Mappings
}}
{{eqn | o = \subseteq
| r = f_g \left({S}\right)
| c = Image of Subset is Subset of Image: Corollary 2
}}
{{eqn | r = \operatorname{Im} \left({g}\right)
| c = Definition of Image of Mapping
}}
{{eqn | o = \subseteq
| r = T
| c = Image is Subset of Codomain: Corollary 1
}}
{{end-eqn}}
So:
:$T \subseteq \operatorname{Im} \left({g}\right) \subseteq T$
and so by definition of set equality:
:$\operatorname{Im} \left({g}\right) = T$
So, by definition, $g$ is a surjection.
{{qed}}
\end{proof}
|
21983
|
\section{Subset equals Preimage of Image implies Injection}
Tags: Subsets, Injections, Subset equals Preimage of Image implies Injection
\begin{theorem}
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.
Let:
:$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$
Then $f$ is an injection.
\end{theorem}
\begin{proof}
Let $g$ be such that:
:$\forall A \in \mathcal P \left({S}\right): A = \left({f_{g^{-1}} \circ f_g}\right) \left({A}\right)$
In particular, it holds for all subsets of $A$ which are singletons.
Now, consider any $x, y \in A$.
We have:
{{begin-eqn}}
{{eqn | l = g \left({x}\right)
| r = g \left({y}\right)
| c =
}}
{{eqn | ll= \implies
| l = \left\{ {g \left({x}\right)}\right\}
| r = \left\{ {g \left({y}\right)}\right\}
| c =
}}
{{eqn | ll= \implies
| l = f_g \left({\left\{ {x}\right\} }\right)
| r = f_g \left({\left\{ {y}\right\} }\right)
| c = Definition of Induced Mapping
}}
{{eqn | ll= \implies
| l = \left\{ {x}\right\}
| r = f_{g^{-1} } \left({f_g \left({\left\{ {x}\right\} }\right)}\right)
| c = by hypothesis: $A = \left({f_{g^{-1} } \circ f_g}\right) \left({A}\right)$
}}
{{eqn | r = f_{g^{-1} } \left({f_g \left({\left\{ {y}\right\} }\right)}\right)
| c =
}}
{{eqn | r = \left\{ {y}\right\}
| c = by hypothesis: $A = \left({f_{g^{-1} } \circ f_g}\right) \left({A}\right)$
}}
{{eqn | ll= \implies
| l = x
| r = y
| c =
}}
{{end-eqn}}
So $g$ is an injection.
{{qed}}
\end{proof}
|
21984
|
\section{Subset has 2 Conjugates then Normal Subgroup}
Tags: Conjugacy, Normal Subgroups
\begin{theorem}
Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $S$ have exactly two conjugates in $G$.
Then $G$ has a proper non-trivial normal subgroup.
\end{theorem}
\begin{proof}
{{MissingLinks}}
Consider the centralizer $\map {C_G} S$ of $S$ in $G$.
From Centralizer of Group Subset is Subgroup, $\map {C_G} S$ is a subgroup of $G$.
If $\map {C_G} S = G$, then $S$ has no conjugate but itself.
{{explain|Link to that result.}}
So, in order for $S$ to have exactly two conjugates in $G$, it is necessary for $\map {C_G} S$ to be a proper subgroup.
Let $e$ be the identity of $G$.
If $\map {C_G} S = \set e$, then for there to be exactly two conjugates of $S$:
:$\forall a \ne b \in G \setminus \set e: b x b^{-1} = a x a^{-1}$
But:
{{begin-eqn}}
{{eqn | l = b x b^{-1}
| r = a x a^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {a^{-1} b} x b^{-1}
| r = x a^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {a^{-1} b} x \paren {a^{-1} b}^{-1}
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = a^{-1} b
| o = \in
| r = \map {C_G} S
| c =
}}
{{end-eqn}}
This implies either that $\map {C_G} S$ is actually nontrivial, or that $a^{-1}b = e \iff a = b$, a contradiction.
Thus $\map {C_G} S$ is a nontrivial proper subgroup of $G$.
We have that there are exactly $2$ conjugacy classes of $S$.
These are in one-to-one correspondence with cosets of $S$.
Thus the index $\index G {\map {C_G} S}$ of the centralizer is:
:$\index G {\map {C_G} S} = 2$
From Subgroup of Index 2 is Normal:
:$\map {C_G} S$ is a proper nontrivial normal subgroup of $G$.
{{qed}}
\end{proof}
|
21985
|
\section{Subset implies Cardinal Inequality}
Tags: Minimal Infinite Successor Set, Cardinals
\begin{theorem}
Let $S$ and $T$ be sets such that $S \subseteq T$.
Furthermore, let:
:$T \sim \card T$
where $\card T$ denotes the cardinality of $T$.
Then:
:$\card S \le \card T$
\end{theorem}
\begin{proof}
For the proof:
:the ordering relation $\le$ for ordinals
and
:the subset relation $\subseteq$
shall be used interchangeably.
Let $f: T \to \card T$ be a bijection.
It follows that $f \restriction_S : S \to \card T$ is an injection.
The image of $S$ under $f$ is a subset of $\card T$ and thus is a subset of an ordinal.
By Unique Isomorphism between Ordinal Subset and Unique Ordinal, there is a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi: x \to f \sqbrk S$ is an order isomorphism.
It follows that $S \sim x$ by the definition of order isomorphism.
Furthermore, $\phi$ is a strictly increasing mapping from ordinals to ordinals.
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = y
| o = \le
| r = \map \phi y
| c = Strictly Increasing Ordinal Mapping Inequality
}}
{{eqn | o = \in
| r = f \sqbrk S
| c = Definition of $\phi$
}}
{{eqn | o = \subseteq
| r = \card T
| c = Image Preserves Subsets
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = \card T
| c = Cardinal Number is Ordinal
}}
{{end-eqn}}
Therefore, $y \in x \implies y \in \card T$ and $x \le \card T$ by the definition of subset.
But $\card S \le x$ by Cardinal Number Less than Ordinal.
So $\card S \le \card T$ by the fact that Subset Relation is Transitive.
{{qed}}
\end{proof}
|
21986
|
\section{Subset is Compatible with Ordinal Addition}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
: $(1): x \le y \implies \left({z + x}\right) \le \left({z + y}\right)$
: $(2): x \le y \implies \left({x + z}\right) \le \left({y + z}\right)$
\end{theorem}
\begin{proof}
The result follows from Subset is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition.
{{qed}}
Category:Ordinal Arithmetic
\end{proof}
|
21987
|
\section{Subset is Compatible with Ordinal Multiplication}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
: $(1): x \le y \implies \left({z \cdot x}\right) \le \left({z \cdot y}\right)$
: $(2): x \le y \implies \left({x \cdot z}\right) \le \left({y \cdot z}\right)$
\end{theorem}
\begin{proof}
The result follows from Subset is Left Compatible with Ordinal Multiplication and Subset is Right Compatible with Ordinal Multiplication.
{{qed}}
Category:Ordinal Arithmetic
\end{proof}
|
21988
|
\section{Subset is Left Compatible with Ordinal Addition}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \paren {z + x} \le \paren {z + y}$
\end{theorem}
\begin{proof}
The result follows from Membership is Left Compatible with Ordinal Addition.
{{qed}}
Category:Ordinal Arithmetic
\end{proof}
|
21989
|
\section{Subset is Left Compatible with Ordinal Multiplication}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \left({z \cdot x}\right) \le \left({z \cdot y}\right)$
\end{theorem}
\begin{proof}
The result follows from Membership is Left Compatible with Ordinal Multiplication.
{{qed}}
Category:Ordinal Arithmetic
\end{proof}
|
21990
|
\section{Subset is Right Compatible with Ordinal Addition}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \paren {x + z} \le \paren {y + z}$
\end{theorem}
\begin{proof}
The proof proceeds by transfinite induction on $z$.
\end{proof}
|
21991
|
\section{Subset is Right Compatible with Ordinal Exponentiation}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies x^z \le y^z$
\end{theorem}
\begin{proof}
The proof shall proceed by Transfinite Induction on $z$.
\end{proof}
|
21992
|
\section{Subset is Right Compatible with Ordinal Multiplication}
Tags: Ordinal Arithmetic
\begin{theorem}
Let $x, y, z$ be ordinals.
Then:
:$x \le y \implies \paren {x \cdot z} \le \paren {y \cdot z}$
\end{theorem}
\begin{proof}
The proof shall proceed by Transfinite Induction on $z$.
\end{proof}
|
21993
|
\section{Subset not necessarily Submagma}
Tags: Magmas, Abstract Algebra
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$.
Then it is not necessarily the case that:
: $\struct {T, \circ} \subseteq \struct {S, \circ}$
That is, it does not always follow that $\struct {T, \circ}$ is a submagma of $\struct {S, \circ}$.
\end{theorem}
\begin{proof}
Let $\struct {\Z, -}$ be the magma which is the set of integers under the operation of subtraction.
We have that the natural numbers $\N$ are a subset of the integers.
Consider $\struct {\N, -}$, the natural numbers under subtraction.
We have that Natural Number Subtraction is not Closed.
For example:
: $1 - 2 = -1 \notin \N$
Thus $\struct {\N, -}$ is not closed.
So $\struct {\N, -}$ is not a submagma of $\struct {\Z, -}$
Hence it is not true to write $\struct {\N, -} \subseteq \struct {\Z, -}$, despite the fact that $\N \subseteq \Z$.
Thus $\struct {\N, -}$ is not a submagma of $\struct {\Z, -}$.
{{qed}}
\end{proof}
|
21994
|
\section{Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup}
Tags: Abelian Groups, Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup
\begin{theorem}
Let $\struct {G, \circ}$ be an abelian group.
Let $S \subset G$ be a non-empty subset of $G$ such that $\struct {S, \circ}$ is closed.
Let $H$ be the set defined as:
:$H := \set {x \circ y^{-1}: x, y \in S}$
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.
\end{theorem}
\begin{proof}
Let $x \in S$.
Then:
:$x \circ x^{-1} \in H$
and so $H \ne \O$.
Now let $a, b \in H$.
Then:
:$a = x_a \circ y_a^{-1}$
and:
:$b = x_b \circ y_b^{-1}$
for some $x_a, y_a, x_b, y_b \in S$.
Thus:
{{begin-eqn}}
{{eqn | l = a \circ b^{-1}
| r = \paren {x_a \circ y_a^{-1} } \circ \paren {x_b \circ y_b^{-1} }^{-1}
| c =
}}
{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {\paren {y_b^{-1} }^{-1} \circ x_b^{-1} }
| c = Inverse of Group Product
}}
{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {y_b \circ x_b^{-1} }
| c = Inverse of Group Inverse
}}
{{eqn | r = x_a \circ \paren {y_a^{-1} \circ y_b} \circ x_b^{-1}
| c = {{GroupAxiom|1}}
}}
{{eqn | r = x_a \circ \paren {y_b\circ y_a^{-1} } \circ x_b^{-1}
| c = {{Defof|Abelian Group}}
}}
{{eqn | r = \paren {x_a \circ y_b} \circ \paren {y_a^{-1} \circ x_b^{-1} }
| c = {{GroupAxiom|1}}
}}
{{eqn | r = \paren {x_a \circ y_b} \circ \paren {x_b \circ y_a}^{-1}
| c = Inverse of Group Product
}}
{{end-eqn}}
As $\struct {S, \circ}$ is closed, both $x_a \circ y_b \in S$ and $x_b \circ y_a \in S$.
Thus $a \circ b$ is in the form $x \circ y^{-1}$ for $x, y \in S$.
Thus $a \circ b \in H$
Hence by the One-Step Subgroup Test, $H$ is a subgroup of $G$.
{{qed}}
\end{proof}
|
21995
|
\section{Subset of Bounded Above Set is Bounded Above}
Tags: Boundedness
\begin{theorem}
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.
Let $B$ be bounded above.
Then $A$ is also bounded above.
\end{theorem}
\begin{proof}
Let $B$ be bounded above.
Then by definition $B$ has an upper bound $U$.
Hence:
:$\forall x \in B: x \le U$
But by definition of subset:
:$\forall x \in A: x \in B$
That is:
:$\forall x \in A: x \le U$
Hence, by definition, $A$ is bounded above by $U$.
{{qed}}
\end{proof}
|
21996
|
\section{Subset of Bounded Below Set is Bounded Below}
Tags: Boundedness
\begin{theorem}
Let $A$ and $B$ be sets of real numbers such that $A \subseteq B$.
