id
stringlengths 1
260
| contents
stringlengths 1
234k
|
|---|---|
22873
|
\section{Two-Person Zero-Sum Game is Non-Cooperative}
Tags: Cooperative Games, Definitions: Game Theory, Game Theory, Two-Person Zero-Sum Games
\begin{theorem}
A two-person zero-sum game necessarily has to be non-cooperative.
\end{theorem}
\begin{proof}
A cooperative game is one where players form coalitions against the other players.
If the players in a two-person zero-sum game were to form a coalition, there would be no other players to form it against.
Further, as the total payoff is zero, there would be no benefit in collaborating on using one strategy over another, as when they pool their payoffs they are back where they started.
{{qed}}
\end{proof}
|
22874
|
\section{Two-Person Zero-Sum Game with Multiple Solutions}
Tags: Examples of Two-Person Zero-Sum Games
\begin{theorem}
There exists a two-person zero-sum game with more than one solution.
\end{theorem}
\begin{proof}
Consider the game defined by the following payoff table:
{{PayoffTable|table = $\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} }
& B_1 & B_2 & B_3 \\
\hline A_1 & 1 & 2 & 4 \\
\hline A_2 & 4 & 2 & 1 \\
\hline \end{array}$}}
This has two solutions:
:$(1): \quad A: \left({1/3, 2/3}\right), B: \left({0, 1, 0}\right)$
:$(2): \quad A: \left({2/3, 1/3}\right), B: \left({0, 1, 0}\right)$
{{explain|The reason why the above are solutions is not indicated.}}
It follows that the strategies:
:$\left({\dfrac t 3, \dfrac {2 \left({1 - t}\right)} 3}\right), B: \left({0, 1, 0}\right)$
for all $0 \le t \le 1$ are also solutions.
{{explain|Again, the explanation of this needs to wait till more learning has been done.}}
\end{proof}
|
22875
|
\section{Two-Step Subgroup Test using Subset Product}
Tags: Subset Products, Subgroups
\begin{theorem}
Let $G$ be a group.
Let $\O\subset H \subseteq G$ be a non-empty subset of $G$.
Then $H$ is a subgroup of $G$ {{iff}}:
:$H H \subseteq H$
:$H^{-1} \subseteq H$
where:
:$H H$ is the product of $H$ with itself
:$H^{-1}$ is the inverse of $H$.
\end{theorem}
\begin{proof}
This is a reformulation of the Two-Step Subgroup Test in terms of subset product.
\end{proof}
|
22876
|
\section{Two-Step Vector Subspace Test}
Tags: Vector Spaces, Linear Algebra, Vector Subspaces
\begin{theorem}
Let $V$ be a vector space over a division ring $K$.
Let $U \subseteq V$ be a non-empty subset of $V$ such that:
:$(1): \qquad \forall u \in U, \lambda \in K: \lambda u \in U$
:$(2): \qquad \forall u, v \in U: u + v \in U$
Then $U$ is a subspace of $V$.
\end{theorem}
\begin{proof}
Suppose that $(1)$ and $(2)$ hold.
From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$.
An application of $(2)$ yields the condition of the One-Step Vector Subspace Test.
Hence $U$ is a vector subspace of $V$.
{{qed}}
\end{proof}
|
22877
|
\section{Two-Valued Functions form Boolean Algebra}
Tags: Boolean Algebras
\begin{theorem}
Let $\mathbf 2$ be the Boolean algebra two, and let $X$ be a set.
Let $\mathbf 2^X$ be the set of all mappings $p: X \to \mathbf 2$.
Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in pointwise fashion thus:
:$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) \vee q (x)$
:$\wedge: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \wedge q}\right) (x) := p (x) \wedge q (x)$
:$\neg: \mathbf 2^X \to \mathbf 2^X, \left({\neg p}\right) (x) := \neg p (x)$
Furthermore, write $\bot$ and $\top$ for the constant mappings with these values, viz:
:$\bot: X \to \mathbf 2, \bot (x) := \bot$
:$\top: X \to \mathbf 2, \top (x) := \top$
Then $\left({\mathbf 2^X, \vee, \wedge, \neg}\right)$ is a Boolean algebra, with $\bot$ and $\top$ as identities for $\vee$ and $\wedge$, respectively.
\end{theorem}
\begin{proof}
Let us verify the axioms for a Boolean algebra in turn.
\end{proof}
|
22878
|
\section{Two-Valued Functions form Boolean Ring}
Tags: Boolean Rings
\begin{theorem}
Let $S$ be a set, and let $2$ be the two ring.
Let $2^S$ be the set of all $2$-valued functions on $S$.
Denote with $+$ and $\cdot$ the pointwise operations induced on $2^S$ by $+_2$ and $\times_2$, respectively.
Then $\struct {2^S, +, \cdot}$ is a Boolean ring.
\end{theorem}
\begin{proof}
By Structure Induced by Ring Operations is Ring, $\struct {2^S, +, \cdot}$ is a ring.
By Unity of Induced Structure, $\struct {2^S, +, \cdot}$ also has a unity.
By Induced Structure is Idempotent, $\cdot$ is an idempotent operation.
Hence $\struct {2^S, +, \cdot}$ is a Boolean ring.
{{qed}}
\end{proof}
|
22879
|
\section{Two Coprime Integers have no Third Integer Proportional}
Tags: Coprime Integers, Ratios, Coprime
\begin{theorem}
Let $a, b \in \Z_{>0}$ be integers such that $a$ and $b$ are coprime.
Then there is no integer $c \in \Z$ such that:
:$\dfrac a b = \dfrac b c$
{{:Euclid:Proposition/IX/16}}
\end{theorem}
\begin{proof}
Suppose such a $c$ exists.
From Coprime Numbers form Fraction in Lowest Terms, $\dfrac a b$ is in canonical form.
From Ratios of Fractions in Lowest Terms:
:$a \divides b$
where $\divides$ denotes divisibility.
This contradicts the fact that $a$ and $b$ are coprime.
Hence such a $c$ cannot exist.
{{qed}}
{{Euclid Note|16|IX}}
\end{proof}
|
22880
|
\section{Two Fifths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
There are $3$ ways $\dfrac 2 5$ can be expressed as a pandigital fraction:
:$\dfrac 2 5 = \dfrac {6894} {17235}$
:$\dfrac 2 5 = \dfrac {8694} {21735}$
:$\dfrac 2 5 = \dfrac {9486} {23715}$
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22881
|
\section{Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution}
Tags: Linear Second Order ODEs, Homogeneous LSOODEs
\begin{theorem}
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Let $y_1$ and $y_2$ be linearly independent.
Then the general solution to $(1)$ is:
:$y = C_1 \map {y_1} x + C_2 \map {y_2} x$
where $C_1 \in \R$ and $C_2 \in \R$ are arbitrary constants.
\end{theorem}
\begin{proof}
Let $\map y x$ be any particular solution to $(1)$ on $\closedint a b$.
It is to be shown that constants $C_1$ and $C_2$ can be found such that:
:$\map y x = C_1 \map {y_1} x + C_2 \map {y_2} x$
for all $x \in \closedint a b$.
By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions:
:a particular solution to $(1)$ over $\closedint a b$ is completely determined by:
::its value
:and:
::the value of its derivative
at a single point.
From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:
:$C_1 \map {y_1} x + C_2 \map {y_2} x$
is a particular solution to $(1)$ over $\closedint a b$
We also have:
:$\map y x$
is a particular solution to $(1)$ over $\closedint a b$
Thus it is sufficient to prove that:
:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
:and:
::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
For this system to be solvable for $C_1$ and $C_2$ it is necessary that:
:$\begin{vmatrix}
\map {y_1} x & \map {y_2} x \\
\map { {y_1}'} x & \map { {y_2}'} x \\
\end{vmatrix} = \map {y_1} x \map { {y_2}'} x - \map {y_2} x \map { {y_1}'} x \ne 0$
That is, that the Wronskian $\map W {y_1, y_2} \ne 0$ at $x_0$.
From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE:
:if $\map W {y_1, y_2} \ne 0$ at $x_0$, then $\map W {y_1, y_2} \ne 0$ for all $x \in \closedint a b$.
Hence it does not matter what point is taken for $x_0$; if the Wronskian is non-zero at one such point, it will be non-zero for all such points.
From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$W \left({y_1, y_2}\right) = 0$ for all $x \in \closedint a b$ {{iff}} $y_1$ and $y_2$ are linearly dependent.
But we have that $y_1$ and $y_2$ are linearly independent.
Hence:
:$\forall x \in \closedint a b: \map W {y_1, y_2} \ne 0$
and so:
:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
:and:
::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
The result follows.
{{qed}}
\end{proof}
|
22882
|
\section{Two Lines Meet at Unique Point}
Tags: Lines, Euclid Book I
\begin{theorem}
Let two straight line segments be constructed on a straight line segment from its endpoints so that they meet at a point.
Then there cannot be two other straight line segments equal to the former two respectively, constructed on the same straight line segment and on the same side of it, meeting at a different point.
{{:Euclid:Proposition/I/7}}
\end{theorem}
\begin{proof}
:400px
Let $AC$ and $CB$ be constructed on $AB$ meeting at $C$.
Let two other straight line segments $AD$ and $DB$ be constructed on $AB$, on the same side of it, meeting at $D$, such that $AC = AD$ and $CB = DB$.
Suppose, with a view to obtaining a contradiction, $C$ and $D$ are different points.
Let $CD$ be joined.
Since $AC = AD$ it follows that $\angle ACD = \angle ADC$.
Therefore $\angle ACD$ is greater than $\angle DCB$ because the whole is greater than the part.
Therefore $\angle CDB$ is much greater than $\angle DCB$.
Now since $CB = DB$, it follows that $\angle CDB = \angle DCB$.
But it was proved much greater than it.
From this contradiction it follows that $C$ and $D$ can not be different points.
Hence the result.
{{qed}}
{{Euclid Note|7|I}}
\end{proof}
|
22883
|
\section{Two Ninths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
There are $2$ ways $\dfrac 2 9$ can be expressed as a pandigital fraction:
:$\dfrac 2 9 = \dfrac {3924} {17658}$
:$\dfrac 2 9 = \dfrac {7596} {34182}$
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22884
|
\section{Two Planes have Line in Common}
Tags: Projective Geometry
\begin{theorem}
Two distinct planes have exactly one (straight) line in common.
\end{theorem}
\begin{proof}
Take two distinct lines in plane $1$.
From Propositions of Incidence: Plane and Line, they each meet plane $2$ in one point each, say at $A$ and $B$.
Thus $A$ and $B$ both lie in both planes.
Thus the line defined by $A$ and $B$ lies in both planes.
{{qed}}
{{Handwaving}}
\end{proof}
|
22885
|
\section{Two Ring is Boolean Ring}
Tags: Boolean Rings
\begin{theorem}
Let $2$ be the two ring.
Then $2$ is a Boolean ring.
\end{theorem}
\begin{proof}
From Ring of Integers Modulo m is Ring, $2$ is a ring with unity.
Furthermore, the identities:
:$0 \cdot 0 = 0$
:$1 \cdot 1 = 1$
show that $2$ is also an idempotent ring.
Hence the result, by definition of Boolean ring.
{{qed}}
\end{proof}
|
22886
|
\section{Two Sevenths as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
There are $6$ ways $\dfrac 2 7$ can be expressed as a pandigital fraction:
:$\dfrac 2 7 = \dfrac {3654} {12789}$
:$\dfrac 2 7 = \dfrac {3674} {12859}$
:$\dfrac 2 7 = \dfrac {5342} {18697}$
:$\dfrac 2 7 = \dfrac {7418} {25963}$
:$\dfrac 2 7 = \dfrac {9786} {34251}$
:$\dfrac 2 7 = \dfrac {9862} {34517}$
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22887
|
\section{Two Thirds as Pandigital Fraction}
Tags: Pandigital Fractions
\begin{theorem}
$\dfrac 2 3$ cannot be expressed as a pandigital fraction.