Let $B$ be bounded below.
Then $A$ is also bounded below.
\end{theorem}
\begin{proof}
Let $B$ be bounded below.
Then by definition $B$ has a lower bound $L$.
Hence:
:$\forall x \in B: x \ge L$
But by definition of subset:
:$\forall x \in A: x \in B$
That is:
:$\forall x \in A: x \ge L$
Hence, by definition, $A$ is bounded below by $L$.
{{qed}}
\end{proof}
|
21997
|
\section{Subset of Cartesian Product}
Tags: Cartesian Product, Axiomatic Set Theory
\begin{theorem}
Let $S$ be a set of ordered pairs.
Then $S$ is the subset of the cartesian product of two sets.
\end{theorem}
\begin{proof}
Let $S$ be a set of ordered pairs.
Let $x \in S$ such that $x = \left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\}$ as defined in Kuratowski Formalization of Ordered Pair.
Since the elements of $S$ are sets, we can form the union $\mathbb S = \bigcup S$ of the sets in $S$.
Since $x \in S$ it follows that the elements of $x$ are elements of $\mathbb S$.
Since $\left\{{a, b}\right\} \in x$ it follows that $\left\{{a, b}\right\} \in \mathbb S$.
Now we can form the union $\mathbb S' = \bigcup \mathbb S$ of the sets in $\mathbb S$.
Since $\left\{{a, b}\right\} \in \mathbb S$ it follows that both $a$ and $b$ are elements of $\mathbb S' = \bigcup \bigcup S$.
Thus from the Kuratowski Formalization of Ordered Pair we have that $S$ is a subset of some $A \times B$.
We can at this stage take both $A$ and $B$ as being equal to $\bigcup \bigcup S$.
Finally, the axiom of specification is applied to construct the sets:
:$A = \left\{{a: \exists b: \left({a, b}\right) \in S}\right\}$
and
:$B = \left\{{b: \exists a: \left({a, b}\right) \in S}\right\}$
$A$ and $B$ are seen to be the first and second projections respectively of $S$.
{{qed}}
\end{proof}
|
21998
|
\section{Subset of Cartesian Product not necessarily Cartesian Product of Subsets}
Tags: Cartesian Product
\begin{theorem}
Let $A$ and $B$ be sets.
Let $A$ and $B$ both have at least two distinct elements.
Then there exists $W \subseteq A \times B$ such that $W$ is not the cartesian product of a subset of $A$ and a subset of $B$.
\end{theorem}
\begin{proof}
Let $a \in A, b \in A, c \in B, d \in B$ be arbitrary elements of $A$ and $B$.
Let:
:$W = \set {\tuple {a, c}, \tuple {a, d}, \tuple {b, d} }$
Then $W \subseteq A \times B$.
Suppose $W = X \times Y$ such that $X \subseteq A, Y \subseteq B$.
Then $a, b \in X$ and $c, d \in Y$.
But $X \times Y$ also contains $\tuple {b, c}$ which is not in $W$.
Hence the result.
{{qed}}
\end{proof}
|
21999
|
\section{Subset of Codomain is Superset of Image of Preimage}
Tags: Induced Mappings, Composite Mappings, Mapping Theory, Subset of Codomain is Superset of Image of Preimage, Preimages under Mappings
\begin{theorem}
Let $f: S \to T$ be a mapping.
Then:
:$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
:$f \sqbrk B$ denotes the image of $B$ under $f$
:$f^{-1}$ denotes the inverse of $f$
:$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
:$\forall B \in \powerset T: \map {\paren {f^\to \circ f^\gets} } B \subseteq B$
\end{theorem}
\begin{proof}
From Image of Preimage of Mapping:
: $B \subseteq T \implies \left({f \circ f^{-1}}\right) \left[{B}\right] = B \cap f \left[{S}\right]$
The result follows from Intersection Subset.
{{qed}}
\end{proof}
|
22000
|
\section{Subset of Countable Set is Countable}
Tags: Countable Sets, Subsets, Subset
\begin{theorem}
A subset of a countable set is countable.
\end{theorem}
\begin{proof}
Let $S$ be a countable set.
Let $T \subseteq S$.
By definition, there exists an injection $f: S \to \N$.
Let $i: T \to S$ be the inclusion mapping.
We have that $i$ is an injection.
Because the composite of injections is an injection, it follows that $f \circ i: T \to \N$ is an injection.
Hence, $T$ is countable.
{{qed}}
\end{proof}
|
22001
|
\section{Subset of Countably Infinite Set is Countable}
Tags: Set Theory, Infinite Sets, Subsets, Countable Sets, Subset
\begin{theorem}
Every subset of a countably infinite set is countable.
\end{theorem}
\begin{proof}
Let $S = \set {a_0, a_1, a_2, \ldots}$ be countably infinite.
Let $T \subseteq S = \set {a_{n_0}, a_{n_1}, a_{n_2}, \ldots}$, where $a_{n_0}, a_{n_1}, a_{n_2}, \ldots$ are the elements of $S$ also in $T$.
If the set of numbers $\set {n_0, n_1, n_2, \ldots}$ has a largest number, then $T$ is finite.
Otherwise, consider the bijection $i \leftrightarrow n_i$.
This leads to the bijection $i \leftrightarrow a_{n_i}$
This latter bijection is the required one-to-one correspondence between the elements of $T$ and those of $\N$, showing that $T$ is indeed countable.
{{finish|formal justification}}
\end{proof}
|
22002
|
\section{Subset of Domain is Subset of Preimage of Image}
Tags: Subset of Domain is Subset of Preimage of Image, Induced Mappings, Mappings, Composite Mappings, Mapping Theory, Preimages under Mappings
\begin{theorem}
Let $f: S \to T$ be a mapping.
Then:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
:$f \sqbrk A$ denotes the image of $A$ under $f$
:$f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
:$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
:$\forall A \in \powerset S: A \subseteq \map {\paren {f^\gets \circ f^\to} } A$
\end{theorem}
\begin{proof}
As a mapping is by definition a left-total relation.
Therefore Preimage of Image under Left-Total Relation is Superset applies:
:$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$
where $\RR$ is a relation.
Hence:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
{{qed}}
\end{proof}
|
22003
|
\section{Subset of Domain is Subset of Preimage of Image/Equality does Not Necessarily Hold}
Tags: Subset of Domain is Subset of Preimage of Image
\begin{theorem}
Let $f: S \to T$ be a mapping.
From Subset of Domain is Subset of Preimage of Image:
:$A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
:$f \sqbrk A$ denotes the image of $A$ under $f$
:$f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
:$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
It is not necessarily the case that:
:$A \subseteq S \implies A = \paren {f^{-1} \circ f} \sqbrk A$
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let:
:$S = \set {0, 1}$
:$T = \set 2$
Let $f: S \to T$ be defined as:
:$\map f 0 = 2$
:$\map f 1 = 2$
Let $A \subseteq S$ be defined as:
:$A = \set 0$
Then we have:
:$f \sqbrk A = \set 2$
but:
:$f^{-1} \circ f \sqbrk A = f^{-1} \sqbrk 2 = \set {0, 1}$
That is:
:$A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
but it is not the case that:
:$A \paren {f^{-1} \circ f} \sqbrk A$
{{qed}}
\end{proof}
|
22004
|
\section{Subset of Empty Set}
Tags: Set Theory, Empty Set
\begin{theorem}
Let $A$ be a class.
Then:
:$A$ is a subset of the empty set $\O$
{{iff}}:
:$A$ is equal to the empty set:
:$A \subseteq \O \iff A = \O$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = A = \O
| o = \leadsto
| r = A \subseteq \O
| c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
Conversely:
{{begin-eqn}}
{{eqn | l = A \subseteq \O
| o = \leadsto
| r = A \subseteq \O \land \O \subseteq A
| c = Empty Set is Subset of All Sets
}}
{{eqn | o = \leadsto
| r = A = \O
| c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
{{qed}}
Category:Empty Set
\end{proof}
|
22005
|
\section{Subset of Empty Set iff Empty}
Tags: Empty Set, Empty Set
\begin{theorem}
Let $S$ be a set.
Let $\O$ denote the empty set.
Then $S \subseteq \O$ {{iff}} $S = \O$.
\end{theorem}
\begin{proof}
Suppose $x \in S$.
Then since $S \subseteq \O$, it follows that $x \in \O$.
Hence $x \notin S$.
That is, $S = \O$.
{{qed}}
Category:Empty Set
\end{proof}
|
22006
|
\section{Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed}
Tags: Euclidean Space, Closed Sets, Real Number Plane with Euclidean Topology
\begin{theorem}
Let $\struct {\R^2, \tau_d}$ be the real number plane with the usual (Euclidean) topology.
Let $A \subseteq R^2$ be the set of all points defined as:
:$A := \set {\tuple {x, y} \in \R^2: x y \ge 1}$
Then $A$ is a closed set in $\struct {\R^2, d}$.
\end{theorem}
\begin{proof}
By definition, $\tau_d$ is the topology induced by the Euclidean metric $d$.
Consider the complement of $A$ in $\R^2$:
:$A' := \R^2 \setminus A$
Thus:
:$A := \set {\tuple {x, y} \in \R^2: x y < 1}$
Let $a = \tuple {x_a, y_a} \in A^2$.
Let $\epsilon = \size {1 - x_a y_a}$.
Then the open $\epsilon$-ball of $a$ in $\R^2$ lies entirely in $A'$.
As $a$ is arbitrary, it follows that any such $a$ has an open $\epsilon$-ball of $a$ in $\R^2$ which lies entirely in $A'$.
Thus, by definition, $A'$ is open in $\R^2$.
So, also by definition, $A$ is closed in $\R^2$.
{{qed}}
\end{proof}
|
22007
|
\section{Subset of Excluded Point Space is not Dense-in-itself}
Tags: Excluded Point Topology, Denseness
\begin{theorem}
Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space such that $S$ is not a singleton.
Let $H \subseteq S$.
Then $H$ is not dense-in-itself.
\end{theorem}
\begin{proof}
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.
So by definition, all points of $H$ are isolated in $H$ except $p$.
So if $H \ne \set p$, $H$ contains at least one point which is isolated in $H$.
As for $p$ itself, from Singleton Point is Isolated we have that $p$ is itself isolated in $\set p$.
So if $H = \set p$, $H$ also contains one point which is isolated in $H$.
Hence the result, by definition of dense-in-itself.
{{qed}}
\end{proof}
|
22008
|
\section{Subset of Finite Set is Finite}
Tags: Set Theory, Subsets, Analysis, Finite Sets, Subset
\begin{theorem}
Let $X$ be a finite set.
If $Y$ is a subset of $X$, then $Y$ is also finite.
\end{theorem}
\begin{proof}
From the definition, $X$ is finite {{iff}} $\exists n \in \N$ such that there exists a bijection:
:$f: X \leftrightarrow \N_n$
where $\N_n$ is the set of all elements of $\N$ less than $n$, that is:
:$\N_n = \set {0, 1, 2, \ldots, n - 1}$
The case in which $X$ is empty is trivial.
We begin proving the following particular case:
:''If $X$ is finite and $a \in X$, then $X \setminus \set a$ is also finite.''
From Bijection between Specific Elements there exists a bijection $f: \N_n \to X$, which satisfies $\map f n = a$.
Next we prove the general case by induction.
\end{proof}
|
22009
|
\section{Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact}
Tags: Hilbert Sequence Space
\begin{theorem}
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Let $H$ be a subset of $\ell^2$ whose interior is non-empty.
Then $H$ is not compact in $\ell^2$.
\end{theorem}
\begin{proof}
Let $x \in H^\circ$, where $H^\circ$ denotes the interior of $H$.
By definition, $H^\circ$ is an open set of $\ell^2$ containing $x$.
Again by definition, $H$ is a neighborhood of $x$.
But from Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood in $\ell^2$.
Thus $H$ cannot be compact in $\ell^2$.
{{qed}}
\end{proof}
|
22010
|
\section{Subset of Indiscrete Space is Compact}
Tags: Compact Spaces, Indiscrete Topology
\begin{theorem}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is compact in $T$.
\end{theorem}
\begin{proof}
The subspace $T_H = \struct {H, \set {\O, S \cap H} }$ is trivially also an indiscrete space.
The only open cover of $T_H$ is $\set H$ itself.
The only subcover of $H$ is, trivially, also $\set H$, which is finite.
So $H$ is (trivially) compact in $T$.
{{qed}}
\end{proof}
|
22011
|
\section{Subset of Indiscrete Space is Compact and Sequentially Compact}
Tags: Sequentially Compact Spaces, Compact Spaces, Indiscrete Topology
\begin{theorem}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
\end{theorem}
\begin{proof}
The subspace $T_H = \left({H, \left\{{\varnothing, S \cap H}\right\}}\right)$ is trivially also an indiscrete space.
The only open cover of $T_H$ is $\left\{{H}\right\}$ itself.
The only subcover of $H$ is, trivially, also $\left\{{H}\right\}$, which is finite.
So $H$ is (trivially) compact in $T$.
From Convergent Sequences in Indiscrete Space, every sequence in $T$ converges to every point of $S$.
So every infinite sequence has a subsequence which converges to every point in $S$.
Hence $H$ is (trivially) sequentially compact in $T$.
{{qed}}
\end{proof}
|
22012
|
\section{Subset of Indiscrete Space is Dense-in-itself}
Tags: Denseness, Indiscrete Topology
\begin{theorem}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ be a subset of $S$ containing more than one point.
Then $H$ is dense-in-itself.
\end{theorem}
\begin{proof}
Let $x \in H$.
Then as $H$ is not singleton, $\exists y \in H: y \ne x$.
Then every neighborhood of $x$ contains $y$, as the only open set of $T$ is $S$, which also contains both $x$ and $y$.
Hence $x$ is not isolated by definition.
$x$ is general, so all points in $H$ are similarly not isolated.
Hence the subset $H$ is dense-in-itself by definition.
{{qed}}
\end{proof}
|
22013
|
\section{Subset of Indiscrete Space is Everywhere Dense}
Tags: Denseness, Indiscrete Topology
\begin{theorem}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$.
Then $H$ is everywhere dense.
\end{theorem}
\begin{proof}
From Limit Points of Indiscrete Space, every point of $T$ is a limit point of $H$.
Hence $H$ is everywhere dense by definition.
{{qed}}
\end{proof}
|
22014
|
\section{Subset of Indiscrete Space is Sequentially Compact}
Tags: Sequentially Compact Spaces, Indiscrete Topology
\begin{theorem}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is sequentially compact in $T$.
\end{theorem}
\begin{proof}
From Sequence in Indiscrete Space converges to Every Point, every sequence in $T$ converges to every point of $S$.
So every infinite sequence has a subsequence which converges to every point in $S$.
Hence $H$ is (trivially) sequentially compact in $T$.
{{qed}}
\end{proof}
|
22015
|
\section{Subset of Linear Code with Even Weight Codewords}
Tags: Linear Codes
\begin{theorem}
Let $C$ be a linear code.
Let $C^+$ be the subset of $C$ consisting of all the codewords of $C$ which have even weight.
Then $C^+$ is a subgroup of $C$ such that either $C^+ = C$ or such that $\order {C^+} = \dfrac {\order C} 2$.
\end{theorem}
\begin{proof}
Note that the zero codeword is in $C^+$ as it has a weight of $0$ which is even.
Let $c$ and $d$ be of even weight, where $c$ and $d$ agree in $k$ ordinates.
Let $\map w c$ denote the weight of $c$.
Then:
{{begin-eqn}}
{{eqn | l = \map w {c + d}
| r = \map w c - k + \map w d - k
| c =
}}
{{eqn | r = \map w c + \map w d - 2 k
| c =
}}
{{end-eqn}}
which is even.
Since the negative of a vector $\mathbf v$ in $\Z_2$ equals $\mathbf v$, it follows that the inverse of $c \in C$ is also in $C$.
It follows from the Two-Step Subgroup Test that $C^+$ is a subgroup of $C$.
Let $C \ne C^+$.
Then $C$ contains a codeword $c$ of odd weight.
Let $C^-$ denote the subset of $C$ consisting of all the codewords of $C$ which have odd weight.
Adding $c$ to each codeword of $C^+$ gives distinct codewords of odd weight, so:
:$\order {C^-} \ge \order {C^+}$
Similarly, adding $c$ to each codeword of $C^-$ gives distinct codewords of even weight, so:
:$\order {C^-} \le \order {C^+}$
As $C = C^+ \cup C^-$ it follows that:
:$\order C = 2 \order {C^+}$
Hence the result.
{{qed}}
\end{proof}
|
22016
|
\section{Subset of Linearly Independent Set is Linearly Independent}
Tags: Unitary Modules, Linear Algebra, Modules
\begin{theorem}
A subset of a linearly independent set is also linearly independent.
\end{theorem}
\begin{proof}
Let $G$ be an unitary $R$-module.
Then $\sequence {a_n}$ is a linearly independent sequence {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
Then clearly $\sequence {a_n}$ is a linearly independent sequence of $G$.
Conversely, let $\sequence {a_n}$ be a linearly independent sequence of $G$.
Let $\sequence {b_m}$ be a sequence of distinct terms of $\set {a_1, a_2, \ldots, a_n}$.
Let $\sequence {\mu_m}$ be a sequence of scalars such that $\ds \sum_{j \mathop = 1}^m \mu_j b_j = 0$.
For each $k \in \closedint 1 n$, let:
:$\lambda_k = \begin{cases}
\mu_j & : j \text { is the unique index such that } a_k = b_j \\
0 & : a_k \notin \set {b_1, b_2, \ldots, b_m}
\end{cases}$
Then:
:$\ds 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
Thus:
:$\forall k \in \closedint 1 n: \lambda_k = 0$
As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that:
:$\forall j \in \closedint 1 m: \mu_j = 0$
and so $\sequence {b_m}$ has been shown to be a linearly independent sequence.
Hence the result.
{{Qed}}
\end{proof}
|
22017
|
\section{Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact}
Tags: Examples of Topologies, Order Topology
\begin{theorem}
Let $X = \hointr 0 1 \cup \openint 2 3 \cup \set 4$.
Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers.
Let $\tau$ be the $\preceq$ order topology on $X$.
Let $Y = \hointr 0 1 \cup \set 4$.
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Then:
:$\struct {Y, \preceq}$ is a complete lattice
:$Y$ is closed in $X$
but:
:$\struct {Y, \tau'}$ is not compact.
\end{theorem}
\begin{proof}
First it is demonstrated that $\struct {Y, \preceq}$ is a complete lattice.
Let $\phi: Y \to \closedint 0 1$ be defined as:
:$\map \phi y = \begin{cases}
y & : y \in \hointr 0 1 \\
1 & : y = 4 \end{cases}$
Then $\phi$ is a order isomorphism.
{{explain|The above needs to be proved.}}
{{qed|lemma}}
We have that $\closedint 0 1$ is a complete lattice.
{{explain|This is probably around here somewhere.}}
Next is is shown that $\struct {Y, \preceq}$ is closed in $X$.
Let $x \in X \setminus Y$.
Then:
:$x \in \openint 2 3$
Thus:
:$x \in \openint {\dfrac {x + 2} 2} {\dfrac {x + 3} 2} \in \tau'$
Since the complement of $Y$ is open, $Y$ is closed.
Finally it is shown that $\struct {Y, \tau'}$ is not compact.
Let:
:$\AA = \set {x^\preceq: x \in \hointr 0 1} \cup \set {\paren {\dfrac 5 2}^\succeq}$
where:
:$x^\preceq$ denotes the lower closure of $x$
:$x^\succeq$ denotes the upper closure of $x$
Then $\AA$ is an open cover of $Y$ with no finite subcover.
{{explain|Prove the above statement.}}
{{qed}}
{{finish|The remaining parts of this proof need to be completed.}}
Category:Order Topology
\end{proof}
|
22018
|
\section{Subset of Meager Set is Meager Set}
Tags: Meager Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be meager in $T$.
Let $B \subseteq A$.
Then $B$ is meager in $T$.
\end{theorem}
\begin{proof}
Since $A$ is meager in $T$:
:there exists a countable collection of sets $\set {U_n: n \in \N}$ nowhere dense in $T$ such that $\ds A = \bigcup_{n \in \N} U_n$.
Then, we have:
{{begin-eqn}}
{{eqn | l = B
| r = A \cap B
| c = Intersection with Subset is Subset
}}
{{eqn | r = \paren {\bigcup_{n \in \N} U_n} \cap B
}}
{{eqn | r = \bigcup_{n \in \N} \paren {U_n \cap B}
| c = Union Distributes over Intersection
}}
{{end-eqn}}
From Intersection is Subset:
:$U_n \cap B \subseteq U_n$
From Subset of Nowhere Dense Subset is Nowhere Dense:
:$U_n \cap B$ is nowhere dense in $T$.
Then, we see that:
:$B$ can be written as the union of nowhere dense sets in $T$.
That is:
:$B$ is meager in $T$.
{{qed}}
Category:Meager Spaces
\end{proof}
|
22019
|
\section{Subset of Metric Space is Subset of its Closure}
Tags: Set Closures
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Then:
:$H \subseteq H^-$
where $H^-$ denotes the closure of $H$.
\end{theorem}
\begin{proof}
By definition of closure:
:$H^- = H' \cup H^i$
where:
:$H'$ denotes the set of limit points of $H$
:$H^i$ denotes the set of isolated points of $H$.
Let $a \in H$.
If $a$ is a limit point of $H$ then $a \in H'$.
Suppose $a \notin H'$.
Then by definition of limit point:
:$\neg \forall \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} \ne \O$
That is:
:$\exists \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} = \O$
and so as $\map d {a, a} = 0$:
:$\exists \epsilon \in \R_{>0}: \set {x \in A: \map d {x, a} < \epsilon} = \set a$
Thus, by definition, $a$ is an isolated point of $H$.
So $a \in H'$ or $a \in H^i$.
Hence by definition of set union:
:$a \in H \subseteq H^-$
By definition of subset:
:$H \subseteq H \subseteq H^-$
and hence the result by definition of closure.
{{qed}}
\end{proof}
|
22020
|
\section{Subset of Module Containing Identity is Linearly Dependent}
Tags: Linear Algebra, Module Theory, Modules
\begin{theorem}
Let $G$ be a group whose identity is $e$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $H \subseteq G$ such that $e \in H$.
Then $H$ is a linearly dependent set.
\end{theorem}
\begin{proof}
From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$.
Let $H \subseteq G$ such that $e \in H$.
Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
So, let $a_j = e$ for some $j \in \closedint 1 n$.
Let $c \in R \ne 0_R$.
Consider the sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as:
:$\lambda_k = \begin{cases}
c & : k \ne j \\
0_R & : k= j
\end{cases}$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k \circ a_k
| r = \lambda_1 \circ a_1 + \lambda_2 \circ a_2 + \cdots + \lambda_j \circ a_j + \cdots + \lambda_n \circ a_n
| c =
}}
{{eqn | r = 0_R \circ a_1 + 0_R \circ a_2 + \cdots + c \circ e + \cdots + 0_R \circ a_n
| c =
}}
{{eqn | r = e + e + \cdots + e + \cdots + e
| c =
}}
{{eqn | r = e
| c =
}}
{{end-eqn}}
Thus there exists a sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$
Hence the result.
{{qed}}
\end{proof}
|
22021
|
\section{Subset of Natural Numbers is Cofinal iff Infinite}
Tags: Natural Numbers, Order Theory
\begin{theorem}
Consider the ordered set $\struct {\N, \le}$, where $\le$ is the usual ordering on the natural numbers.
Let $S \subseteq \N$.
Then $S$ is cofinal {{iff}} it is infinite.
\end{theorem}
\begin{proof}
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive.
Therefore, the following suffices:
\end{proof}
|
22022
|
\section{Subset of Natural Numbers under Max Operation is Monoid}
Tags: Examples of Monoids, Max Operation, Monoid Examples, Max and Min Operations, Natural Numbers, Monoids
\begin{theorem}
Let $S \subseteq \N$ be a subset of the natural numbers $\N$.
Let $\struct {S, \max}$ denote the algebraic structure formed from $S$ and the max operation.
Then $\struct {S, \max}$ is a monoid.
Its identity element is the smallest element of $S$.
\end{theorem}
\begin{proof}
By the Well-Ordering Principle, $\N$ is a well-ordered set.
By definition, every subset of a well-ordered set is also well-ordered.
Thus $S$ is a well-ordered set.
The result follows from Max Operation on Woset is Monoid.
{{qed}}
\end{proof}
|
22023
|
\section{Subset of Naturals is Finite iff Bounded}
Tags: Set Theory, Natural Numbers, Subsets, Subset
\begin{theorem}
Let $X$ be a subset of the natural numbers $\N$.