\end{theorem}
\begin{proof}
Can be verified by brute force.
Category:Pandigital Fractions
\end{proof}
|
22888
|
\section{Two divides Power Plus One iff Odd}
Tags: Number Theory
\begin{theorem}
Let $q, n \in \Z_{>0}$.
Then:
:$2 \divides \paren {q^n + 1}$
{{iff}} $q$ is odd.
In the above, $\divides$ denotes divisibility.
\end{theorem}
\begin{proof}
By Parity of Integer equals Parity of Positive Power, $q^n$ is even {{iff}} $q$ is even.
Thus it follows that $q^n + 1$ is even {{iff}} $q$ is odd.
The result follows by definition of even integer.
{{qed}}
\end{proof}
|
22889
|
\section{Two is Boolean Algebra}
Tags: Boolean Algebras
\begin{theorem}
Let $\mathbf 2$ denote two.
Then $\mathbf 2$ is a Boolean algebra.
\end{theorem}
\begin{proof}
It is useful to first state the Cayley tables for the three logical operations $\lor$, $\land$ and $\neg$:
:$\begin{array}{c|cc}
\lor & \bot & \top \\ \hline
\bot & \bot & \top \\
\top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
\land & \bot & \top \\ \hline
\bot & \bot & \bot \\
\top & \bot & \top
\end{array} \qquad \begin{array}{c|cc}
& \bot & \top \\ \hline
\neg & \top & \bot
\end{array}$
Let us now verify the axioms for a Boolean algebra in turn.
\end{proof}
|
22890
|
\section{Tychonoff's Theorem/General Case}
Tags: Topology
\begin{theorem}
Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact.
\end{theorem}
\begin{proof}
First assume that $X$ is compact.
From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.
It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges.
Thus let $\FF$ be an ultrafilter on $X$.
From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.
Each $X_i$ is compact by assumption.
So by Equivalent Definitions of Compactness, each $\map {\pr_i} \FF$ converges.
By Filter on Product Space Converges iff Projections Converge, $\FF$ converges.
{{explain|Find the appropriate link for "converges" and explain why this proves the result.}}
{{qed}}
{{AoC|Filter on Product Space Converges iff Projections Converge}}
\end{proof}
|
22891
|
\section{Tychonoff's Theorem for Hausdorff Spaces}
Tags: Hausdorff Spaces, Topology
\begin{theorem}
Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty Hausdorff spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact.
== Proof ==
First assume that $X$ is compact.
From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.
From Continuous Image of Compact Space is Compact, it follows that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges.
Thus let $\FF$ be an ultrafilter on $X$.
From Image of Ultrafilter is Ultrafilter, for each $i \in I$, the image filter $\map {\pr_i} \FF$ is an ultrafilter on $X_i$.
Each $X_i$ is compact by assumption.
So by Equivalent Definitions of Compactness, each $\map {\pr_i} \FF$ converges.
From Filter on Product of Hausdorff Spaces Converges iff Projections Converge, $\FF$ converges.
{{explain|Find the appropriate link for "converges" and explain why this proves the result.}}
{{qed}}
{{BPI|Equivalent Definitions of Compactness}}
\end{theorem}
\begin{proof}
First assume that $X$ is compact.
Since the projections $\operatorname{pr}_i : X \to X_i$ are continuous, it follows that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
By Equivalent Definitions of Compactness it is enough to show that every ultrafilter on $X$ converges.
Thus let $\mathcal F$ be an ultrafilter on $X$.
For each $i \in I$, the image filter $\operatorname{pr}_i \left({\mathcal F}\right)$ then is an ultrafilter on $X_i$.
Since each $X_i$ is compact by assumption, by Equivalent Definitions of Compactness, each $\operatorname{pr}_i \left({\mathcal F}\right)$ converges.
By Filter on Product of Hausdorff Spaces Converges iff Projections Converge, $\mathcal F$ converges.
{{qed}}
{{BPI|Filter on Product of Hausdorff Spaces Converges iff Projections Converge}}
Category:Topology
123104
123102
2013-01-02T18:05:32Z
Dfeuer
1672
123104
wikitext
text/x-wiki
{{refactor|shares too much with the general form of Tychonoff's Theorem}}
{{tidy}}
\end{proof}
|
22892
|
\section{Tychonoff Space is Preserved under Homeomorphism}
Tags: Tychonoff Spaces, Separation Axioms, Homeomorphisms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a Tychonoff (completely regular) space, then so is $T_B$.
\end{theorem}
\begin{proof}
We have that $\struct {S, \tau}$ is a Tychonoff space {{iff}}:
:$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
Hence the result.
{{qed}}
\end{proof}
|
22893
|
\section{Tychonoff Space is Regular, T2 and T1}
Tags: Hausdorff Spaces, Tychonoff Spaces, T1 Spaces, Regular Spaces, Separation Axioms
\begin{theorem}
Let $\struct {S, \tau}$ be a Tychonoff space.
Then $\struct {S, \tau}$ is also:
:a regular space
:a $T_2$ (Hausdorff) space
:a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a Tychonoff space.
From the definition of Tychonoff space:
:$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
We have that a $T_{3 \frac 1 2}$ space is a $T_3$ space.
From the definition, a regular space is:
:a $T_3$ space
:a $T_0$ (Kolmogorov) space.
So a Tychonoff space is a regular space.
Then we have a regular space is a $T_2$ (Hausdorff) space.
Then we have a $T_2$ (Hausdorff) space is a $T_1$ (Fréchet) space.
{{qed}}
\end{proof}
|
22894
|
\section{Tychonoff Space is Urysohn Space}
Tags: Tychonoff Spaces, Urysohn Spaces, Separation Axioms
\begin{theorem}
Let $\struct {S, \tau}$ be a Tychonoff space.
Then $\struct {S, \tau}$ is also an Urysohn space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a Tychonoff space.
Let $x, y \in S: x \ne y$.
From the definition of Tychonoff space:
:$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
From Tychonoff Space is Regular, $T_2$ and $T_1$ we use the fact that $T$ is a $T_1$ (Fréchet) space.
Clearly $\set x$ and $\set y$ are disjoint.
From the definition of $T_1$ (Fréchet) space, we have that $\set x$ and $\set y$ are closed.
So by definition of a $T_{3 \frac 1 2}$ space, there exists an Urysohn function for $\set x$ and $\set y$.
This is exactly the definition of an an Urysohn space.
{{qed}}
\end{proof}
|
22895
|
\section{Type Space is Compact}
Tags: Model Theory
\begin{theorem}
Let $\MM$ be an $\LL$-structure.
Let $A$ be a subset of the universe of $\MM$.
The type space $\map {S_n^\MM} A$ of $n$-types over $A$ is compact.
\end{theorem}
\begin{proof}
It will suffice to show that every open cover of $\map {S_n^\MM} A$ by the basic open sets $[\phi]$ of the topology has a finite subcover.
Let $\UU = \set { [\phi_i] : i \in I}$ be a cover of $\map {S_n^\MM} A$ by basic open sets.
This means that every complete $n$-type over $A$ contains some $\phi_i$.
We will find a finite subcover of $\UU$.
Let $\Gamma = \set {\neg \phi_i: i \in I}$.
Then $\map {\operatorname{Th}_\AA} M \cup \Gamma$ cannot be satisfied, since if $\NN \models \map {\operatorname{Th}_\AA} M \cup \map \Gamma {\bar b}$, then the type $\map {\operatorname{tp}_\NN} {\bar b / A}$ is a complete $n$-type in $\map {S_n^\MM} A$ which does not contain any $\phi_i$.
By the Compactness Theorem, $\map {\operatorname{Th}_\AA} M \cup \Gamma$ must have a finite subset $\Delta$ which is not satisfiable.
Since $\Delta$ is not satisfiable but $\map {\operatorname{Th}_\AA} M$ is, $\Delta$ must contain some of the $\neg \phi_i$ from $\Gamma$.
Furthermore, we must have that every model of $\map {\operatorname{Th}_\AA} M$ fails to satisfy at least one of these finitely many $\neg \phi_i$ in $\Delta$.
We claim that the finite set $\FF = \set {[\phi_i] : \neg \phi_i \in \Delta}$ is a finite subcover of $\UU$.
Since $\FF$ is clearly a subset of $\UU$, we need only show that every $p \in \map {S_n^\MM} A$ is contained in some $[\phi_i] \in \FF$.
That is, we must show that each $p \in \map {S_n^\MM} A$ contains one of these $\phi_i$.
Let $p \in \map {S_n^\MM} A$.
By definition, this means that $p \cup \map {\operatorname{Th}_\AA} M$ is satisfiable by some $\NN$.
But, as mentioned above, since $\NN \models \map {\operatorname{Th}_\AA} M$, we have that $\NN \not\models \neg \phi_i$ for some $\neg \phi_i \in \Delta$.
Since $p$ is complete, it must contain either $\phi_i$ or $\neg \phi_i$.
However, if $p$ contained $\neg \phi_i$, then $\NN \models \neg \phi_i$, which contradicts $\NN \not\models \neg \phi_i$.
Thus, $p$ contains $\phi_i$.
This demonstrates that $\FF$ is a finite subcover for $\UU$.
Thus, $\map {S_n^\MM} A$ is compact.
{{qed}}
Category:Model Theory
\end{proof}
|
22896
|
\section{Type is Realized in some Elementary Extension}
Tags: Model Theory
\begin{theorem}
Let $\MM$ be an $\LL$-structure.
Let $A$ be a subset of the universe of $\MM$.
Let $p$ be an $n$-type over $A$.
There exists an elementary extension of $\MM$ which realizes $p$.
\end{theorem}
\begin{proof}
The idea is to work in a language with constant symbols for all elements of $\MM$ and show that the union of $p$ and the elementary diagram of $\MM$ is satisfiable.
Since $\MM$ naturally embeds into any model of such a theory, this will prove the theorem.
Let $\LL_\MM$ be the language obtained by adding to $\LL$ constant symbols for each element of $\MM$.
Denote by $\map {\operatorname {Diag}_{\mathrm {el} } } \MM$ the elementary diagram of $\MM$.
Let $T$ be $p \cup \map {\operatorname {Diag}_{\mathrm{el} } } \MM$.
We will show that $T$ is finitely satisfiable.
It will follow by the Compactness Theorem that $T$ is satisfiable.
To this end, let $\Delta$ be a finite subset of $T$.
We have that $\Delta$ is finite.
Thus it consists of:
:finitely many $\LL_A$-sentences $\phi_0, \dotsc, \phi_n$ from $p$ (which are $\LL_\MM$ sentences since $A \subseteq \MM$)
along with:
:finitely many $\LL_\MM$-sentences $\psi_0,\dots,\psi_k$ from $\map {\operatorname{Diag}_{\mathrm{el} } } \MM$.
By definition, $p$ is satisfiable by some $\LL_A$-structure $\NN$ such that:
:$\NN \models p \cup \map {\operatorname{Th}_A} \MM$
Thus, since $\phi_0, \dotsc, \phi_n \in p$:
:$\NN$ satisfies $\phi_0, \dotsc, \phi_n$.
We will show that the same $\NN$ also satisfies $\psi_0, \dotsc, \psi_k$.
The obstacle to overcome is that the $\psi_i$ are $\LL_\MM$-formulas, and we only know $\NN$ as an $\LL_A$-structure which satisfies sentences with parameters from $A$.
The $\psi_i$ may have parameters from $\MM$ outside of $A$.
The idea is to quantify away the excess parameters and appropriately select the interpretation of new symbols so that $\NN$ is a good $\LL_\MM$-structure.
Explicitly:
Let $\psi$ be the conjunction $\psi_0 \wedge \cdots \wedge \psi_k$.
Note that since $\psi$ is an $\LL_\MM$-sentence, it can be written as an $\LL_A$-formula $\map \psi {\bar b}$, where $\bar b$ is a tuple of parameters from $\MM$ not in $A$.
By existentially quantifying away the tuple $\bar b$, we obtain an $\LL_A$-sentence $\exists \bar x \map \psi {\bar x}$.
Now, since $\MM \models \map \psi {\bar b}$, we have:
:$\MM \models \exists \bar x: \map \psi {\bar x}$
Hence $\exists \bar x: \map \psi {\bar x}$ is in $\map {\operatorname{Th}_A} \MM$.
By choice of $\NN$, it follows that:
:$\NN \models \exists \bar x: \map \psi {\bar x}$
and thus there must be some tuple $\bar c$ of elements from $\NN$ such that:
:$\NN \models \map \psi {\bar c}$
Now, by interpreting the $\LL_\MM$-symbols $\bar b$ as the elements $\bar c$, we can view $\NN$ as an $\LL_\MM$-structure which satisfies:
:$\phi_0 \wedge \cdots \wedge \phi_n \wedge \psi_0 \wedge \cdots \wedge \psi_k$.
Thus $\NN$ satisfies all of $\Delta$.
This demonstrates that $T$ is finitely satisfiable and hence satisfiable by the Compactness Theorem.
This means that there is an $\LL_\MM$-structure $\MM^*$ which satisfies:
:$p \cup \map {\operatorname{Diag}_{\mathrm {el} } } \MM$
Since $\MM^*$ interprets a symbol for each element of $\MM$, there is an obvious embedding of $\MM$ into $\MM^*$.
This embedding is elementary since $\MM^*$ satisfies the elementary diagram of $\MM$.
Thus $\MM^*$ is an elementary extension of $\MM$.
Finally, since $\MM^*$ satisfies $p$, there must be a tuple of elements $\bar d$ such that $\MM^* \models \map \phi d$ for each $\map \phi {\bar v} \in p$.
Thus $\MM^*$ realizes $p$.
{{qed}}
Category:Model Theory
\end{proof}
|
22897
|
\section{UFD is GCD Domain}
Tags: Ring Theory, Factorization, Unique Factorization Domains, GCD Domains
\begin{theorem}
Let $A$ be a unique factorisation domain.
Then $A$ is a GCD domain.
\end{theorem}
\begin{proof}
Let $x \divides y$ denote $x$ divides $y$.
Let $x, y \in A$, with complete factorizations:
:$x = u x_1 \cdots x_r$
:$y = v y_1 \cdots y_s$
where:
:$u, v$ are units
:the $x_i$, $y_i$ irreducible.
We arrange the complete factorizations as follows:
:$x = u \paren {x_1 \cdots x_t} x_{t + 1} \cdots x_r$
:$y = v \paren {y_1 \cdots y_t} y_{t + 1} \cdots y_s$
where:
:$t \le \min \set {r, s}$
:For $i = 1, \ldots, t$, $x_i$ and $y_i$ are associates
:For any $i \in \set {t + 1, \ldots, r}$, $j \in \set {t + 1, \ldots, s}$, $x_i$ and $y_j$ are not associates.
Let $d = x_1 \cdots x_t$ (recall that the empty product is $1$, i.e. $d = 1$ when $t = 0$).
We claim that $d$ is a greatest common divisor for $x$ and $y$.
Certainly $d \divides x$ and $d \divides y$.
So, let $f$ be another common divisor of $x$ and $y$.
We can find $w, z \in A$ such that $x = f w$, and $y = f z$.
If $f$ is a unit, then $f \divides d$ by definition.
{{AimForCont}} $f \nmid d$.
Then the complete factorization of $f$ must contain an irreducible element that does not divide $d$.
Call this irreducible element $g$.
We have that:
:$g$ must divide some $x_j$ where $j > t$
and
:$g$ must divide some $y_k$ where $k > t$.
Either:
:$g$ is a unit, contradicting its irreducibility
or:
:$x_j$ and $y_k$ are not irreducible, which is a contradiction also.
Hence by Proof by Contradiction:
:$f \divides d$
and so $x$ and $y$ have a greatest common divisor.
{{qed}}
Category:Unique Factorization Domains
Category:GCD Domains
Category:Factorization
\end{proof}
|
22898
|
\section{URM Computable Function is Recursive}
Tags: Recursive Functions, URM Programs
\begin{theorem}
Every URM computable function is recursive.
\end{theorem}
\begin{proof}
Let $f: \N^k \to \N$ be a URM computable function.
Then by hypothesis there is a URM program that computes $f$.
Let $P$ be the URM program with the smallest code number that computes $f$.
Let $e = \gamma \left({P}\right)$ be the code number of $P$.
Consider the function $g: \N^k \to \N$ given by:
:$g \left({n_1, n_2, \ldots, n_k}\right) \approx \mu t \left({\left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1 > \operatorname{len} \left({e}\right)}\right)$
where:
* $\operatorname{len} \left({e}\right)$ is the length of $e$;
* $\mu t$ is the minimization operation on $S_k$;
* $\approx$ denotes partial function equality;
* $S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)$ is the state code at stage $t$ of the computation of $P$ with input $\left({n_1, n_2, \ldots, n_k}\right)$;
* the number $\left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1$ is the code number of the instruction about to be carried out at stage $t$.
So the inequality:
:$\left({S_k \left({e, n_1, n_2, \ldots, n_k, t}\right)}\right)_1 > \operatorname{len} \left({e}\right)$
expresses the fact that at stage $t$ the computation has halted.
So the value of $g \left({n_1, n_2, \ldots, n_k}\right)$ is the state code of the first stage at which computation has halted, if there is one, and undefined otherwise.
Since the functions in this inequality, and the sign $>$ itself, are all primitive recursive, it follows that the inequality expresses a primitive recursive relation on $e, n_1, n_2, \ldots, n_k, t$.
Thus $g$ is a recursive function by definition, as it can be obtained by minimization on a recursive relation.
Now consider the function $h: \N^k \to \N$ given by:
:$h \left({n_1, n_2, \ldots, n_k}\right) \approx S_k \left({e, n_1, n_2, \ldots, n_k, g \left({n_1, n_2, \ldots, n_k}\right)}\right)$.
This is recursive because it was obtained by substitution from known recursive functions.
Now $h \left({n_1, n_2, \ldots, n_k}\right)$ is defined iff the computation halts, and it gives the value of the state code when it has halted.
The output of this computation, which gives the value of $f$, is the number in register $R_1$.
But the number in $R_1$ is the exponent of $p_2 = 3$ in the expression of the state code $h \left({n_1, n_2, \ldots, n_k}\right)$ in the form $p_1^a p_2^{r_1} p_3^{r_2} \cdots p_{k+1}^{r_k}$.
Thus the function $f$ is given by:
:$f \left({n_1, n_2, \ldots, n_k}\right) \approx \left({h \left({n_1, n_2, \ldots, n_k}\right)}\right)_2$.
It follows that $f$ is a recursive function, since it is obtained by substitution from known recursive functions.
{{qed}}
\end{proof}
|
22899
|
\section{URM Computable Functions of One Variable is Countably Infinite}
Tags: Countable Sets, Infinite Sets, URM Programs
\begin{theorem}
The set $\mathbf U$ of all URM computable functions of $1$ variable is countably infinite.
\end{theorem}
\begin{proof}
Let $\mathbf U$ be the set of all URM computable functions.
For each $f \in \mathbf U$, let $P_f$ be a URM program which computes $f$.
Such a program is very probably not unique, so in order to be definite about it, we can pick $P_f$ to be the URM program with the smallest code $\gamma \left({P_f}\right)$.
This is possible from the Well-Ordering Principle.
Let us define the function $h: \mathbf U \to \N$ as:
:$h \left({f}\right) = \gamma \left({P_f}\right)$
Since the same URM program can not compute two different functions of $1$ variable, it can be seen that $h$ is injective.
The result follows from Domain of Injection to Countable Set is Countable.
{{qed}}
Category:URM Programs
Category:Countable Sets
\end{proof}
|
22900
|
\section{URM Instructions are Countably Infinite}
Tags: Countable Sets, Infinite Sets, URM Programs
\begin{theorem}
The set $\Bbb I$ of all basic URM instructions is countably infinite.
\end{theorem}
\begin{proof}
We can immediately see that $\Bbb I$ is infinite as, for example, $\phi: \N \to \Bbb I$ defined as:
:$\phi \left({n}\right) = Z \left({n}\right)$
is definitely injective.
From Unique Code for URM Instruction, we see that $\beta: \Bbb I \to \N$ is also an injection.
The result follows from Domain of Injection to Countable Set is Countable.
{{qed}}
Category:URM Programs
Category:Countable Sets
\end{proof}
|
22901
|
\section{URM Programs are Countably Infinite}
Tags: Countable Sets, Infinite Sets, URM Programs
\begin{theorem}
The set $\mathbf P$ of all URM programs is countably infinite.
\end{theorem}
\begin{proof}
We can immediately see that $\mathbf P$ is infinite as the number of URM instructions is infinite.
From Unique Code for URM Program, we see that $\gamma: \mathbf P \to \N$ is also an injection.
The result follows from Domain of Injection to Countable Set is Countable.
{{qed}}
Category:URM Programs
Category:Countable Sets
\end{proof}
|
22902
|
\section{Ultraconnected Space is Connected}
Tags: Connected Spaces, Ultraconnected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is connected.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
From Ultraconnected Space is Path-Connected, $T$ is path-connected.
The result follows from Path-Connected Space is Connected.
{{qed}}
\end{proof}
|
22903
|
\section{Ultraconnected Space is Path-Connected}
Tags: Ultraconnected Space, Path-Connected Space, Path-Connected Spaces, Ultraconnectedness, Ultraconnected Spaces, Path-Connectedness, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is path-connected.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be ultraconnected.
Let $a, b \in S$.
Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the closure of $\set a$.
Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\set a^- \cap \set b^- \ne \O$.
Consider the mapping $f: \closedint 0 1 \to X$ such that:
:$\map f x = \begin{cases}
a & : x \in \hointr 0 {\dfrac 1 2} \\
p & : x = \dfrac 1 2 \\
b & : x \in \hointl {\dfrac 1 2} 1 \\
\end{cases}$
Then $f$ is continuous.
{{explain|The above is a nontrivial claim}}
The result follows from the definition of path-connectedness.
{{qed}}
\end{proof}
|
22904
|
\section{Ultraconnected Space is T4}
Tags: Ultraconnectedness, Separation Axioms, Ultraconnected Spaces, Connectedness, T4 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is a $T_4$ space.
\end{theorem}
\begin{proof}
Recall the definition of a $T_4$ space:
$T = \struct {S, \tau}$ is a $T_4$ space {{iff}}:
:for any two disjoint closed sets $A, B \subseteq S$, there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
As no two closed sets of an ultraconnected space are actually disjoint, it follows that $T_4$-ness follows vacuously.
{{qed}}
\end{proof}
|
22905
|
\section{Ultraconnected Space is not necessarily Arc-Connected}
Tags: Arc-Connected Spaces, Ultraconnected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is not necessarily arc-connected.
\end{theorem}
\begin{proof}
Let $T$ be an excluded point space.
From Excluded Point Space is Ultraconnected, $T$ is an ultraconnected space.
From Excluded Point Space is not Arc-Connected, $T$ is not arc-connected.
Hence the result.