Then $X$ is finite {{iff}} it is bounded.
\end{theorem}
\begin{proof}
A subset of the natural numbers is also a subset of the real numbers $\R$.
By definition, a bounded subset of $\R$ is bounded below in $\R$ and bounded above in $\R$.
By the Well-Ordering Principle, $X$ is bounded below
Thus $X$ is bounded {{iff}} $X$ is bounded above.
That is, {{iff}}:
:$\exists p \in \N: \forall x \in X: x \le p$
\end{proof}
|
22024
|
\section{Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition}
Tags: Set Closures, Normed Vector Spaces, Denseness
\begin{theorem}
Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D$ be dense.
Then:
:$D^- = X$
\end{theorem}
\begin{proof}
Let $x \in X \setminus D$.
Suppose $D$ is dense in $X$.
Then:
:$\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$
where $\ds \map {B_{\frac 1 n}} x$ is an open ball.
Let $\sequence {d_n}_{n \mathop \in \N}$ be a sequence in $D$.
Then:
:$\forall n \in \N : \norm {x - d_n} < \frac 1 n$
Hence, $x$ is a limit point of $D$.
In other words, $x \in D^-$.
We have just shown that:
:$x \in X \setminus D \implies x \in D^-$
Hence:
:$X \setminus D \subseteq D^-$.
By definition of closure:
:$D \subseteq D^-$
Therefore:
{{begin-eqn}}
{{eqn | l = X
| r = D \cup \paren {X \setminus D}
}}
{{eqn | o = \subseteq
| r = D^-
}}
{{eqn | o = \subseteq
| r = X
}}
{{end-eqn}}
Thus:
:$X = D^-$.
{{qed}}
\end{proof}
|
22025
|
\section{Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition}
Tags: Set Closures, Normed Vector Spaces, Denseness
\begin{theorem}
Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D^- = X$.
Then $D$ is dense.
\end{theorem}
\begin{proof}
Let $X = D^-$.
We have to show, that for every $x \in X$ there is an open ball with an element from $D^-$.
We have that $X = D \cup \paren {X \setminus D}$.
Suppose $x \in X \setminus D$.
Then $x \in D^- \setminus D$.
Hence, $x$ is a limit point of $D$.
Therefore, there is a sequence $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which converges to $x$.
Thus:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$
In other words:
:$\ds d_N \in D \implies d_N \in \map {B_\epsilon} x$
Therefore:
:$d_N \in D \cap \map {B_\epsilon} x$.
Suppose $x \in D$.
Let $\epsilon > 0$.
Then $x \in \map {B_\epsilon} x \cap D$.
From both parts and definition we conclude that $D^-$ is dense in $X$.
{{qed}}
\end{proof}
|
22026
|
\section{Subset of Nowhere Dense Subset is Nowhere Dense}
Tags: Denseness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be nowhere dense in $T$.
Let $B \subseteq A$.
Then $B$ is nowhere dense in $T$.
\end{theorem}
\begin{proof}
{{AimForCont}} it is not the case that $B$ is nowhere dense in $T$.
Then by definition of nowhere dense:
:$B^-$ contains some open set of $T$ which is non-empty.
From Set Closure Preserves Set Inclusion, we have:
:$B^- \subseteq A^-$
So:
:$A^-$ contains some open set of $T$ which is non-empty.
So $A$ is not nowhere dense in $T$.
This contradicts our assertion that $A$ is nowhere dense in $T$:
The result follows by Proof by Contradiction.
{{qed}}
Category:Denseness
\end{proof}
|
22027
|
\section{Subset of Open Reciprocal-N Balls forms Neighborhood Basis in Real Number Line}
Tags: Real Number Line with Euclidean Metric, Real Intervals, Real Number Space
\begin{theorem}
Let $\R$ denote the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $\R$.
Let $k \in \Z$ be some fixed integer.
Let $\BB_a$ be defined as:
:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$
that is, the set of all open $\epsilon$-balls of $a$ for $\epsilon$ which are reciprocals of integers greater than $k$.
Then the $\BB_a$ is a basis for the neighborhood system of $a$.
\end{theorem}
\begin{proof}
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition:
:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
From Between two Real Numbers exists Rational Number:
:$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$
Let $\epsilon''$ be expressed in canonical form:
:$\epsilon'' = \dfrac p q$
Let $\epsilon''' = \dfrac 1 q$
Then $\epsilon''' \le \epsilon'' < \epsilon'$
If $q \le k$, let $\epsilon = \dfrac 1 {k + 1}$
Otherwise, let $\epsilon = \epsilon'''$.
Then:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
From Open Real Interval is Open Ball
:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
is the open $\epsilon$-ball at $a$.
By its method of construction:
:$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$
From Subset Relation is Transitive:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$.
Hence the result by definition of basis for the neighborhood system of $a$.
{{qed}}
\end{proof}
|
22028
|
\section{Subset of Ordinal implies Cardinal Inequality}
Tags: Cardinals
\begin{theorem}
Let $S$ be a set.
Let $x$ be an ordinal such that $S \subseteq x$.
Then:
:$\card S \le \card x$
where $\card S$ denotes the cardinality of $S$.
\end{theorem}
\begin{proof}
Since $x$ is an ordinal, it follows that $x \sim \card x$ by Ordinal Number Equivalent to Cardinal Number.
This satisfies the hypothesis for Subset implies Cardinal Inequality.
Therefore:
:$\card S \le \card x$
{{qed}}
\end{proof}
|
22029
|
\section{Subset of Ordinals has Minimal Element}
Tags: Ordinals, Class Theory
\begin{theorem}
Let $A$ be an ordinal (we shall allow $A$ to be a proper class).
Let $B$ be a nonempty subset of $A$.
Then $B$ has an $\Epsilon$-minimal element.
{{explain|$\Epsilon$}}
That is:
:$\exists x \in B: B \cap x = \O$
\end{theorem}
\begin{proof}
We have that $\Epsilon$ creates a well-ordering on any ordinal.
Also, the initial segments of $x$ are sets.
Therefore from Proper Well-Ordering Determines Smallest Elements:
:$B$ has an $\Epsilon$-minimal element.
\end{proof}
|
22030
|
\section{Subset of Particular Point Space is either Open or Closed}
Tags: Open Sets, Particular Point Topology, Closed Sets, Clopen Sets
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $H \subseteq S$ be any subset of $T$.
Then $H$ is either open or closed in $T$.
The only sets which are both closed and open in $T$ are $S$ and $\O$.
\end{theorem}
\begin{proof}
Let $H \subseteq S$.
There are two cases to consider:
:$p \in H$
:$p \notin H$
If $p \in H$ then by definition of a particular point topology, $H$ is open.
If $p \notin H$, then $p \in \relcomp S H$, where $\relcomp S H$ is the relative complement of $H$ in $S$.
So $\relcomp S H$ is open by definition of a particular point topology.
From the definition of a closed set, $\relcomp S {\relcomp S H}$ is closed in $T$.
From Relative Complement of Relative Complement we have that $\relcomp S {\relcomp S H} = H$ and so $H$ is closed in $T$.
Now suppose $H \subseteq T$ is both closed and open in $T$.
From Open and Closed Sets in Topological Space, if $H = \O$ or $H = S$ then H is both closed and open in $T$.
So, suppose $H \ne \O$ or $H \ne S$.
As $H \subseteq T$ is both closed and open in $T$, $\relcomp S H$ is also both closed and open in $T$.
From Boundary is Intersection of Closure with Closure of Complement we have:
:$\partial H = H^- \cap \paren {\relcomp S H}^-$
where $H^-$ is the closure of $H$.
By Closure of Open Set of Particular Point Space we have that $H^- = S$, and of course $\paren {\relcomp S H}^- = S$.
So:
:$\partial H = H^- \cap \paren {\relcomp S H}^- = S$
However, from Set Clopen iff Boundary is Empty we have that $\partial H = \O$.
Hence if $H$ is both closed and open in $T$, then $H = \O$ or $H = S$.
{{qed}}
Category:Open Sets
Category:Closed Sets
Category:Clopen Sets
Category:Particular Point Topology
\end{proof}
|
22031
|
\section{Subset of Preimage under Relation is Preimage of Subset}
Tags: Preimages under Relations, Subsets, Subset, Relation Theory, Mappings, Mapping Theory, Relations
\begin{theorem}
Let $\RR \subseteq S \times T$ be a relation.
Let $X \subseteq S, Y \subseteq T$.
Then:
:$X \subseteq \RR^{-1} \sqbrk Y \iff \RR \sqbrk X \subseteq Y$
In the language of direct image mappings, this can be written:
:$X \subseteq \map {\RR^\gets} Y \iff \map {\RR^\to} X \subseteq Y$
\end{theorem}
\begin{proof}
As $\RR$ is a relation, then so is its inverse $\RR^{-1}$.
Let $\RR \sqbrk X \subseteq Y$.
Thus:
{{begin-eqn}}
{{eqn | l = \RR \sqbrk X
| o = \subseteq
| r = Y
| c =
}}
{{eqn | ll= \leadsto
| l = \RR^{-1} \sqbrk {\RR \sqbrk X}
| o = \subseteq
| r = \RR^{-1} \sqbrk Y
| c = Image of Subset under Relation is Subset of Image: Corollary 1
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = \RR^{-1} \sqbrk Y
| c = Image of Preimage under Relation is Subset, Subset Relation is Transitive
}}
{{end-eqn}}
So:
:$\RR \sqbrk X \subseteq Y \implies X \subseteq \RR^{-1} \sqbrk Y$
{{qed|lemma}}
Now let $X \subseteq \RR^{-1} \sqbrk Y$.
The same argument applies:
{{begin-eqn}}
{{eqn | l = X
| o = \subseteq
| r = \RR^{-1} \sqbrk Y
| c =
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk X
| o = \subseteq
| r = \RR \sqbrk {\RR^{-1} \sqbrk Y}
| c = Image of Subset under Relation is Subset of Image
}}
{{eqn | ll= \leadsto
| l = \RR \sqbrk X
| o = \subseteq
| r = Y
| c = Image of Preimage under Relation is Subset, Subset Relation is Transitive
}}
{{end-eqn}}
So:
:$X \subseteq \RR^{-1} \sqbrk Y \implies \RR \sqbrk X \subseteq Y$
{{qed|lemma}}
Thus we have:
:$X \subseteq \RR^{-1} \sqbrk Y \implies \RR \sqbrk X \subseteq Y$
:$\RR \sqbrk X \subseteq Y \implies X \subseteq \RR^{-1} \sqbrk Y$
Hence the result.
{{qed}}
Category:Subsets
Category:Preimages under Relations
\end{proof}
|
22032
|
\section{Subset of Real Numbers is Interval iff Connected}
Tags: Connected Spaces, Analysis, Real Intervals, Topology, Connectedness
\begin{theorem}
Let the real number line $\R$ be considered as a topological space.
Let $S$ be a subspace of $\R$.
Then $S$ is connected {{iff}} $S$ is an interval of $\R$.
That is, the only subspaces of $\R$ that are connected are intervals.
\end{theorem}
\begin{proof}
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive.
Therefore, the following suffices:
\end{proof}
|
22033
|
\section{Subset of Satisfiable Set is Satisfiable}
Tags: Formal Semantics
\begin{theorem}
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-satisfiable set of formulas from $\LL$.
Let $\FF'$ be a subset of $\FF$.
Then $\FF'$ is also $\mathscr M$-satisfiable.
\end{theorem}
\begin{proof}
Since $\FF$ is $\mathscr M$-satisfiable, there exists some model $\MM$ of $\FF$:
:$\MM \models_{\mathscr M} \FF$
Thus for every $\psi \in \FF$:
:$\MM \models_{\mathscr M} \psi$
Now, for every $\psi$ in $\FF'$:
:$\psi \in \FF$
by definition of subset.
Hence:
:$\forall \psi \in \FF': \MM \models_{\mathscr M} \psi$
that is, $\MM$ is a model of $\FF'$.
Hence $\FF'$ is $\mathscr M$-satisfiable.
{{qed}}
Category:Formal Semantics
\end{proof}
|
22034
|
\section{Subset of Set Difference iff Disjoint Set}
Tags: Disjoint Sets, Set Difference
\begin{theorem}
Let $S, T$ be sets.
Let $A \subseteq S$
Then:
:$A \cap T = \O \iff A \subseteq S \setminus T$
where:
:$A \cap T$ denotes set intersection
:$\O$ denotes the empty set
:$S \setminus T$ denotes set difference.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = A \cap \paren {S \setminus T}
| r = \paren {A \cap S} \setminus T
| c = Intersection with Set Difference is Set Difference with Intersection
}}
{{eqn | r = A \setminus T
| c = Intersection with Subset is Subset
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = S \setminus T
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = A
| r = A \cap \paren {S \setminus T}
| c = Intersection with Subset is Subset
}}
{{eqn | ll= \leadstoandfrom
| l = A
| r = A \setminus T
| c = As $A \cap \paren {S \setminus T} = A \setminus T$
}}
{{eqn | ll = \leadstoandfrom
| l = A \cap T
| r = \O
| c = Set Difference with Disjoint Set
}}
{{end-eqn}}
{{qed}}
Category:Set Difference
Category:Disjoint Sets
\end{proof}
|
22035
|
\section{Subset of Set is Coarser than Set}
Tags: Preorder Theory
\begin{theorem}
Let $\left({S, \preceq}\right)$ be a preordered set.
Let $A, B$ be subset of $S$ such that
:$A \subseteq B$
Then $A$ is coarser than $B$.
\end{theorem}
\begin{proof}
Let $x \in A$.
By definition of subset:
:$x \in B$
By definition of reflexivity:
:$x \preceq x$
Thus
:$\exists y \in B: y \preceq x$
{{qed}}
\end{proof}
|
22036
|
\section{Subset of Standard Discrete Metric Space is Neighborhood of Each Point}
Tags: Discrete Metrics, Neighborhoods
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space where $d$ is the standard discrete metric.
Let $S \subseteq A$.
Let $a \in S$.
Then $S$ is a neighborhood of $a$.
That is, every subset of $A$ is a neighborhood of each of its points.
\end{theorem}
\begin{proof}
Let $S \subseteq A$.
Let $a \in S$.
From Neighborhoods in Standard Discrete Metric Space, $\set a$ is a neighborhood of $a$.
As $a \in S$ it follows from Singleton of Element is Subset that $\set a \subseteq S$.
The result follows from Superset of Neighborhood in Metric Space is Neighborhood.
{{qed}}
\end{proof}
|
22037
|
\section{Subset of Standard Discrete Metric Space is Open}
Tags: Discrete Metrics, Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a standard discrete metric space.
Let $S \subseteq A$ be a subset of $A$.
Then $S$ is an open set of $M$.
\end{theorem}
\begin{proof}
From the definition of standard discrete metric:
:$\forall x, y \in A: \map d {x, y} = \begin {cases}
0 & : x = y \\
1 & : x \ne y
\end {cases}$
Let $\epsilon \in \R_{>0}$ be such that $0 < \epsilon \le 1$.
Let $x \in S$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$.
Then by definition of $\epsilon$ and $d$:
:$\map {B_\epsilon} x = \set x$
Thus:
:$\forall x \in S: \map {B_\epsilon} x \subseteq S$
Hence the result by definition of open set.
{{qed}}
\end{proof}
|
22038
|
\section{Subset of Subset Product}
Tags: Subset Products, Abstract Algebra
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\powerset S$ be the power set of $S$.
Let $X, Y, Z \in \powerset S$.
Then:
:$X \subseteq Y \implies \paren {X \circ Z} \subseteq \paren {Y \circ Z}$
:$X \subseteq Y \implies \paren {Z \circ X} \subseteq \paren {Z \circ Y}$
where $X \circ Z$ etc. denotes subset product.
\end{theorem}
\begin{proof}
Let $x \in X, z \in Z$.
Then:
:$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$
Now:
:$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$
:$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$
But by the definition of a subset:
:$x \in X \implies x \in Y$
Thus:
:$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$
and the result follows.
{{Qed}}
\end{proof}
|
22039
|
\section{Subset of Toset is Toset}
Tags: Total Orderings
\begin{theorem}
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $T \subseteq S$.
Then $\left({T, \preceq \restriction_T}\right)$ is also a totally ordered set.
In the above, $\preceq \restriction_T$ denotes the restriction of $\preceq$ to $T$.
\end{theorem}
\begin{proof}
As $\left({S, \preceq}\right)$ is a totally ordered set, the relation $\preceq$ is a total ordering, and is by definition:
* reflexive
* antisymmetric
* transitive
* connected
From Properties of Restriction of Relation, a restriction of a relation which has all those properties inherits them all.
Thus $\preceq \restriction_T$ is also:
* reflexive
* antisymmetric
* transitive
* connected
and so is also a total ordering.
Hence the result, by definition of totally ordered set.
{{Qed}}
\end{proof}
|
22040
|
\section{Subset of Well-Founded Relation is Well-Founded}
Tags: Well-Founded Relations
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation on $S$.
Let $\QQ$ be a subset of $\RR$.
Then $\QQ$ is also a well-founded relation on $S$.
\end{theorem}
\begin{proof}
{{AimForCont}} $\struct {S, \QQ}$ is not a well-founded set.
By Infinite Sequence Property of Well-Founded Relation there exists an infinite sequence $\sequence {x_n}$ in $S$ such that:
:$\forall n \in \N: \tuple {x_{n + 1}, x_n} \in \QQ \text { and } x_{n + 1} \ne x_n$
But then because $\QQ \subseteq \RR$, it follows that:
:$\forall n \in \N: \tuple {x_{n + 1}, x_n} \in \QQ \implies \tuple {x_{n + 1}, x_n} \in \RR$
and it follows that:
:$\forall n \in \N: \tuple {x_{n + 1}, x_n} \in \RR \text { and } x_{n + 1} \ne x_n$
Hence by Infinite Sequence Property of Well-Founded Relation it follows that $\RR$ is not a well-founded relation on $S$.
The result follows by Proof by Contradiction.
{{Qed}}
\end{proof}
|
22041
|
\section{Subset of Well-Ordered Set is Well-Ordered}
Tags: Well-Orderings, Subset of Well-Ordered Set is Well-Ordered, Orderings
\begin{theorem}
Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then the relational structure $\struct {T, \preceq'}$ is a well-ordered set.
\end{theorem}
\begin{proof}
Let $\struct {S, \preceq}$ be a well-ordered set.
Let $T \subseteq S$.
Let $X \subseteq T$.
By Subset Relation is Transitive, $X \subseteq S$.
By the definition of a well-ordered set, $X$ has a smallest element.
It follows by definition that $T$ is well-ordered.
Hence the result.
{{Qed}}
\end{proof}
|
22042
|
\section{Subsets in Increasing Union}
Tags: Subsets, Set Union, Unions, Union, Subset
\begin{theorem}
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a nested sequence of sets, that is:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
Then:
:$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$
\end{theorem}
\begin{proof}
Let $k \in \N$.
Let $j \ge k$.
Then by as many applications as necessary of Subset Relation is Transitive, we have:
:$S_k \subseteq S_j$
Now $s \in S$ means, by definition of set union, that:
:$\exists S_k \subseteq S: s \in S_k$
Then from above:
:$j \ge k \implies S_k \subseteq S_j$
it follows directly that:
:$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$
from the definition of subset.
{{qed}}
\end{proof}
|
22043
|
\section{Subsets of Disjoint Sets are Disjoint}
Tags: Disjoint Sets, Subsets, Subset
\begin{theorem}
Let $S$ and $T$ be disjoint sets.
Let $S' \subseteq S$ and $T' \subseteq T$.
Then $S'$ and $T'$ are disjoint.
\end{theorem}
\begin{proof}
Let $S \cap T = \O$.
Let $S' \subseteq S$ and $T' \subseteq T$.
{{AimForCont}} $S' \cap T' \ne \O$.
Then:
{{begin-eqn}}
{{eqn | l = \exists x
| o = \in
| r = S' \cap T'
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S'
| c = {{Defof|Set Intersection}}
}}
{{eqn | lo= \land
| l = x
| o = \in
| r = T'
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | lo= \land
| l = x
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \cap T
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| l = S \cap T
| o = \ne
| r = \O
| c = {{Defof|Set Intersection}}
}}
{{end-eqn}}
From this contradiction:
:$S' \cap T' = \O$
Hence the result by definition of disjoint sets.
{{qed}}
\end{proof}
|
22044
|
\section{Subsets of Equidecomposable Subsets are Equidecomposable}
Tags: Topology
\begin{theorem}
Let $A, B \subseteq \R^n$ be equidecomposable.
Let $S \subseteq A$.
Then there exists $T \subseteq B$ such that $S$ and $T$ are equidecomposable.
\end{theorem}
\begin{proof}
Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \ldots, \mu_m, \nu_1, \ldots, \nu_m: \R^n \to \R^n$ such that:
:$\ds A = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$
and
:$\ds B = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$
Define:
:$Y_i = \mu_i^{-1} \paren {S \cap \map {\mu_i} {X_i} }$
Then:
{{begin-eqn}}
{{eqn | l = \bigcup_{i \mathop = 1}^m \map {\mu_i} {Y_i}
| r = \bigcup_{i \mathop = 1}^m \paren {S \cap \map {\mu_i} {X_i} }
| c =
}}
{{eqn | r = S \cap \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}
| c =
}}
{{eqn | r = S \cap A
| c =
}}
{{eqn | r = S
| c =
}}
{{end-eqn}}
and so $\sequence {Y_i}_{i \mathop = 1}^m$ forms a decomposition of $S$.
But for each $i$:
:$\paren {S \cap \map {\mu_i} {X_i} } \subseteq \map {\mu_i} {X_i}$
and so:
{{begin-eqn}}
{{eqn | l = Y_i
| r = \map {\mu_i^{-1} } {S \cap \map {\mu_i} {X_i} }
| c =
}}
{{eqn | o = \subseteq
| r = \map {\mu_i^{-1} } {\map {\mu_i} {X_i} }
| c =
}}
{{eqn | r = X_i
| c =
}}
{{end-eqn}}
Hence:
:$\map {\nu_i} {Y_i} \subseteq \map {\nu_i} {X_i}$
and so:
{{begin-eqn}}
{{eqn | l = \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i}
| o = \subseteq
| r = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}
| c =
}}
{{eqn | r = B
| c =
}}
{{end-eqn}}
Define:
:$\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i} = T$
Hence the result.
{{qed}}
Category:Topology
\end{proof}
|
22045
|
\section{Subspace Topology is Initial Topology with respect to Inclusion Mapping}
Tags: Inclusion Mappings, Initial Topology, Topological Subspaces
\begin{theorem}
Let $\struct {X, \tau}$ be a topological space.
Let $Y$ be a non-empty subset of $X$.
Let $\iota: Y \to X$ be the inclusion mapping.
Let $\tau_Y$ be the initial topology on $Y$ with respect to $\iota$.
Then $\struct {Y, \tau_Y}$ is a topological subspace of $\struct {X, \tau}$.
That is:
:$\tau_Y = \set {U \cap Y: U \in \tau}$
\end{theorem}
\begin{proof}
By Initial Topology with respect to Mapping equals Set of Preimages, it follows that:
:$\tau_Y = \set {\iota^{-1} \sqbrk U: U \in \tau}$
From Preimage of Subset under Inclusion Mapping, we have:
:$\forall S \subseteq X: \iota^{-1} \sqbrk S = S \cap Y$
Hence the result.
{{qed}}
Category:Topological Subspaces
Category:Inclusion Mappings
Category:Initial Topology
\end{proof}
|
22046
|
\section{Subspace of Complete Metric Space is Closed iff Complete}
Tags: Metric Subspaces, Complete Metric Spaces, Metric Spaces
\begin{theorem}
Let $\struct {M, d}$ be a complete metric space.
Let $\struct {S, d}$ be a subspace of $\struct {M, d}$.
Then $S$ is closed {{iff}} $S$ is complete.
\end{theorem}
\begin{proof}
This will be proved by demonstrating the contrapositive:
:$S$ is not complete {{iff}} $S$ is not closed.
\end{proof}
|
22047
|
\section{Subspace of Either-Or Space less Zero is not Lindelöf}
Tags: Either-Or Topology, Lindelöf Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be the either-or space.
Let $H = S \setminus \set 0$ be the set $S$ without zero.
Then the topological subspace $T_H = \struct {H, \tau_H}$ is not a Lindelöf space.
\end{theorem}
\begin{proof}
By definition of topological subspace, $U \subseteq H$ is open in $T_H$ {{iff}}:
:$(1): \quad \set 0 \nsubseteq U$
or:
:$(2): \quad \openint {-1} 1 \subseteq U$
But for all $U \subseteq H$, condition $(1)$ holds as $0 \notin H$.