{{qed}}
\end{proof}
|
22906
|
\section{Ultrafilter Lemma}
Tags: Set Theory, Filter Theory, Named Theorems, Topology, Axiom of Choice
\begin{theorem}
Let $S$ be a set.
Every filter on $S$ is contained in an ultrafilter on $S$.
\end{theorem}
\begin{proof}
Let $\Omega$ be the set of filters on $S$.
From Subset Relation is Ordering, the subset relation "$\subseteq$" makes $\Omega$ a partially ordered set.
If $C \subseteq \Omega$ is a non-empty chain, then $\bigcup C$ is again a filter on $S$ and thus an upper bound of $C$.
{{explain|That the union is again a filter is not trivial}}
For any $\mathcal F \in \Omega$ there is therefore by Zorn's Lemma a maximal element $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$.
The maximality of $\mathcal F'$ is in this context equivalent to $\mathcal F'$ being an ultrafilter.
{{qed}}
{{AoC|Zorn's Lemma|4}}
{{BPIT}}
Category:Set Theory
Category:Filter Theory
Category:Named Theorems
119898
119463
2012-12-16T05:43:39Z
Dfeuer
1672
119898
wikitext
text/x-wiki
\end{proof}
|
22907
|
\section{Ultrafilter Lemma/Corollary}
Tags: Set Theory, Filter Theory
\begin{theorem}
Let $S$ be a non-empty set.
Let $\AA$ be a set of subsets of $S$.
Suppose that $\AA$ has the finite intersection property.
Then there is an ultrafilter $\UU$ on $S$ such that $\AA \subseteq \UU$.
\end{theorem}
\begin{proof}
Let $\II$ be the set of intersections of non-empty finite subsets of $\AA$.
Let $\FF = \set {T \in \powerset S: \exists B \in \II: B \subseteq T}$.
Note that $\AA \subseteq \II \subseteq \FF$.
$\FF$ is a filter on $S$:
Because $\AA$ has the finite intersection property:
: $\O \notin \II$
Because each element of $\FF$ is a superset of some element of $\II$:
: $\O \notin \FF$
Let $P \in \FF$ and $P \subseteq Q \subseteq S$.
Then:
:$\exists B \in \II: B \subseteq P$
and so:
: $B \subseteq Q \subseteq S$
Thus:
:$Q \in \FF$
Let $P \in \FF$ and $Q \in \FF$.
Then:
:$\exists B, C \in \II: B \subseteq P$ and $C \subseteq Q$
Then:
:$B \cap C \in \II$
and:
: $B \cap C \subseteq P \cap Q \subseteq S$
Thus:
: $P \cap Q \in \FF$
Thus, by definition, $\FF$ is a filter on $S$.
By the Ultrafilter Lemma, there exists an ultrafilter $\UU$ on $S$ such that:
: $\FF \subseteq \UU$
Because $\AA \subseteq \FF$:
:$\AA \subseteq \UU$
{{qed}}
Category:Set Theory
Category:Filter Theory
\end{proof}
|
22908
|
\section{Ultraproduct is Well-Defined}
Tags: Definitions: Model Theory, Model Theory
\begin{theorem}
Ultraproduct is well-defined.
\end{theorem}
\begin{proof}
Let:
{{begin-eqn}}
{{eqn | l = S
| o = :=
| r = \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } \in R^{\MM_i} }
}}
{{eqn | l = S'
| o = :=
| r = \set {i \in I: \tuple {m'_{1, i}, \dotsc, m'_{n, i} } \in R^{\MM_i} }
}}
{{eqn | l = I^*
| o = :=
| r = \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } }
}}
{{eqn | l = T
| o = :=
| r = I^* \cap S
}}
{{eqn | l = T'
| o = :=
| r = I^* \cap S'
}}
{{end-eqn}}
From the lemma:
:$I^* \in \UU$
therefore:
:$S \in \UU \implies T \in \UU$
Note that:
:$\tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} }$ for $i \in I^*$
we have:
:$T = T'$
Hence:
:$T' \in \UU$
and:
:$S' \in \UU$ since $S' \supseteq T'$
So far we have proved:
:$S \in \UU \implies S' \in \UU$
By symmetry:
:$S' \in \UU \implies S \in \UU$
{{qed}}
Category:Model Theory
\end{proof}
|
22909
|
\section{Unary Product for Object is Itself}
Tags: Category Theory
\begin{theorem}
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Then $\ds \prod \set C = C$, where $\ds \prod$ denotes product.
\end{theorem}
\begin{proof}
Follows directly from Limit of Singleton.
{{qed}}
Category:Category Theory
\end{proof}
|
22910
|
\section{Unary Truth Functions}
Tags: Truth Functions
\begin{theorem}
There are $4$ distinct unary truth functions:
:$(1): \quad$ The constant function $\map f p = \F$
:$(2): \quad$ The constant function $\map f p = \T$
:$(3): \quad$ The identity function $\map f p = p$
:$(4): \quad$ The logical not function $\map f p = \neg p$
\end{theorem}
\begin{proof}
From Count of Truth Functions there are $2^{\paren {2^1} } = 4$ distinct truth functions on $1$ variable.
These can be depicted in a truth table as follows:
:$\begin{array}{|c|cccc|} \hline
p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
\hline
\T & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\hline
\end{array}$
$\circ_1$: Whether $p = \T$ or $p = \F$, $\map {\circ_1} p = \T$.
Thus $\circ_1$ is the constant function $\map {\circ_1} p = \T$.
$\circ_2$: We have:
:$(1): \quad p = \T \implies \map {\circ_2} p = \T$
:$(2): \quad p = \F \implies \map {\circ_2} p = \F$
Thus $\circ_2$ is the identity function $\map {\circ_2} p = p$.
$\circ_3$: We have:
:$(1): \quad p = \T \implies \map {\circ_3} p = \F$
:$(2): \quad p = \F \implies \map {\circ_3} p = \T$
Thus $\circ_3$ is the logical not function $\map {\circ_3} p = \neg p$.
$\circ_4$: Whether $p = \T$ or $p = \F$, $\map {\circ_4} p = \F$.
Thus $\circ_4$ is the constant function $\map {\circ_4} p = \F$.
All four have been examined, and there are no other unary truth functions.
{{qed}}
\end{proof}
|
22911
|
\section{Unbounded Monotone Sequence Diverges to Infinity}
Tags: Unbounded Monotone Sequence Diverges to Infinity, Limits of Sequences, Analysis
\begin{theorem}
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be monotone, that is either increasing or decreasing.
\end{theorem}
\begin{proof}
Let $\sequence {x_n}$ be increasing and unbounded above.
Let $H > 0$.
As $\sequence {x_n}$ is unbounded above, $\exists N: x_N > H$.
As $\sequence {x_n}$ is increasing, then $\forall n \ge N: x_n \ge x_N > H$.
It follows from the definition of divergence to infinity that $x_n \to \infty$ as $n \to \infty$.
{{qed|lemma}}
The same argument can be used for the case where $\sequence {x_n}$ is decreasing and unbounded below.
{{Qed}}
\end{proof}
|
22912
|
\section{Unbounded Monotone Sequence Diverges to Infinity/Increasing}
Tags: Limits of Sequences, Unbounded Monotone Sequence Diverges to Infinity
\begin{theorem}
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be increasing and unbounded above.
Then $x_n \to +\infty$ as $n \to \infty$.
\end{theorem}
\begin{proof}
Let $H > 0$.
As $\sequence {x_n}$ is unbounded above:
:$\exists N: x_N > H$
As $\sequence {x_n}$ is increasing:
:$\forall n \ge N: x_n \ge x_N > H$
It follows from the definition of divergence to $+\infty$ that $x_n \to +\infty$ as $n \to \infty$.
{{qed}}
\end{proof}
|
22913
|
\section{Uncountable Closed Ordinal Space is Countably Compact}
Tags: Countably Compact Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a countably compact space.
\end{theorem}
\begin{proof}
We have:
:Closed Ordinal Space is Compact
:Compact Space is Countably Compact
{{qed}}
\end{proof}
|
22914
|
\section{Uncountable Closed Ordinal Space is Lindelöf}
Tags: Ordinal Spaces, Lindelöf Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a Lindelöf space.
\end{theorem}
\begin{proof}
We have:
:Closed Ordinal Space is Compact
:Compact Space is Lindelöf
{{qed}}
\end{proof}
|
22915
|
\section{Uncountable Closed Ordinal Space is Sigma-Compact}
Tags: Ordinal Spaces, Sigma-Compact Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a $\sigma$-compact space.
\end{theorem}
\begin{proof}
We have:
:Closed Ordinal Space is Compact
:Compact Space is $\sigma$-Compact
{{qed}}
\end{proof}
|
22916
|
\section{Uncountable Closed Ordinal Space is not First-Countable}
Tags: Ordinal Spaces, First-Countable Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a first-countable space.
\end{theorem}
\begin{proof}
From Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.
Thus, by definition, $\Omega$ does not have a countable local basis.
Hence the result by definition of first-countable space.
{{qed}}
\end{proof}
|
22917
|
\section{Uncountable Closed Ordinal Space is not Perfectly Normal}
Tags: Perfectly Normal Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a perfectly normal space.
\end{theorem}
\begin{proof}
From Omega is Closed in Uncountable Closed Ordinal Space but not $G_\delta$ Set, $\set \Omega$ is not a $G_\delta$ set.
From Ordinal Space is Completely Normal, $\closedint 0 \Omega$ is a $T_1$ (Fréchet) space.
Thus by definition $\set \Omega$ is closed in $\closedint 0 \Omega$.
Thus we have that $\set \Omega$ is a closed set of $\closedint 0 \Omega$ which is not a $G_\delta$ set.
The result follows by definition of perfectly normal space.
{{qed}}
\end{proof}
|
22918
|
\section{Uncountable Closed Ordinal Space is not Separable}
Tags: Separable Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a separable space.
\end{theorem}
\begin{proof}
Let $H \subseteq \closedint 0 \Omega$ be a countable subset of $\closedint 0 \Omega$.
From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$.
Thus the closure of $H$ in $\closedint 0 \Omega$ does not equal $\closedint 0 \Omega$.
Thus $H$ is not everywhere dense in $\closedint 0 \Omega$.
Hence, by definition, $\closedint 0 \Omega$ is not separable.
{{qed}}
\end{proof}
|
22919
|
\section{Uncountable Discrete Space is not Lindelöf}
Tags: Uncountable Sets, Lindelöf Spaces, Separable Spaces, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not a Lindelöf space.
\end{theorem}
\begin{proof}
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover of $S$ which is finer than any other open cover of $S$.
That is, $\CC$ is an open cover of $S$ which is uncountable and has no countable subcover.
(Note that a subcover is a refinement of a cover.)
So by definition $T$ can not be a Lindelöf space.
{{qed}}
\end{proof}
|
22920
|
\section{Uncountable Discrete Space is not Second-Countable}
Tags: Uncountable Sets, Second-Countable Spaces, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not second-countable.
\end{theorem}
\begin{proof}
We have that an Uncountable Discrete Space is not Separable.
From Second-Countable Space is Separable, it follows that $T$ can not be second-countable.
{{qed}}
\end{proof}
|
22921
|
\section{Uncountable Discrete Space is not Separable}
Tags: Uncountable Sets, Separable Spaces, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not separable.
\end{theorem}
\begin{proof}
By definition, $T$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$.
Let $H \subseteq S$ be everywhere dense in $T$.
Then by definition of everywhere dense, $H^- = S$ where $H^-$ denotes the closure of $H$.
However, as $T$ is a discrete space, $H^- = H$ from Interior Equals Closure of Subset of Discrete Space.
So $H^- = S \implies H = S$.
But $S$ is uncountable.
So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense.
Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable.
{{qed}}
\end{proof}
|
22922
|
\section{Uncountable Discrete Space is not Sigma-Compact}
Tags: Uncountable Sets, Compact Spaces, Sigma-Compact Spaces, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not $\sigma$-compact.
\end{theorem}
\begin{proof}
We have that an Uncountable Discrete Space is not Lindelöf.
But a $\sigma$-compact space is Lindelöf.
So an uncountable discrete space can not be $\sigma$-compact.
{{qed}}
\end{proof}
|
22923
|
\section{Uncountable Excluded Point Space is not Second-Countable}
Tags: Excluded Point Topology, Uncountable Excluded Point Space is not Second-Countable, Second-Countable Spaces
\begin{theorem}
Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not second-countable.
\end{theorem}
\begin{proof}
Let $H = S \setminus \left\{{p}\right\}$ where $\setminus$ denotes set difference.
By definition, $H$ is an uncountable discrete space.
The result follows from Uncountable Discrete Space is Not Second-Countable.
{{qed}}
\end{proof}
|
22924
|
\section{Uncountable Excluded Point Space is not Separable}
Tags: Separable Spaces, Excluded Point Topology
\begin{theorem}
Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not separable.
\end{theorem}
\begin{proof}
Let $H \subseteq S$ such that $H$ is countable.
Then $H \ne S$ as $S$ is uncountable by hypothesis.
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.
So, by definition, the closure of $H$ is $H \cup \set p$.
From Countable Union of Countable Sets is Countable we have that $H \cup \set p$ is countable.
So $H \cup \set p \ne S$.
So $H$ is not everywhere dense in $T$.
Thus $T$ can have no countable subset of $S$ which is everywhere dense in $T$.
Hence the result by definition of separable space.
{{qed}}
\end{proof}
|
22925
|
\section{Uncountable Finite Complement Space is not First-Countable}
Tags: First-Countable Spaces, Finite Complement Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then $T$ is not first-countable.
\end{theorem}
\begin{proof}
{{AimForCont}} some $x \in S$ has a countable local basis.
That means:
:there exists a countable set $\BB_x \subseteq \tau$
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
Let $y \in S$ with $y \ne x$.
Because its complement relative to $S$ is finite, we have that $S \setminus \set y$ is an open set containing $x$.
Because $\BB_x$ is a neighborhood basis at $x$, there exists some $B \in \BB_x$ such that:
:$B \subseteq S \setminus \set y$
whence $y \notin B$.
So:
:$y \notin \bigcap \BB_x$
So:
{{begin-eqn}}
{{eqn | l = \bigcap \BB_x
| r = \set x
| c =
}}
{{eqn | ll= \leadsto
| l = S \setminus \set x
| r = S \setminus \bigcap \BB_x
| c =
}}
{{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}
| c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
By definition, each of $S \setminus B$ is finite.
From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is countable.
So $S \setminus \set x$ and therefore $S$ is also countable.
From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.
Hence by definition $T$ can not be first-countable.
{{qed}}
\end{proof}
|
22926
|
\section{Uncountable Finite Complement Topology is not Perfectly T4}
Tags: Perfectly T4 Spaces, Finite Complement Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then $T$ is not a perfectly $T_4$ space.
\end{theorem}
\begin{proof}
Recall the definition of a perfectly $T_4$ space
:Every closed set in $T$ can be written as a countable intersection of open sets of $T$.
Let $V$ be a closed set in $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set.
The result follows by definition of perfectly $T_4$ space.
{{qed}}
\end{proof}
|
22927
|
\section{Uncountable Fort Space is not First-Countable}
Tags: Fort Space, Fort Spaces, First-Countable Spaces
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a first-countable space.
\end{theorem}
\begin{proof}
Let $\UU$ be a countable set of open neighborhoods of $p$.
Let $U \in \UU$.
Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be open it must follow that $\relcomp S U$ is finite.
From De Morgan's Laws: Complement of Intersection we have:
:$\ds H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$
From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.
So $\ds \bigcap \UU = \relcomp S H$ must be uncountable.
So (trivially):
:$H \ne \set p$
So:
:$\exists q \ne p: q \in \ds \bigcap \UU$
So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$.
So $\UU$ is not a countable local basis of $p$.
Hence by definition $T$ is not a first-countable space.
{{qed}}
\end{proof}
|
22928
|
\section{Uncountable Fort Space is not Perfectly Normal}
Tags: Fort Space, Perfectly Normal Spaces, Fort Spaces
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a perfectly normal space.
\end{theorem}
\begin{proof}
From Clopen Points in Fort Space, $\set p$ is closed in $T$.
Consider a countable intersection of open sets of $T$ which contain $p$.
By definition, all these are cofinite in $S$ and so uncountable.
So this intersection must itself contain all but a countable number of points of $S$.
So $\set p$ is not a $G_\delta$ set.
Hence $T$ is not a perfectly normal space as not all its closed sets is a $G_\delta$ set.
{{qed}}
\end{proof}
|
22929
|
\section{Uncountable Fort Space is not Separable}
Tags: Separable Spaces, Fort Space, Fort Spaces
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a separable space.
\end{theorem}
\begin{proof}
Let $C \subseteq S$ be a countable set.
Since $S$ is uncountable, by Uncountable Set less Countable Set is Uncountable, so is $\relcomp S C$.
Thus there exists some point $x \in \relcomp S C$ and $x \ne p$.
By Clopen Points in Fort Space, $\set x \in \tau_p$.
By Empty Intersection iff Subset of Complement, we have $C \cap \set x = \O$.
Therefore $C$ is not everywhere dense.
Since $C$ is arbitrary, $T$ is not a separable space.
{{qed}}
\end{proof}
|
22930
|
\section{Uncountable Open Ordinal Space is Countably Compact}
Tags: Countably Compact Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a countably compact space.
\end{theorem}
\begin{proof}
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.
{{LinkWanted|sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$}}
But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$.
{{LinkWanted|$\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$}}
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$.
This means that $\hointr 0 \Omega$ is countably compact.
{{qed}}
\end{proof}
|
22931
|
\section{Uncountable Open Ordinal Space is Sequentially Compact}
Tags: Sequentially Compact Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a sequentially compact space.
\end{theorem}
\begin{proof}
We have that:
:Uncountable Open Ordinal Space is First-Countable
:Uncountable Open Ordinal Space is Countably Compact
The result follows from First-Countable Space is Sequentially Compact iff Countably Compact.
{{qed}}
\end{proof}
|
22932
|
\section{Uncountable Open Ordinal Space is not Lindelöf}
Tags: Ordinal Spaces, Paracompact Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a Lindelöf space.
\end{theorem}
\begin{proof}
{{AimForCont}} $\hointr 0 \Omega$ is a Lindelöf space.
From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.
From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.
From Lindelöf $T_3$ Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
22933
|
\section{Uncountable Open Ordinal Space is not Metacompact}
Tags: Metacompact Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a metacompact space.
\end{theorem}
\begin{proof}
{{AimForCont}} $\hointr 0 \Omega$ is a metacompact space.
From Open Ordinal Space is not Compact in Closed Ordinal Space we have that $\hointr 0 \Omega$ is a countably compact space.
From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space.
But from Open Ordinal Space is not Compact in Closed Ordinal Space this contradicts the fact that $\hointr 0 \Omega$ is not a compact space.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
22934
|
\section{Uncountable Open Ordinal Space is not Paracompact}
Tags: Ordinal Spaces, Paracompact Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a paracompact space.
\end{theorem}
\begin{proof}
{{AimForCont}} $\hointr 0 \Omega$ is a paracompact space.
From Paracompact Space is Metacompact, it follows that $\hointr 0 \Omega$ is a metacompact space.
But from Uncountable Open Ordinal Space is not Metacompact this contradicts the fact that $\hointr 0 \Omega$ is not a metacompact space.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
22935
|
\section{Uncountable Open Ordinal Space is not Separable}
Tags: Separable Spaces, Ordinal Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a separable space.
\end{theorem}
\begin{proof}
Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable.
Let $H \subseteq \hointr 0 \Omega$ be a countable subset of $\hointr 0 \Omega$.
Let $\sigma$ be the supremum of $H$.
As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable.
Thus $\sigma$ strictly precedes $\Omega$.
The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$.
By definition of the closure $H^-$ of $H$ as the smallest closed set of $\hointr 0 \Omega$ containing $H$, it follows that $H^- \subseteq \closedint 0 \sigma$.
Therefore, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$.
Thus the closure of $H$ does not equal $\hointr 0 \Omega$.
Thus $H$ is not everywhere dense in $\hointr 0 \Omega$.
Hence, by definition, $\hointr 0 \Omega$ is not separable.
{{qed}}
\end{proof}
|
22936
|
\section{Uncountable Open Ordinal Space is not Sigma-Compact}
Tags: Ordinal Spaces, Sigma-Compact Spaces
\begin{theorem}
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a $\sigma$-compact space.
\end{theorem}
\begin{proof}
{{AimForCont}} $\hointr 0 \Omega$ is a $\sigma$-compact space.
From Sigma-Compact Space is Lindelöf, $\hointr 0 \Omega$ is a Lindelöf space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf space.
Hence the result by Proof by Contradiction.
{{qed}}
\end{proof}
|
22937
|
\section{Uncountable Particular Point Space is not Lindelöf}
Tags: Particular Point Topology, Lindelöf Spaces
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not a Lindelöf space.
\end{theorem}
\begin{proof}
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is uncountable, then so is $\CC$, as we can set up a bijection $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ has no countable subcover.
The result follows by definition of Lindelöf space.
{{qed}}
\end{proof}
|
22938
|
\section{Uncountable Particular Point Space is not Second-Countable}
Tags: Particular Point Topology, Second-Countable Spaces
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not second-countable.
\end{theorem}
\begin{proof}
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Every subset $V \subseteq H$ is a closed set from Subset of Particular Point Space is either Open or Closed.
Thus we can consider $H$ as an uncountable discrete space.
The result follows from Uncountable Discrete Space is not Second-Countable.
{{qed}}
\end{proof}
|
22939
|
\section{Uncountable Product of First-Countable Spaces is not always First-Countable}
Tags: Product Spaces, First-Countable Spaces, Product Topology
\begin{theorem}
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a first-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a first-countable space.
\end{theorem}
\begin{proof}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Discrete Space is First-Countable, $T$ is a first-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not First-Countable, $T$ is not a first-countable space.
Hence the result.
{{qed}}
\end{proof}
|
22940
|
\section{Uncountable Product of Second-Countable Spaces is not always Second-Countable}
Tags: Second-Countable Spaces, Product Spaces, Product Topology
\begin{theorem}
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a second-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a second-countable space.
\end{theorem}
\begin{proof}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Second-Countable, $T$ is a second-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
{{qed}}
\end{proof}
|
22941
|
\section{Uncountable Product of Separable Spaces is not always Separable}
Tags: Separable Spaces, Product Spaces, Product Topology
\begin{theorem}
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a separable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a separable space.
\end{theorem}
\begin{proof}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $\ds T' = \struct {\prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Separable, $T$ is a separable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Separable, $T$ is not a separable space.
Hence the result.
{{qed}}
\end{proof}
|
22942
|
\section{Uncountable Product of Sequentially Compact Spaces is not always Sequentially Compact}
Tags: Sequentially Compact Spaces, Product Spaces, Product Topology
\begin{theorem}
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be sequentially compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also sequentially compact.
\end{theorem}
\begin{proof}
Let $\mathbb I$ denote the closed unit interval.
Let $T = \struct {\mathbb I, \tau}$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology.
Let $T' = \mathbb I^{\mathbb I} = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\mathbb I, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\mathbb I, \tau}$ indexed by $\mathbb I$ with the product topology $\tau'$.