So $T_H$ is by definition a discrete space.
As $T_H$ is uncountable, we have that Uncountable Discrete Space is not Lindelöf holds.
Hence the result.
{{qed}}
\end{proof}
|
22048
|
\section{Subspace of Finite Complement Topology is Compact}
Tags: Compact Spaces, Finite Complement Topology, Compactness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then every topological subspace of $T$, including $T$ itself, is a compact space.
\end{theorem}
\begin{proof}
Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\CC$ be an open cover of $T_H$.
Let $U \in \CC$ be any set in $C$.
$U$ covers all but a finite number of points of $T_H$.
So for each of those points we pick an element of $\CC$ which covers each of those points.
Hence we have a finite subcover of $T_H$.
So by definition $T_H$ is a compact space.
{{qed}}
\end{proof}
|
22049
|
\section{Subspace of Metric Space is Metric Space}
Tags: Metric Subspaces, Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Let $d_H: H \times H \to \R$ be the restriction $d \restriction_{H \times H}$ of $d$ to $H$.
Let $\struct {H, d_H}$ be a metric subspace of $\struct {A, d}$.
Then $d_H$ is a metric on $H$.
\end{theorem}
\begin{proof}
By definition of restriction:
:$\forall x, y \in H: \map {d_H} {x, y} = \map d {x, y}$
As $d$ is a metric, the metric space axioms are all fulfilled by all $x, y \in A$ under $d$.
As $H \subseteq A$, by definition of subset, all $x, y \in H$ are also elements of $A$.
Therefore the metric space axioms are all fulfilled by all $x, y \in H$ under $d_H$.
{{qed}}
\end{proof}
|
22050
|
\section{Subspace of Noetherian Space is Noetherian}
Tags: Noetherian Spaces
\begin{theorem}
Let $X$ be a Noetherian topological space.
Let $Y \subseteq X$ be a subspace.
Then $Y$ is Noetherian.
\end{theorem}
\begin{proof}
Let $Y_1 \subset Y_2 \subset \ldots \subset$ be an ascending chain of open sets in $Y$.
{{explain|If we are going to use the term "ascending chain", then we need to have a page that defines exactly that.}}
By definition of subspace topology, there exists open sets $X_1, X_2, \ldots$ such that
:$X_i \cap Y = Y_i$
for all $i$.
By taking the intersection $Z_i := \ds \bigcap_{j \mathop = i}^{\infty} X_j$, we have:
:$Z_i \cap Y = Y_i$
:$Z_1 \subset Z_2 \subset \dots$
Since $X$ is Noetherian, every ascending chain of open sets is eventually constant.
Hence $Z_i$ is eventually constant.
Then $Y_i = Z_i \cap Y$ is eventually constant.
Hence $Y$ is also Noetherian.
\end{proof}
|
22051
|
\section{Subspace of Product Space is Homeomorphic to Factor Space}
Tags: Homeomorphisms, Subspace of Product Space Homeomorphic to Factor Space, Topological Subspaces, Subspace of Product Space is Homeomorphic to Factor Space, Product Spaces
\begin{theorem}
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.
\end{theorem}
\begin{proof}
Take any factor space $X_i$, and $z_j \in X_j$ where $j \ne i$.
Then $\displaystyle Y_i = X_i \times \prod_{i\ne j\in I} \{z_j\} \subseteq X$ is a subspace of $(X,\tau)$.
Even more, from:
* Projection from Product Topology is Continuous
* Projection from Product Topology is Open
we have that $\operatorname{pr}_i \restriction_{Y_i}: Y_i\to X_i$ is a homeomorphism because it is open, continuous and bijective.
{{qed}}
Category:Product Spaces
Category:Topological Subspaces
Category:Homeomorphisms
127675
82903
2013-01-19T03:21:39Z
Dfeuer
1672
Make the mathematics more precise and improve the theorem conclusion to make it more specific and hence useful. The "subspace" described wasn't actually a subspace (though it was homeomorphic to one), so nothing quite worked.
127675
wikitext
text/x-wiki
\end{proof}
|
22052
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Product Spaces, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $T_1$ and $T_2$ be non-empty topological spaces.
Let $b \in T_2$.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the product space of $T_2$ and $T_1$.
Then:
:$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$
:$T_1$ is homeomorphic to the subspace $\set b \times T_1$ of $T_2 \times T_1$
\end{theorem}
\begin{proof}
The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set{b}$ be defined as:
:$\map f x = \tuple {x, b}$
\end{proof}
|
22053
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
For all for all $j \in I$ let:
::$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$
Then:
:$Y_i = \prod_{j \mathop \in I} Z_j$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | r = x \in Y_i
| o =
}}
{{eqn | ll = \leadstoandfrom
| q = \forall j \in I
| l = x_j
| r = \begin {cases} z_j & j \ne i \\ x_i \in X_i & i = j \end {cases}
| c = Definition of $Y_i$
}}
{{eqn | ll = \leadstoandfrom
| q = \forall j \in I
| l = x_j
| o = \in
| r = Z_j
| c = Definition of $Z_j$ for all $j \in I$
}}
{{eqn | ll = \leadstoandfrom
| l = x
| o = \in
| r = \prod_{j \mathop \in I} Z_j
| c = {{Defof|Cartesian Product}}
}}
{{end-eqn}}
The result follows by definition of set equality.
{{qed}}
Category:Subspace of Product Space is Homeomorphic to Factor Space
\end{proof}
|
22054
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 2}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $\pr_i : X \to X_i$ be the $i$th-projection from $X$.
For all for all $j \in I$ let:
:$Z_j = \begin{cases} X_i & i = j \\ \set{z_j} & j \ne i \end{cases}$
Let $Y_i = \prod_{j \mathop \in I} Z_j$
Let $p_i : Y_i \to X_i$ be the $i$th-projection from $Y_i$.
Then:
:$\pr_i {\restriction_{Y_i} } = p_i$
\end{theorem}
\begin{proof}
For all $y \in Y_i$:
{{begin-eqn}}
{{eqn | l = \map {\pr_i {\restriction_{Y_i} } } y
| r = \map {\pr_i} y
| c = {{Defof|Restriction of Mapping}}: $\pr_i {\restriction_{Y_i} } : Y_i \to X_i$
}}
{{eqn | r = y_i
| c = {{Defof|Projection}}: $\pr_i: X \to X_i$
}}
{{eqn | r = \map {p_i} y
| c = {{Defof|Projection}}: $p_i: Y_i \to X_i$
}}
{{end-eqn}}
By Equality of Mappings:
:$\pr_i {\restriction_{Y_i} } = p_i$
{{qed}}
Category:Subspace of Product Space is Homeomorphic to Factor Space
\end{proof}
|
22055
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Continuous Mapping}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Let $p_i = \pr_i {\restriction_{Y_i}}$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
:$p_i$ is continuous.
\end{theorem}
\begin{proof}
Let $V \in \tau_i$.
Let $\ds U = \prod_{i \mathop \in I} U_i$ where:
:$U_j = \begin{cases} X_j & j \ne i \\ V & j = i \end{cases}$
From Natural Basis of Product Topology, $U$ is an element of the the natural basis.
By definition of the product topology $\tau$ on the product space $\struct {X, \tau}$ the natural basis is a basis for the product topology.
It follows that:
:$U$ is open in $\struct {X, \tau}$
Let $x \in Y_i$.
Now:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {p_i^\gets} V
}}
{{eqn | ll= \leadstoandfrom
| l = \map {p_i} x
| o = \in
| r = V
| c = {{Defof|Inverse Image Mapping of Mapping}}: $p_i^\gets$
}}
{{eqn | ll= \leadstoandfrom
| l = \map {\pr_i} x
| o = \in
| r = V
| c = {{Defof|Restriction of Mapping}} $p_i$
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = U
| c = Definition of $U$
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = U \cap Y_i
| c = as $x \in Y_i$
}}
{{end-eqn}}
By set equality:
:$\map {p_i^\gets} V = U \cap Y_i$
By definition of the subspace topology on $Y_i$:
:$\map {p_i^\gets} V \in \upsilon_i$
It follows that $p_i$ is continuous by definition.
{{qed}}
Category:Subspace of Product Space is Homeomorphic to Factor Space
\end{proof}
|
22056
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Injection}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $p_i = \pr_i {\restriction_{Y_i}}$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
:$p_i$ is an injection.
\end{theorem}
\begin{proof}
Note that by definitions of a restriction and a projection then:
:$\forall y \in Y_i: \map {p_i} y = y_i$
Let $x, y \in Y_i$.
Then for all $j \in I \setminus \set i$:
:$x_j = z_j = y_j$
Let $\map {p_i} x = \map {p_i} y$.
Then:
:$x_i = y_i$
Thus:
:$x = y$
It follows that $p_i$ is an injection by definition.
{{qed}}
Category:Subspace of Product Space is Homeomorphic to Factor Space
\end{proof}
|
22057
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Lemma 1}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
For all $k \in I$, let $\pr_k$ denote the projection from $X$ to $X_k$.
Let $z \in X$.
Let $i, k \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $p_i = \pr_i {\restriction_{Y_i} }$ be the restriction of $\pr_i$ to $Y_i$.
Let $V_k \in \tau_k$.
Let $\map {\pr_k^\gets } {V_k} \cap Y_i \ne \O$.
Then:
:$\map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}$ is open in $\struct{X_i, \tau_i}$
\end{theorem}
\begin{proof}
We have that $p_i$ is a bijection from the lemmas:
:$p_i$ is an injection
:$p_i$ is a surjection
Let $x \in X_i$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}
}}
{{eqn | ll= \leadstoandfrom
| l = \map {p_i^{-1} } x
| o = \in
| r = \map {\pr_k^\gets} {V_k} \cap Y_i
| c = {{Defof|Direct Image Mapping of Mapping}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map {p_i^{-1} } x
| o = \in
| r = \map {\pr_k^\gets} {V_k}
| c = as $\map {p_i^{-1} } x \in Y_i$
}}
{{eqn | ll= \leadstoandfrom
| l = \map {\pr_k} {\map {p_i^{-1} } x}
| o = \in
| r = V_k
| c = {{Defof|Direct Image Mapping of Mapping}}
}}
{{end-eqn}}
By definition of $p_i$:
:$\map {p_i^{-1} } x = y$
where:
:$\forall j \in I : y_j = \begin {cases} z_j & j \ne i \\ x & j = i \end {cases}$
\end{proof}
|
22058
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Open Mapping}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Let $p_i = \pr_i {\restriction_{Y_i} }$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
:$p_i$ is an open mapping.
\end{theorem}
\begin{proof}
Let $U \in \upsilon_i$.
Let $x \in \map {p_i^\to} U$.
Then by definition of the direct image mapping:
:$\exists y \in U : x = \map {p_i} y$
By the definition of the subspace topology:
:$\exists U' \in \tau: U = U' \cap Y_i$
For all $k \in I$ let $\pr_k$ denote the projection from $X$ to $X_k$.
By definition of the natural basis of the product topology $\tau$:
:there exists a finite subset $J$ of $I$
and:
:for each $k \in J$, there exists a $V_k \in \tau_k$
such that:
:$\ds y \in \bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} \subseteq U'$
Then:
:$\ds y \in \paren {\bigcap_{k \mathop \in J} \map{\pr_k^\gets} {V_k} } \cap Y_i \subseteq U' \cap Y_i = U$
By definition of direct image mapping:
:$\ds x = \map {p_i} y \in \map {p_i^\to} {\paren {\bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} } \cap Y_i} \subseteq \map {p_i^\to} U$
Recall that $p_i$ is an injection.
Then:
{{begin-eqn}}
{{eqn | l = \map {p_i^\to} {\paren {\bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} } \cap Y_i}
| r = \map {p_i^\to} {\bigcap_{k \mathop \in J} \paren {\map {\pr_k^\gets} {V_k} \cap Y_i} }
| c = Intersection Distributes over Intersection
}}
{{eqn | r = \bigcap_{k \mathop \in J} \map {p_i^\to} {\map {\pr_k^\gets } {V_k} \cap Y_i}
| c = Image of Intersection under Injection
}}
{{end-eqn}}
Let $k \in J$.
\end{proof}
|
22059
|
\section{Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Surjection}
Tags: Subspace of Product Space Homeomorphic to Factor Space, Subspace of Product Space is Homeomorphic to Factor Space
\begin{theorem}
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $p_i = \pr_i {\restriction_{Y_i} }$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
:$p_i$ is a surjection.
\end{theorem}
\begin{proof}
Note that by definitions of a restriction and a projection then:
:$\forall y \in Y_i: \map {p_i} y = y_i$
Let $x \in X_i$.
Let $y \in Y_i$ be defined by:
:$\forall j \in I: y_j = \begin{cases}
z_j & j \ne i \\
x & j = i
\end{cases}$
Then:
:$\map {p_i} y = y_i = x$
It follows that $p_i$ is an surjection by definition.
{{qed}}
Category:Subspace of Product Space is Homeomorphic to Factor Space
\end{proof}
|
22060
|
\section{Subspace of Real Continuous Functions}
Tags: Linear Algebra, Vector Subspaces, Analysis
\begin{theorem}
Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a closed interval of the real number line $\R$.
Let $\map \CC {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$.
Then $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
\end{theorem}
\begin{proof}
By definition, $\map \CC {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map \CC {\mathbb J}$.
By Two-Step Vector Subspace Test, it needs to be shown that:
:$(1): \quad f + g \in \map \CC {\mathbb J}$
:$(2): \quad \lambda f \in \map \CC {\mathbb J}$ for any $\lambda \in \R$
$(1)$ follows by Sum Rule for Continuous Real Functions.
$(2)$ follows by Multiple Rule for Continuous Real Functions.
Hence $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
{{qed}}
\end{proof}
|
22061
|
\section{Subspace of Real Differentiable Functions}
Tags: Differential Calculus, Differentiation, Vector Subspaces
\begin{theorem}
Let $\mathbb J$ be an open interval of the real number line $\R$.
Let $\map \DD {\mathbb J}$ be the set of all differentiable real functions on $\mathbb J$.
Then $\struct {\map \DD {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
\end{theorem}
\begin{proof}
Note that by definition, $\map \DD {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map \DD {\mathbb J}$.
Let $\lambda \in \R$.
From Linear Combination of Derivatives, we have that:
:$f + \lambda g$ is differentiable on $\mathbb J$.
That is:
:$f + \lambda g \in \map \DD {\mathbb J}$
So, by One-Step Vector Subspace Test:
:$\struct {\map \DD {\mathbb J}, +, \times}_\R$ is a subspace of $\R^{\mathbb J}$.
{{qed}}
\end{proof}
|
22062
|
\section{Subspace of Real Functions of Differentiability Class}
Tags: Vector Subspaces, Analysis
\begin{theorem}
Let $\mathbb J = \set {x \in \R: a < x < b}$ be an open interval of the real number line $\R$.
Let $\map {\CC^m} {\mathbb J}$ be the set of all continuous real functions on $\mathbb J$ in differentiability class $m$.
Then $\struct {\map {\CC^m} {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
\end{theorem}
\begin{proof}
Note that by definition, $\map {\CC^m} {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map {\CC^m} {\mathbb J}$.
Let $\lambda \in \R$.
Applying Linear Combination of Derivatives $m$ times we have:
:$f + \lambda g$ is $m$-times differentiable on $\mathbb J$ with $m$th derivative $f^{\paren m} + \lambda g^{\paren m}$.
Since both $f$ and $g$ are of differentiability class $m$:
:$f^{\paren m}$ and $g^{\paren m}$ are continuous on $\mathbb J$.
From Combined Sum Rule for Continuous Real Functions:
:$f^{\paren m} + \lambda g^{\paren m} = \paren {f + \lambda g}^{\paren m}$ is continuous on $\mathbb J$.
So:
:$f + \lambda g \in \map {\CC^m} {\mathbb J}$
Therefore, by One-Step Vector Subspace Test:
:$\struct {\map {\CC^m} {\mathbb J}, +, \times}_\R$ is a subspace of $\struct {\R^{\mathbb J}, +, \times}_\R$.
{{qed}}
\end{proof}
|
22063
|
\section{Subspace of Riemann Integrable Functions}
Tags: Vector Subspaces, Analysis
\begin{theorem}
Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a closed interval of the real number line $\R$.
Let $\map \RR {\mathbb J}$ be the set of all Riemann integrable functions on $\mathbb J$.
Then $\struct {\map \RR {\mathbb J}, +, \times}_\R$ is a subspace of the $\R$-vector space $\struct {\R^{\mathbb J}, +, \times}_\R$.
\end{theorem}
\begin{proof}
Note that by definition, $\map \RR {\mathbb J} \subseteq \R^{\mathbb J}$.
Let $f, g \in \map \RR {\mathbb J}$.
Let $\lambda \in \R$.
By Linear Combination of Definite Integrals:
:$f + \lambda g$ is Riemann integrable on $\mathbb J$.
That is:
:$f + \lambda g \in \map \RR {\mathbb J}$
So by One-Step Vector Subspace Test:
:$\struct {\map \RR {\mathbb J}, +, \times}_\R$ is a subspace of $\struct {\R^{\mathbb J}, +, \times}_\R$.
{{qed}}
\end{proof}
|
22064
|
\section{Subspace of Subspace is Subspace}
Tags: Topological Subspaces
\begin{theorem}
Let $T = \struct{S, \tau}$ be a topological space.
Let $H \subseteq S$ and $\tau_H$ be the subspace topology on $H$.
Let $K\subseteq H$.
Then the subspace topology on $K$ induced by $\tau$ equals the subspace topology on $K$ induced by $\tau_H$.
\end{theorem}
\begin{proof}
Let $\tau_K$ be the subspace topology on $K$ induced by $\tau$.
Let $\tau’_K$ be the subspace topology on $K$ induced by $\tau_H$.
Then
{{begin-eqn}}
{{eqn | l = V \in \tau’_K
| o = \leadstoandfrom
| r = \exists U’ \in \tau_H : V = U’ \cap K
| c = {{Defof|Subspace Topology}} $\tau’_K$
}}
{{eqn | o = \leadstoandfrom
| r = \exists U \in \tau : V = \paren {U \cap H} \cap K
| c = {{Defof|Subspace Topology}} $\tau_H$
}}
{{eqn | o = \leadstoandfrom
| r = \exists U \in \tau : V = U \cap \paren {H \cap K}
| c = Intersection is Associative
}}
{{eqn | o = \leadstoandfrom
| r = \exists U \in \tau : V = U \cap K
| c = Intersection with Subset is Subset
}}
{{eqn | o = \leadstoandfrom
| r = V \in \tau_K
| c = {{Defof|Subspace Topology}} $\tau_K$
}}
{{end-eqn}}
Category:Topological Subspaces
\end{proof}
|
22065
|
\section{Subspaces of Dimension 2 Real Vector Space}
Tags: Subspaces of Dimension 2 Real Vector Space, Linear Algebra
\begin{theorem}
Take the $\R$-vector space $\left({\R^2, +, \times}\right)_\R$.
Let $S$ be a subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ is one of:
: $(1): \quad \left({\R^2, +, \times}\right)_\R$
: $(2): \quad \left\{{0}\right\}$
: $(3): \quad$ A line through the origin.
\end{theorem}
\begin{proof}
* Let $S$ be a non-zero subspace of $\left({\R^2, +, \times}\right)_\R$.
Then $S$ contains a non-zero vector $\left({\alpha_1, \alpha_2}\right)$.
Hence $S$ also contains $\left\{{\lambda \times \left({\alpha_1, \alpha_2}\right), \lambda \in \R}\right\}$.
From Equation of a Straight Line, this set may be described as a line through the origin.
* Suppose $S$ also contains a non-zero vector $\left({\beta_1, \beta_2}\right)$ which is not on that line.
Then $\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1 \ne 0$.
Otherwise $\left({\beta_1, \beta_2}\right)$ would be $\zeta \times \left({\alpha_1, \alpha_2}\right)$, where either $\zeta = \beta_1 / \alpha_1$ or $\zeta = \beta_2 / \alpha_2$ according to whether $\alpha_1 \ne 0$ or $\alpha_2 \ne 0$.
But then $S = \left({\R^2, +, \times}\right)_\R$.
Because, if $\left({\gamma_1, \gamma_2}\right)$ is any vector at all, then:
: $\left({\gamma_1, \gamma_2}\right) = \lambda \times \left({\alpha_1, \alpha_2}\right) + \mu \times \left({\beta_1, \beta_2}\right)$
where $\lambda = \dfrac {\gamma_1 \times \beta_2 - \gamma_2 \times \beta_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}, \mu = \dfrac {\alpha_1 \times \gamma_2 - \alpha_2 \times \gamma_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}$
which we get by solving the simultaneous eqns:
{{begin-eqn}}
{{eqn | l=\alpha_1 \times \lambda + \beta_1 \times \mu
| r=0
| c=
}}
{{eqn | l=\alpha_2 \times \lambda + \beta_2 \times \mu
| r=0
| c=
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
22066
|
\section{Substitution Instance of Term is Term}
Tags: Predicate Logic
\begin{theorem}
Let $\beta, \tau$ be terms of predicate logic.
Let $x \in \operatorname {VAR}$ be a variable.
Let $\map \beta {x \gets \tau}$ be the substitution instance of $\beta$ substituting $\tau$ for $x$.
Then $\map \beta {x \gets \tau}$ is a term.
\end{theorem}
\begin{proof}
Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$.
If $\beta = y$ for some variable $y$, then:
:$\map \beta {x \gets \tau} = \begin{cases} \tau & : \text {if $y = x$} \\ y &: \text {otherwise} \end{cases}$
In either case, $\map \beta {x \gets \tau}$ is a term.
If $\beta = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then:
:$\map \beta {x \gets \tau} = \map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$
By the induction hypothesis, each $\map {\tau_i} {x \gets \tau}$ is a term.
Hence so is $\map \beta {x \gets \tau}$.
The result follows by the Principle of Structural Induction.
{{qed}}
\end{proof}
|
22067
|
\section{Substitution Instance of WFF is WFF}
Tags: Predicate Logic
\begin{theorem}
Let $\mathbf A$ be a WFF of predicate logic.
Let $\tau$ be a term of predicate logic.
Let $x \in \mathrm{VAR}$ be a variable.
Let $\mathbf A \left({x \gets \tau}\right)$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$.
Then $\mathbf A \left({x \gets \tau}\right)$ is a WFF.
\end{theorem}
\begin{proof}
Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$.
If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
:$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\right)$
where $\tau_i \left({x \gets \tau}\right)$ is the substitution instance of $\tau_i$.
By Substitution Instance of Term is Term, each such $\tau_i \left({x \gets \tau}\right)$ is again a term.
It follows that $\mathbf A \left({x \gets \tau}\right)$ is again a WFF.
If $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$, then:
:$\mathbf A \left({x \gets \tau}\right) = \neg \mathbf B \left({x \gets \tau}\right)$
and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic.
If $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$:
:$\mathbf A \left({x \gets \tau}\right) = \mathbf B \left({x \gets \tau}\right) \circ \mathbf B' \left({x \gets \tau}\right)$
and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic.
If $\mathbf A = \exists x: \mathbf B$, and the induction hypothesis applies to $\mathbf B$:
:$\mathbf A \left({x \gets \tau}\right) = \exists x: \mathbf B \left({x \gets \tau}\right)$
and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic.
If $\mathbf A = \forall x : \mathbf B$, and the induction hypothesis applies to $\mathbf B$:
:$\mathbf A \left({x \gets \tau}\right) = \forall x: \mathbf B \left({x \gets \tau}\right)$
and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a WFF of predicate logic.
The result follows by the Principle of Structural Induction.
{{qed}}
\end{proof}
|
22068
|
\section{Substitution Rule for Matrices}
Tags: Matrix Algebra
\begin{theorem}
Let $\mathbf A$ be a square matrix of order $n$.
Then:
:$(1): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{j k} = a_{i k}$
:$(2): \quad \ds \sum_{j \mathop = 1}^n \delta_{i j} a_{k j} = a_{k i}$
where:
:$\delta_{i j}$ is the Kronecker delta
:$a_{j k}$ is element $\tuple {j, k}$ of $\mathbf A$.
\end{theorem}
\begin{proof}
By definition of Kronecker delta:
:$\delta_{i j} = \begin {cases}
1 & : i = j \\
0 & : i \ne j \end {cases}$
Thus:
:$\delta_{i j} a_{j k} = \begin {cases}
a_{i k} & : i = j \\
0 & : i \ne j \end {cases}$
and:
:$\delta_{i j} a_{k j} = \begin {cases}
a_{k i} & : i = j \\
0 & : i \ne j \end {cases}$
from which the result follows.