From Closed Real Interval is Sequentially Compact, $T$ is sequentially compact.
{{explain|The above has not been proven yet, but I think it is: $\mathbb I$ is a metric space, and a compact metric space is sequentially compact.}}
But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.
Hence the result.
{{qed}}
\end{proof}
|
22943
|
\section{Uncountable Set less Countable Set is Uncountable}
Tags: Uncountable Sets, Countable Sets
\begin{theorem}
Let $S$ be an uncountable set.
Let $T \subseteq S$ be a countable subset of $S$.
Then:
:$S \setminus T$ is uncountable
where $\setminus$ denotes set difference.
\end{theorem}
\begin{proof}
{{AimForCont}} $S \setminus T$ were countable.
By definition of relative complement:
:$S \setminus T = \relcomp S T$
Thus from Union with Relative Complement:
:$\paren {S \setminus T} \cup T = S$
But from Finite Union of Countable Sets is Countable it follows that $S$ is countable.
From this contradiction it follows that $S \setminus T$ is uncountable.
{{qed}}
{{LEM|Union with Relative Complement}}
Category:Uncountable Sets
Category:Countable Sets
\end{proof}
|
22944
|
\section{Uncountable Subset of Countable Complement Space Intersects Open Sets}
Tags: Countable Complement Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is uncountable.
\end{theorem}
\begin{proof}
Let $U \in \tau$ be any non-empty open set of $T$.
Then $\relcomp S U$ is countable.
Suppose $H \cap U = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable.
So if $H$ is uncountable it is bound to have a non-empty intersection with every open set in $T$.
{{finish|It still needs to be shown that $U \cap H$ is uncountable {{iff}} $H$ is uncountable.}}
\end{proof}
|
22945
|
\section{Uncountable Sum as Series}
Tags: Uncountable Sum as Series, Real Analysis
\begin{theorem}
Let $X$ be an uncountable set.
Let $f: X \to \closedint 0 {+\infty}$ be an extended real-valued function.
The uncountable sum:
:$\ds \sum_{x \mathop \in X} \map f x = \sup \set {\sum_{x \mathop \in F} \map f x : F \subseteq X, F \text{ finite} }$
is $+\infty$, or can be expressed as a (possibly divergent) series.
\end{theorem}
\begin{proof}
Define:
:$A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$
Then $\sequence {A_n}_{n \mathop \in \N} \uparrow A$ is an exhausting sequence of sets, where:
:$A = \set {x \in X: \map f x > 0}$
Suppose $A$ is uncountable.
From Countable Union of Countable Sets is Countable, necessarily there is some $A_{n_0}$ which is uncountable.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{x \mathop \in X} \map f x
| r = \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq X, F \text{ finite} }
}}
{{eqn | o = \ge
| r = \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq A_{n_0}, F \text{ finite} }
}}
{{eqn | o = \ge
| r = \sup \set {\sum_{x \mathop \in F} \frac 1 {n_0} : F \subseteq A_{n_0}, F \text{ finite} }
}}
{{eqn | r = \sum_{n \mathop = 1}^{\infty} \frac 1 {n_0}
}}
{{eqn | r = +\infty
}}
{{end-eqn}}
Thus:
:$\ds \sum_{x \mathop \in X} \map f x = +\infty$
Otherwise, suppose then that $A$ is countably infinite.
Then there is a bijection $g: \N \leftrightarrow A$
Define $B_N = g \sqbrk {\set {1, 2, \ldots, N - 1, N} }$
Then every finite subset $F$ of $A$ is contained in some $B_N$.
This implies the inequality:
:$\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^N \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
for each $F$ finite.
Taking the supremum of this inequality over $N$:
:$\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
Taking the supremum of this inequality over $F$:
:$\ds \sum_{x \mathop \in X} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
Thus:
:$\ds \sum_{x \mathop \in X} \map f x = \sum_{n \mathop = 1}^\infty \map f {\map g n}$
for some bijection $g: \N \leftrightarrow A$
{{qed}}
\end{proof}
|
22946
|
\section{Undecidability Theorem}
Tags: Mathematical Logic
\begin{theorem}
Let $T$ be the set of theorems of some consistent theory in the language of arithmetic which contains minimal arithmetic $Q$.
$T$ is not recursive.
\end{theorem}
\begin{proof}
Let $\Theta$ be the set of Gödel numbers of the theorems in $T$.
We have that $T$ is a consistent extension of $Q$.
By Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic, $\Theta$ is not a definable set in $T$.
Since Recursive Sets are Definable in Arithmetic, this means that $\Theta$ is not recursive.
{{AimForCont}} $T$ were recursive.
Then by Gödel Numbering is Recursive, we could recursively go from $\Theta$ to $T$.
Then using the recursivity of $T$, we could then recursively go back to $\Theta$.
Thus $\Theta$ would be recursive.
This would be a contradiction.
Thus, by Proof by Contradiction, $T$ is not recursive.
{{qed}}
\end{proof}
|
22947
|
\section{Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual}
Tags: Evaluation Linear Transformations (Module Theory), Linear Transformations
\begin{theorem}
Let $\struct {R, +, \times}$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G$.
For each $x \in G$, let $x^\wedge: G^* \to R$ be defined as:
:$\forall t \in G^*: \map {x^\wedge} t = \map t x$
Then:
:$x^\wedge \in G^{**}$
\end{theorem}
\begin{proof}
We have that $x^\wedge$ is a mapping from $G^* \to R$.
It remains to be demonstrates that $x^\wedge$ is in fact a linear transformation.
Hence we need to show that:
:$(1): \quad \forall u, v \in G^*: \map {x^\wedge} {u + v} = \map {x^\wedge} u + \map {x^\wedge} v$
:$(2): \quad \forall u \in G^*: \forall \lambda \in R: \map {x^\wedge} {\lambda \times u} = \lambda \times \map {x^\wedge} u$
Hence:
{{begin-eqn}}
{{eqn | n = 1
| l = \map {x^\wedge} {u + v}
| r = \map {\paren {u + v} } t
| c = Definition of $x^\wedge$
}}
{{eqn | r = \map u t + \map v t
| c = {{Defof|Pointwise Addition of Linear Transformations}}
}}
{{eqn | r = \map {x^\wedge} u + \map {x^\wedge} v
| c = Definition of $x^\wedge$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | n = 2
| l = \map {x^\wedge} {\lambda \times u}
| r = \map {\paren {\lambda \times u} } t
| c = Definition of $x^\wedge$
}}
{{eqn | r = \lambda \times \map u t
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \lambda \times \map {x^\wedge} u
| c = Definition of $x^\wedge$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22948
|
\section{Underlying Set of Topological Space is Clopen}
Tags: Clopen Sets
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is both open and closed in $T$.
\end{theorem}
\begin{proof}
From the definition of topology, $S$ is open in $T$.
From Underlying Set of Topological Space is Closed $S$ is closed in $T$.
Hence the result.
{{qed}}
\end{proof}
|
22949
|
\section{Underlying Set of Topological Space is Closed}
Tags: Closed Sets
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is closed in $T$.
\end{theorem}
\begin{proof}
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Set Difference with Empty Set is Self:
:$S \setminus \O = S$
Hence $S$ is closed in $T$.
{{qed}}
\end{proof}
|
22950
|
\section{Underlying Set of Topological Space is Everywhere Dense}
Tags: Denseness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is everywhere dense in $T$.
\end{theorem}
\begin{proof}
From Underlying Set of Topological Space is Closed, $S$ is closed in $T$.
From Closed Set Equals its Closure, $S = S^-$.
The result follows from definition of everywhere dense.
{{qed}}
Category:Denseness
\end{proof}
|
22951
|
\section{Uniform Continuity on Metric Space does not imply Compactness}
Tags: Uniformly Continuous Mappings, Continuity, Compact Spaces, Uniform Continuity, Metric Spaces
\begin{theorem}
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a uniformly continuous mapping on $A_1$.
Then $M_1$ does not necessarily have to be a compact metric space.
\end{theorem}
\begin{proof}
Let $M_1 = \struct {A_1, d_1}$ be any metric space which is not compact.
Let $I_{M_1}: M_1 \to M_1$ be the identity mapping.
From Identity Mapping is Uniformly Continuous, $I_{M_1}$ is uniformly continuous on $M_1$.
Hence the result.
{{qed}}
\end{proof}
|
22952
|
\section{Uniform Contraction Mapping Theorem}
Tags: Named Theorems, Fixed Point Theorems, Metric Spaces, Implicit Functions
\begin{theorem}
Let $M$ and $N$ be metric spaces.
Let $M$ be complete.
Let $f : M \times N \to M$ be a continuous uniform contraction.
Then for all $t \in N$ there exists a unique $\map g t \in M$ such that $\map f {\map g t, t} = \map g t$, and the mapping $g: N \to M$ is continuous.
\end{theorem}
\begin{proof}
For every $t\in N$, the mapping:
:$f_t: M \to M : x \mapsto \map f {x, t}$ is a contraction.
By the Banach Fixed-Point Theorem, there exists a unique $\map g t \in M$ such that $\map {f_t} {\map g t} = \map g t$.
We show that $g$ is continuous.
Let $K < 1$ be a uniform Lipschitz constant for $f$.
Let $s, t \in N$.
Then
{{begin-eqn}}
{{eqn | l = \map d {\map g s, \map g t}
| r = \map d {\map f {\map g s, s}, \map f {\map g t, t} }
| c = Definition of $g$
}}
{{eqn | o = \le
| r = \map d {\map f {\map g s, s}, \map f {\map g t, s} } + \map d {\map f {\map g t, s}, \map f {\map g t, t} }
| c = {{Defof|Metric}}
}}
{{eqn | o = \le
| r = K \cdot \map d {\map g s, \map g t} + \map d {\map f {\map g t, s}, \map f {\map g t, t} }
| c = $f$ is a uniform contraction
}}
{{end-eqn}}
and thus:
:$\map d {\map g s, \map g t} \le \dfrac 1 {1 - K} \map d {\map f {\map g t, s}, \map f {\map g t, t} }$
The continuity of $g$ now follows from that of $f$ using the definition of product metric
{{qed}}
\end{proof}
|
22953
|
\section{Uniform Convergence is Hereditary}
Tags: Metric Spaces, Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings defined on $A$.
Let $\sequence {f_n}$ be uniformly convergent on $S \subseteq A$.
Then $\sequence {f_n}$ is uniformly convergent on every metric subspace of $S$.
That is, uniform convergence is a hereditary property of a metric space.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Metric Spaces
\end{proof}
|
22954
|
\section{Uniform Convergence of General Dirichlet Series}
Tags: General Dirichlet Series
\begin{theorem}
Let $\map \arg z$ denote the argument of the complex number $z \in \C$.
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converges uniformly for all $s$ such that:
:$\cmod {\map \arg {s - s_0} } \le a < \dfrac \pi 2$
\end{theorem}
\begin{proof}
Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges.