{{qed}}
\end{proof}
|
22069
|
\section{Substitution Theorem for Terms}
Tags: Predicate Logic, Named Theorems
\begin{theorem}
Let $\beta, \tau$ be terms.
Let $x \in \mathrm {VAR}$ be a variable.
Let $\map \beta {x \gets \tau}$ be the substitution instance of $\beta$ substituting $\tau$ for $x$.
Let $\AA$ be a structure for predicate logic.
Let $\sigma$ be an assignment for $\beta$ and $\tau$.
Suppose that:
:$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = a$
where $\map {\operatorname{val}_\AA} \tau \sqbrk \sigma$ is the value of $\tau$ under $\sigma$.
Then:
:$\map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma = \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }$
where $\sigma + \paren {x / a}$ is the extension of $\sigma$ by mapping $x$ to $a$.
\end{theorem}
\begin{proof}
Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$.
If $\beta = y$ for some variable $y$, then:
:$\map \beta {x \gets \tau} = \begin {cases} \tau & : \text{if $y = x$} \\ y & : \text {otherwise} \end {cases}$
In the first case:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma
| r = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma
}}
{{eqn | r = a
| c = Definition of $a$
}}
{{eqn | r = \map {\paren {\sigma + \paren {x / a} } } x
| c = {{Defof|Extension of Assignment}}
}}
{{eqn | r = \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }
| c = {{Defof|Value of Term under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
In the second case:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma
| r = \map {\operatorname{val}_\AA} y \sqbrk \sigma
}}
{{eqn | r = \map \sigma y
| c = {{Defof|Value of Term under Assignment|Value under $\sigma$}}
}}
{{eqn | r = \map {\paren {\sigma + \paren {x / a} } } y
| c = {{Defof|Extension of Assignment}}
}}
{{eqn | r = \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }
| c = {{Defof|Value of Term under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
as desired.
If $\beta = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then:
:$\map \beta {x \gets \tau} = \map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$
Now:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} {\map \beta {x \gets \tau} } \sqbrk \sigma
| r = \map {\operatorname{val}_\AA} {\map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} } } \sqbrk \sigma
}}
{{eqn | r = \map {f_\AA} {\map {\operatorname{val}_\AA} {\map {\tau_1} {x \gets \tau} } \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\map {\tau_n} {x \gets \tau} } \sqbrk \sigma}
| c = {{Defof|Value of Term under Assignment|Value under $\sigma$}}
}}
{{eqn | r = \map {f_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk {\sigma + \paren {x / a} }, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk {\sigma + \paren {x / a} } }
| c = Induction Hypothesis
}}
{{eqn | r = \map {\operatorname{val}_\AA} \beta \sqbrk {\sigma + \paren {x / a} }
| c = {{Defof|Value of Term under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
as desired.
The result follows by the Principle of Structural Induction.
{{qed}}
\end{proof}
|
22070
|
\section{Substitution Theorem for Well-Formed Formulas}
Tags: Predicate Logic, Named Theorems
\begin{theorem}
Let $\mathbf A$ be a WFF of predicate logic.
Let $x \in \mathrm{VAR}$ be a variable.
Let $\tau$ be a term of predicate logic which is freely substitutable for $x$ in $\mathbf A$.
Let $\map {\mathbf A} {x \gets \tau}$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$.
Let $\AA$ be a structure for predicate logic.
Let $\sigma$ be an assignment for $\mathbf A$ and $\tau$.
Suppose that:
:$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = a$
where $\map {\operatorname{val}_\AA} \tau \sqbrk \sigma$ is the value of $\tau$ under $\sigma$.
Then:
:$\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$
where $\sigma + \paren {x / a}$ is the extension of $\sigma$ by mapping $x$ to $a$.
\end{theorem}
\begin{proof}
Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$.
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\mathbf A} {x \gets \tau} = \map p {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$
where $\map {\tau_i} {x \gets \tau}$ is the substitution instance of $\tau_i$.
Now:
{{begin-eqn}}
{{eqn|l = \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma
|r = \map {\operatorname{val}_\AA} {\map p {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} } } \sqbrk \sigma
}}
{{eqn|r = \map {p_\AA} {\map {\operatorname{val}_\AA} {\map {\tau_1} {x \gets \tau} } \sqbrk \sigma, \ldots, \map {\operatorname{val}_\AA} {\map {\tau_n} {x \gets \tau} } \sqbrk \sigma}
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma$}}
}}
{{eqn|r = \map {p_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk {\sigma + \paren {x / a} }, \ldots, \map {\operatorname{val}_\AA} {\tau_n} \sqbrk {\sigma + \paren {x / a} } }
|c = Substitution Theorem for Terms
}}
{{eqn|r = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
Suppose $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$.
Then since $\tau$ is also free for $x$ in $\mathbf B$:
{{begin-eqn}}
{{eqn|l = \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma
|r = \map {f^\neg} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma$}}
}}
{{eqn|r = \map {f^\neg} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} } }
|c = Induction Hypothesis
}}
{{eqn|r = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
Suppose $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$.
Then since $\tau$ is also free for $x$ in $\mathbf B$ and $\mathbf B'$:
{{begin-eqn}}
{{eqn|l = \map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma
|r = \map {f^\circ} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk \sigma}
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma$}}
}}
{{eqn|r = \map {f^\circ} {\map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} }, \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} } }
|c = Induction Hypothesis
}}
{{eqn|r = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }
|c = {{Defof|Value of Formula under Assignment|Value under $\sigma + \paren {x / a}$}}
}}
{{end-eqn}}
Suppose $\mathbf A = \exists y: \mathbf B$ or $\mathbf A = \forall y: \mathbf B$, and the induction hypothesis applies to $\mathbf B$.
Because $\tau$ is free for $x$ in $\mathbf A$, it must be that either $x$ does not occur freely in $\mathbf A$, or $y$ does not occur in $\tau$.
In the first case:
:$\map {\mathbf A} {x \gets \tau} = \mathbf A$
and by Value of Formula under Assignment Determined by Free Variables:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$
Now consider the case where $y$ does not occur in $\tau$.
From the definition of value under $\sigma$, $\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma$ is determined by:
:$\map {\operatorname{val}_\AA} {\map {\mathbf B} {x \gets \tau} } \sqbrk {\sigma + \paren {y / a'} }$
where $a'$ ranges over $\AA$.
Now from Value of Term under Assignment Determined by Variables, since $y$ does not occur in $\tau$:
:$\map {\operatorname{val}_\AA} \tau \sqbrk {\sigma + \paren {y / a'} } = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma = a$
for all $a'$.
Hence the induction hypothesis also applies to the assignment $\sigma + \paren {y / a'}$.
Thus, for all $a'$:
{{begin-eqn}}
{{eqn|l = \map {\operatorname{val}_\AA} {\map {\mathbf B} {x \gets \tau} } \sqbrk {\sigma + \paren {y / a'} }
|r = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {y / a'} + \paren {x / a} }
|c = Induction Hypothesis
}}
{{eqn|r = \map {\operatorname{val}_\AA} {\mathbf B} \sqbrk {\sigma + \paren {x / a} + \paren {y / a'} }
|c = {{Defof|Extension of Assignment}}
}}
{{end-eqn}}
from which we infer:
:$\map {\operatorname{val}_\AA} {\map {\mathbf A} {x \gets \tau} } \sqbrk \sigma = \map {\operatorname{val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a} }$
as desired.
The result follows by the Principle of Structural Induction.
{{qed}}
\end{proof}
|
22071
|
\section{Substitution for Equivalent Subformula is Equivalent}
Tags: Propositional Logic, Boolean Interpretations
\begin{theorem}
Let $\mathbf B$ a WFF of propositional logic.
Let $\mathbf A, \mathbf A'$ be equivalent WFFs.
Let $\mathbf A$ be a subformula of $\mathbf B$.
Let $\mathbf B' = \map {\mathbf B} {\mathbf A \,//\, \mathbf A'}$ be the substitution of $\mathbf A'$ for $\mathbf A$ in $\mathbf B$.
Then $\mathbf B$ and $\mathbf B'$ are equivalent.
\end{theorem}
\begin{proof}
Let $v$ be an arbitrary boolean interpretation.
Then $\map v {\mathbf A} = \map v {\mathbf A'}$.
It is to be shown that $\map v {\mathbf B} = \map v {\mathbf B'}$.
We proceed by induction.
Let $\map n {\mathbf B}$ be the number of WFFs $\mathbf C$ such that:
:$\mathbf A$ is a subformula of $\mathbf C$, and $\mathbf C$ is a subformula of $\mathbf B$.
Note that $\map n {\mathbf B} \ne 0$, for $\mathbf C = \mathbf A$ is a valid choice.
Suppose now that $\map n {\mathbf B} = 1$.
Because we have the valid choices $\mathbf C = \mathbf A$ and $\mathbf C = \mathbf B$, it follows that these choices must be identical, i.e. $\mathbf A = \mathbf B$.
Hence $\mathbf B' = \mathbf A'$, and so:
:$\map v {\mathbf B} = \map v {\mathbf B'}$
Suppose now that the assertion is true for all $\mathbf B$ with $\map n {\mathbf B} \le n$.
Let $\map n {\mathbf B} = n + 1$.
Suppose $\mathbf B = \neg \mathbf B_1$.
Then obviously $\map n {\mathbf B_1} = n$, so by hypothesis:
:$\map v {\mathbf B_1} = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }$
Also, by definition of substitution:
:$\mathbf B' = \neg \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'}$
Now, by definition of boolean interpretation:
{{begin-eqn}}
{{eqn | l = \map v {\mathbf B}
| r = \map {f^\neg} {\map v {\mathbf B_1} }
}}
{{eqn | r = \map {f^\neg} {\map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} } }
}}
{{eqn | r = \map v {\neg \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }
}}
{{eqn | r = \map v {\mathbf B'}
}}
{{end-eqn}}
Suppose now that $\mathbf B = \mathbf B_1 \mathbin{\mathsf B} \mathbf B_2$ for a binary connective $\mathsf B$.
Then $\map n {\mathbf B_1}, \map n {\mathbf B_2} \le n$, so:
:$\map v {\mathbf B_1} = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }$
:$\map v {\mathbf B_2} = \map v {\map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} }$
This follows by either the induction hypothesis, or when $\mathbf A$ is not a subformula of $\mathbf B_1$ or $\mathbf B_2$, is entirely trivial, considering the substitution does not change anything.
Also, by definition of substitution:
:$\mathbf B' = \map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} \mathbin {\mathsf B} \map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'}$
Now, by definition of boolean interpretation:
{{begin-eqn}}
{{eqn | l = \map v {\mathbf B}
| r = \map {f^{\mathsf B} } {\map v {\mathbf B_1}, \map v {\mathbf B_2} }
}}
{{eqn | r = \map {f^{\mathsf B} } {\map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} }, \map v {\map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} } }
}}
{{eqn | r = \map v {\map {\mathbf B_1} {\mathbf A \,//\, \mathbf A'} \mathbin {\mathsf B} \map {\mathbf B_2} {\mathbf A \,//\, \mathbf A'} }
}}
{{eqn | r = \map v {\mathbf B'}
}}
{{end-eqn}}
By definition of the language of propositional logic, $\mathbf B$ must have either of the above forms.
Hence the result, from the Second Principle of Mathematical Induction.
{{qed}}
\end{proof}
|
22072
|
\section{Substitution in Big-O Estimate/General Result}
Tags: Asymptotic Notation
\begin{theorem}
Let $X$ and $Y$ be topological spaces.
Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\norm {\,\cdot\,}$.
Let $x_0 \in X$ and $y_0 \in Y$.
Let $f: X \to Y$ be a function with $\map f {x_0} = y_0$ that is continuous at $x_0$.
Let $g, h: Y \to V$ be functions.
Suppose $\map g y = \map O {\map h y}$ as $y \to y_0$, where $O$ denotes big-O notation.
Then $\map {\paren {g \circ f} } x = \map O {\map {\paren {h \circ f} } x}$ as $x \to x_0$.
\end{theorem}
\begin{proof}
Because $g = \map O h$, there exists a neighborhood $V$ of $y_0$ and a real number $c$ such that:
:$\norm {\map g x} \le c \cdot \norm {\map h x}$ for all $y \in V$.
By definition of continuity, there exists a neighborhood $U$ of $x_0$ with $\map f U \subset V$.
For $x \in U$, we have:
:$\norm {\map g {\map f x} } \le c \cdot \norm {\map h {\map f x} }$
Thus $g \circ f = \map O {h \circ f}$ as $x \to x_0$.
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.