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at $0$ by writing:
{{begin-eqn}}
{{eqn | l = \map g s
| r = \map f {s + s_0}
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n \paren {s + s_0} }
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty a_n e^{-\lambda_n s_0} e^{-\lambda_n s}
}}
{{end-eqn}}
Thus it suffices to show $\map g s$ converges uniformly for for $\cmod {\map \arg s} \le a < \frac \pi 2$
By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon > 0$ there exists an $N$ independent of $s$ such that for all $m, n > N$:
:$\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$
By Abel's Lemma: Formulation 2 we may write:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0} e^{-\lambda_k s} }
| r = \cmod {\sum_{k \mathop = n}^m \paren {\map S {k, n} - \map S {k - 1, n} } e^{-\lambda_k s} }
}}
{{eqn | r = \cmod {\map S {m, n} e^{-\lambda_m s} + \sum_{k \mathop = n}^{m - 1} \map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }
}}
{{eqn | o = \le
| r = \cmod {\map S {m, n} e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }
| c = Triangle Inequality
}}
{{end-eqn}}
Because $\map S {k, j}$ is the difference of two terms of a convergent, and thus cauchy, sequence, we may pick $N$ large enough so that for $j > N$:
:$\ds \cmod {\map S {k, j} } < \frac {\epsilon \cos a} 3$
which gives us:
{{begin-eqn}}
{{eqn | l = \cmod {\map S {m, n} e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\map S {k, n} \paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } }
| o = \le
| r = \frac {\epsilon \cos a} 3 \paren {\cmod {e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } } }
}}
{{end-eqn}}
We see that:
{{begin-eqn}}
{{eqn | l = \cmod {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} }
| r = \cmod {\int_{\lambda_k}^{\lambda_{k + 1} } -s e^{-x s} \rd x}
}}
{{eqn | o = \le
| r = \int_{\lambda_k}^{\lambda_{k + 1} } \cmod {-s e^{-x s} } \rd x
| c = Modulus of Complex Integral
}}
{{eqn | r = \int_{\lambda_k}^{\lambda_{k + 1} } \cmod s e^{-x \sigma} \rd x
| c =
}}
{{eqn | r = \cmod s \int_{\lambda_k}^{\lambda_{k + 1} } e^{-x \sigma} \rd x
| c =
}}
{{eqn | r = \frac {\cmod s} \sigma \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }
| c =
}}
{{eqn | r = \map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }
| c =
}}
{{end-eqn}}
From Shape of Secant Function, we have that on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
:$\cmod {\map \arg s} \le a \implies \map \sec {\map \arg s} \le \sec a$
which gives us:
:$\map \sec {\map \arg s} \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} } \le \sec a \paren {e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }$
Hence:
{{begin-eqn}}
{{eqn | l = \frac {\epsilon \cos a} 3 \paren {\cmod {e^{-\lambda_m s} } + \sum_{k \mathop = n}^{m - 1} \cmod {\paren {e^{-\lambda_k s} - e^{-\lambda_{k + 1} s} } } }
| o = \le
| r = \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m \sigma} + \sec a \sum_{k \mathop = n}^{m - 1} e^{-\lambda_k \sigma} - e^{-\lambda_{k + 1} \sigma} }
}}
{{eqn | r = \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m \sigma} + \sec a \paren {e^{-\lambda_n \sigma} - e^{-\lambda_m \sigma} } }
| c = Telescoping Sum
}}
{{end-eqn}}
Because $\sigma>0$, we have that $ - \lambda_k \sigma <0$ and hence:
:$e^{-\lambda_k \sigma} < 1 \le \sec a$
which gives us:
{{begin-eqn}}
{{eqn | l = \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} }
| o = \le
| r = \frac {\epsilon \cos a} 3 \paren {e^{-\lambda_m\ sigma} + \sec a \paren {e^{-\lambda_n \sigma} - e^{-\lambda_{m}\sigma} } }
}}
{{eqn | o = \le
| r = \frac {\epsilon \cos a} 3 \paren {\sec a + \sec a \paren {1 + 1} }
}}
{{eqn | r = \frac {\epsilon \cos a} 3 \paren {3 \sec a}
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22955
|
\section{Uniform Limit Theorem}
Tags: Continuity, Continuous Mappings, Analysis, Named Theorems, Metric Spaces, Functional Analysis
\begin{theorem}
Let $\struct {M, d_M}$ and $\struct {N, d_N}$ be metric spaces.
Let $\sequence {f_n}$ be a sequence of mappings from $M$ to $N$ such that:
:$(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
:$(2): \quad \sequence {f_n}$ converges uniformly to $f$
Then:
:$f$ is continuous at every point of $M$.
\end{theorem}
\begin{proof}
Let $a \in M$.
We are given that $d_N$ is a metric on $N$.
By applying {{Metric-space-axiom|2}} twice:
{{begin-eqn}}
{{eqn | n = 3
| q = \forall n \in \N, \forall x \in M
| l = \map {d_N} {\map f x, \map f a}
| o = \le
| r = \map {d_N} {\map f x, \map {f_n} x} + \map {d_N} {\map {f_n} x, \map {f_n} a} + \map {d_N} {\map {f_n} a, \map f a}
}}
{{end-eqn}}
Let $\epsilon \in \R_{>0}$.
Since $\sequence {f_n}$ converges uniformly to $f$:
{{begin-eqn}}
{{eqn | n = 4 a
| q = \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}: \forall x \in M
| l = \map {d_N} {\map f x, \map {f_n} x}
| o = <
| r = \frac \epsilon 3
}}
{{eqn | n = 4 b
| ll= \leadsto
| q = \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}
| l = \map {d_N} {\map f a, \map {f_n} a}
| o = <
| r = \frac \epsilon 3
| c = Universal Instantiation of $x$
}}
{{end-eqn}}
We are given that $\forall n \in \N: f_n$ is continuous.
Hence:
{{begin-eqn}}
{{eqn | n = 5
| q = \forall n \in \N: \exists \delta \in \R_{>0}: \forall x \in M
| l = \map {d_M} {x, a}
| o = <
| r = \delta
}}
{{eqn | ll= \implies
| l = \map {d_N} {\map {f_n} x, \map {f_n} a}
| o = <
| r = \frac \epsilon 3
}}
{{end-eqn}}
Combining $(3)$, $\text {(4 a)}$, $\text {(4 b)}$ and $(5)$:
{{begin-eqn}}
{{eqn | q = \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}: \exists \delta \in \R_{>0}: \forall x \in M
| l = \map {d_M} {x, a}
| o = <
| r = \delta
}}
{{eqn | ll= \implies
| l = \map {d_N} {\map f x, \map f a}
| o = <
| r = \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
As $a$ and $\epsilon$ are arbitrary, it follows by Universal Instantiation of $n$ that:
{{begin-eqn}}
{{eqn | q = \forall a \in M: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in M
| l = \map {d_M} {x, a}
| o = <
| r = \delta
}}
{{eqn | ll= \implies
| l = \map {d_N} {\map f x, \map f a}
| o = <
| r = \epsilon
}}
{{end-eqn}}
Hence, $f$ is continuous at every point of $M$.
{{qed}}
\end{proof}
|
22956
|
\section{Uniform Limit of Analytic Functions is Analytic}
Tags: Complex Analysis, Uniform Convergence
\begin{theorem}
Let $U$ be an open subset of $\C$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.
Let $\sequence {f_n}$ converge locally uniformly to $f$ on $U$.
Then $f$ is analytic.
\end{theorem}
\begin{proof}
By Equivalence of Local Uniform Convergence and Compact Convergence, $f_n$ converges to $f$ locally uniformly on $U$.
Then for any $z \in U$, there is an $\epsilon > 0$ so that:
:$\map {B_\epsilon} z \subset U$
and $f_n$ converges uniformly on $\map {B_\epsilon} z$.
Let $\gamma$ be any simple closed curve in $\map {B_\epsilon} z$.
Since $f_n \to f$ uniformly on $\gamma$ (because $\gamma \subset \map {B_\epsilon} z$), we have:
:$\ds \lim_{n \mathop \to \infty} \int_\gamma \map {f_n} z \rd z = \int_\gamma \map f z \rd z$
Since each $f_n$ is analytic, we have that:
:$\ds \forall n \in \N: \int_\gamma \map {f_n} z \rd z = 0$
So we conclude also that:
:$\ds \int_\gamma \map f z \rd z = 0$
Since $\gamma$ was arbitrary, we have by Morera's Theorem that $f$ is analytic in $\map {B_\epsilon} z$.
Since $z$ was arbitrary, $f$ is analytic on all of $U$.
{{qed}}
\end{proof}
|
22957
|
\section{Uniform Matroid is Matroid}
Tags: Definitions: Matroid Theory, Matroid Theory
\begin{theorem}
Let $S$ be a finite set of cardinality $n$.
Let $0 \le k \le n$.
Let $U_{k, n} = \struct{S, \mathscr I}$ be the uniform matroid of rank $k$.
Then $U_{k, n}$ is a matroid.
\end{theorem}
\begin{proof}
It needs to be shown that $\mathscr I$ satisfies the matroid axioms $(I1)$, $(I2)$ and $(I3)$.
\end{proof}
|
22958
|
\section{Uniform Product of Continuous Functions is Continuous}
Tags: Uniform Convergence, Uniform Product of Continuous Functions is Continuous, Infinite Products
\begin{theorem}
Let $X$ be a metric space.
Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.
Let $\sequence {f_n}$ be a sequence of bounded continuous mappings $f_n: X \to \mathbb K$.
Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly to $f$.
Then $f$ is continuous.
\end{theorem}
\begin{proof}
Follows directly from:
:Partial Products of Uniformly Convergent Product Converge Uniformly
:Uniform Limit Theorem
{{qed}}
\end{proof}
|
22959
|
\section{Uniform Space whose Topology is Metrizable is not necessarily Metrizable}
Tags: Uniformities, Metrizable Topologies
\begin{theorem}
Let $\UU$ be a uniformity on a set $S$.
Let $\struct {\struct {S, \UU}, \tau}$ be the uniform space generated from $\UU$.
Let $T = \struct {S, \tau}$ be the uniformizable space yielded by $\struct {\struct {S, \UU}, \tau}$.
Let $T$ be a metrizable space.
Then it is not necessarily the case that $\UU$ is itself a metrizable uniformity.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be an uncountable discrete ordinal space.
From Uncountable Discrete Ordinal Space is Metrizable, $T$ is a metrizable space.
However, from Uncountable Discrete Ordinal Space has Unmetrizable Uniformity, there exists a uniformity $\UU$ which yields the uniformizable space $T = \struct {S, \tau}$ which is not itself a metrizable uniformity.
{{qed}}
\end{proof}
|
22960
|
\section{Uniformity iff Quasiuniformity has Symmetric Basis}
Tags: Uniformities
\begin{theorem}
Let $S$ be a set.
Let $\UU$ be a quasiuniformity on $S$.
Then $\UU$ is a uniformity {{iff}} $\UU$ has a symmetric filter basis.
\end{theorem}
\begin{proof}
Let $\UU$ be a quasiuniformity on $S$ which has a symmetric filter basis $\BB$.
From the definition of filter basis, all the elements of $\UU$ can be formed from intersections of elements of $\BB$.
But from Intersection of Symmetric Relations is Symmetric, it follows that all elements of $\UU$ are symmetric.
Now suppose $\UU$ is a uniformity.
If $\BB$ is a filter basis of $\UU$ then all the elements of $\BB$ are also elements of $\UU$.
Hence $\BB$ is a symmetric filter basis of $\UU$.
{{qed}}
\end{proof}
|
22961
|
\section{Uniformly Absolutely Convergent Product is Uniformly Convergent}
Tags: Uniform Convergence, Infinite Products
\begin{theorem}
Let $X$ be a set.
Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let $\mathbb K$ be complete.
Let $\sequence {f_n}$ be a sequence of bounded mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly absolutely on $X$.
Then it converges uniformly.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Uniform Convergence
Category:Infinite Products
\end{proof}
|
22962
|
\section{Uniformly Continuous Function Preserves Uniform Convergence}
Tags: Uniform Convergence, Uniform Continuity
\begin{theorem}
Let $X$ be a set.
Let $M$ and $N$ be metric spaces.
Let $\sequence {g_n}$ be a sequence of mappings $g_n: X \to M$.
Let $g_n$ converge uniformly to $g: X \to M$.
Let $f: M \to N$ be uniformly continuous.
Then $\sequence {f \circ g_n}$ converges uniformly to $f \circ g$.
\end{theorem}
\begin{proof}
Let $\epsilon \in \R_{>0}$.
Because $f$ is uniformly continuous, there exist $\delta \in \R_{>0}$ such that $\map d {\map f x, \map f y} < \epsilon$ for $\map d {x, y} < \delta$.
Because $\sequence {g_n}$ converges uniformly, there exist $N > 0$ such that $\map d {\map {g_n} x, \map g x} < \delta$ for $n > N$ and all $x \in X$.
Thus $\map d {\map f {\map {g_n} x}, \map f {\map g x} } < \epsilon$ for $n > N$ and all $x \in X$.
Thus $\sequence {f \circ g_n}$ converges uniformly to $f \circ g$.
{{qed}}
Category:Uniform Convergence
Category:Uniform Continuity
\end{proof}
|
22963
|
\section{Uniformly Continuous Function is Continuous/Metric Space}
Tags: Uniformly Continuous Mappings, Continuity, Continuous Mappings in Metric Spaces, Continuous Mappings on Metric Spaces, Continuous Mappings, Uniformly Continuous Mapping, Metric Spaces
\begin{theorem}
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_1, d_1}\right)$ be metric spaces.
Let the mapping $f: M_1 \to M_2$ be uniformly continuous on $M_1$.
Then $f$ is continuous on $M_1$.
\end{theorem}
\begin{proof}
Let $f$ be uniformly continuous on $M_1$.
Let $x \in M_1$.
Let $\epsilon > 0$.
As $f$ is uniformly continuous, $\exists \delta > 0$ such that:
:$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$
Thus by definition $f$ is continuous at $x$.
{{qed}}
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Category:Uniformly Continuous Mappings
\end{proof}
|
22964
|
\section{Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 1}
Tags: Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
\begin{theorem}
Let $\tuple {X, d}$ be a metric space.
Let $\tuple {Y, d'}$ be a complete metric space.
Let $A \subseteq X$ be a set that is not closed.
Let $f : A \to Y$ be a uniformly continuous function.
Let $\sequence {a_n}$ be a sequence in $A$ convergent to $a \in A^-$.
Then $\sequence {\map f {a_n} }$ converges.
\end{theorem}
\begin{proof}
From Set is Closed iff Equals Topological Closure, $A^- \setminus A$ is non-empty.
Let $a \in A^- \setminus A$.
From Point in Closure of Subset of Metric Space iff Limit of Sequence:
:there exists a sequence $\sequence {a_n}$ in $A$ converging to $a$.
Consider now the sequence $\sequence {\map f {a_n} }$ in $Y$.
Note that since $Y$ is complete, if we can show that $\sequence {\map f {a_n} }$ is a Cauchy sequence, then we know that it converges.
We therefore want to show that there exists an $N$ such that for $n, m > N$ we have:
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
Since $f$ is uniformly continuous, we can find $\delta > 0$ such that:
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
whenever:
:$\map d {a_n, a_m} < \delta$
Since $\sequence {a_n}$ is a Cauchy sequence, we can find $N$ such that for $n, m > N$ we have:
:$\map d {a_n, a_m} < \delta$
giving:
:$\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$
for $n, m > N$.
So $\sequence {\map f {a_n} }$ is a Cauchy sequence, and so is convergent.
{{qed}}
Category:Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
\end{proof}
|
22965
|
\section{Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 2}
Tags: Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
\begin{theorem}
Let $\struct {X, d}$ be a metric space.
Let $\struct {Y, d'}$ be a complete metric space.
Let $A \subseteq X$.
Let $f : A \to Y$ be a uniformly continuous function.
Let $\sequence {a_n}$ be a convergent sequence in $A$.
Then the limit of $\sequence {\map f {a_n} }$ is dependent only on the limit of $\sequence {a_n}$.
That is, there exists a function $L : A^- \to Y$ such that:
:$\ds \lim_{n \mathop \to \infty} \map f {a_n} = \map L {\lim_{n \mathop \to \infty} a_n}$
for every convergent sequence $\sequence {a_n}$.
\end{theorem}
\begin{proof}
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $A$ such that $a_n \to a$ and $b_n \to a$, with:
:$\map f {a_n} \to L_1$
and:
:$\map f {b_n} \to L_2$
We have, by the Triangle Inequality:
:$\map {d'} {\map f {a_n}, L_2} \le \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, L_2}$
We want to show that we can find $K$ such that:
:$\map {d'} {\map f {a_n}, L_2} < \epsilon$
for $n > K$.
Note that since $f$ is uniformly continuous, we can find $\delta_1$ such that for:
:$\map d {a_n, b_n} < \delta_1$
we have:
:$\map {d'} {\map f {a_n}, \map f {b_n} } < \dfrac \epsilon 2$
By the Triangle Inequality, we can write:
:$\map d {a_n, b_n} \le \map d {a_n, a} + \map d {a, b_n}$
Since $a_n \to a$ and $b_n \to a$, we can find $K_1, K_2$ such that for $n > K_1$ we have:
:$\map d {a_n, a} < \dfrac {\delta_1} 2$
and for $n > K_2$ we have:
:$\map d {b_n, a} < \dfrac {\delta_1} 2$
So, for $n > \max \set {K_1, K_2}$, we have:
:$\map d {a_n, b_n} < \delta$
and so:
:$\map {d'} {\map f {a_n}, \map f {b_n} } < \dfrac \epsilon 2$
for $n > \max \set {K_1, K_2}$.
Since $\map f {b_n} \to L_2$, we can pick $K_3$ such that:
:$\map {d'} {\map f {b_n}, L_2} < \dfrac \epsilon 2$
for $n > K_3$.
So, setting $K = \max \set {K_1, K_2, K_3}$, we have:
:$\map {d'} {\map f {b_n}, L_2} < \dfrac \epsilon 2$
for $n > K$.
So:
:$\map {d'} {\map f {a_n}, L_2} < \epsilon$
for $n > K$.
So $\map f {a_n} \to L_2$.
From Convergent Sequence in Metric Space has Unique Limit, we have:
:$L_1 = L_2$
So, if $a_n \to a$, the sequence $\sequence {\map f {a_n} }$ converges to some limit $\map L a$ dependent only on $a$.
{{qed}}
Category:Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain
\end{proof}
|
22966
|
\section{Uniformly Convergent Sequence Evaluated on Convergent Sequence}
Tags: Uniform Convergence, Metric Spaces, Convergence
\begin{theorem}
Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be metric spaces.
{{explain|Definition:Complete Metric Space is invoked, but it is called just "metric space". Analyse to determine whether it needs to be complete here or not.}}
Let $K$ be a subspace of $X$.
Let $\FF = \sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to Y$ uniformly convergent on $K$.
Let $\sequence {a_n}$ be a convergent sequence in $K$ with limit $a \in K$.
Then $\sequence {\map {f_n} {a_n} }$ is convergent such that:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$
\end{theorem}
\begin{proof}
We want to show that:
:$\size {\map {f_n} {a_n} - \map f a} \to 0$ as $n \to \infty$
From the Triangle Inequality:
:$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a}$
Now fix $\epsilon \in \R^{>0}$.
Since $f$ is continuous and $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:
:$\exists M \in \N: n \ge M \implies \size {\map {f_n} {a_n} - \map f {a_n} } < \dfrac \epsilon 2$
Since $\sequence {f_n}$ is uniformly convergent on $K$:
:$\exists N \in \N: n \ge N \implies \size {\map {f_n} {a_n} \to \map f {a_n} } < \dfrac \epsilon 2$
So if $n \ge \max \set {M, N}$:
:$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a} < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$
This completes the proof.
{{qed}}
Category:Metric Spaces
Category:Uniform Convergence
\end{proof}
|
22967
|
\section{Uniformly Convergent Sequence Multiplied with Function}
Tags: Uniform Convergence
\begin{theorem}
Let $X$ be a set.
Let $V$ be a normed vector space over $\mathbb K$.
Let $\sequence {f_n}$ be a sequence of mappings $f_n: X \to V$.
Let $\sequence {f_n}$ be uniformly convergent.
Let $g: X \to \mathbb K$ be bounded.
Then $\sequence {f_n g}$ is uniformly convergent.
\end{theorem}
\begin{proof}
Denote $\norm {\, \cdot \,}$ as the norm on $V$.
Let $\epsilon > 0$.
By boundedness of $g$:
:$\exists M \in \R: \forall x \in X: \norm {\map g x} < M$
By uniformly convergence of $\sequence {f_n}$:
:$\exists f: X \to \mathbb K: \exists N \in \R: \forall x \in X: \norm {\map {f_n} x - \map f x} < \dfrac \epsilon M$
Pick any $x \in X$. Then:
{{begin-eqn}}
{{eqn | l = \norm {\map {f_n} x \map g x - \map f x \map g x}
| r = \norm {\map g x \paren {\map {f_n} x - \map f x} }
}}
{{eqn | r = \norm {\map g x} \norm {\map {f_n} x - \map f x}
| c = {{NormAxiomMult|2}}
}}
{{eqn | o = <
| r = M \cdot \dfrac \epsilon M
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
As the choice of $x$ is arbitrary, $\sequence {f_n g}$ uniformly converges to $\sequence {f g}$.
{{qed}}
\end{proof}
|
22968
|
\section{Uniformly Convergent Sequence of Bounded Functions is Uniformly Bounded}
Tags: Metric Spaces
\begin{theorem}
Let $X = \left({A, d}\right)$ and $Y = \left({B, \rho}\right)$ be metric spaces.
Let $\left \langle{f_i}\right \rangle_{i \in I}$ be a uniformly convergent sequence of mappings $f_i: X \to Y$.
$\forall i \in I$, let $f_i$ be bounded.
Then $\left \langle{f_i}\right \rangle$ is uniformly bounded.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Metric Spaces
\end{proof}
|
22969
|
\section{Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function}
Tags: Continuous Functions, Uniform Convergence
\begin{theorem}
Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f: S \to \R$.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Then $f$ is continuous at $x$.
\end{theorem}
\begin{proof}
Let $\epsilon \in \R_{> 0}$.
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:
:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$
for all $x \in S$ and $n \ge N$.
Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that:
:for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$
Then for $y$ with $\size {x - y} < \delta$ we have:
{{begin-eqn}}
{{eqn | l = \size {\map f x - \map f y}
| r = \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y}
}}
{{eqn | r = \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} }
}}
{{eqn | o = \le
| r = \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y}
| c = Triangle Inequality for Real Numbers
}}
{{eqn | o = <
| r = 3 \times \frac \epsilon 3
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
Since $\epsilon$ was arbitrary, $f$ is continuous at $x$.
{{qed}}
\end{proof}
|
22970
|
\section{Uniformly Convergent Sequence on Dense Subset}
Tags: Uniform Convergence
\begin{theorem}
Let $X$ be a metric space.
Let $Y \subset X$ be dense.
Let $V$ be a Banach space.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n : X\to V$.
Let $\sequence {f_n}$ be uniformly convergent on $Y$.
Then $\sequence {f_n}$ is uniformly convergent on $X$.
\end{theorem}
\begin{proof}
Let $\epsilon>0$.
Let $N \in\N$ be such that $\norm {f_n - f_m} < \epsilon$ for $n, m > N$ on $Y$.
Let $x \in X$.
Then there exists a sequence $\sequence {y_n} \in Y$ with $y_n \to x$.
By continuity:
:$\norm {\map {f_n} x - \map {f_m} x} \le \epsilon$
Because $V$ is complete, $\sequence {f_n}$ is uniformly convergent on $X$.
{{qed}}
Category:Uniform Convergence
\end{proof}
|
22971
|
\section{Uniformly Convergent Series of Continuous Functions Converges to Continuous Function}
Tags: Continuous Functions, Continuous Real Functions, Uniform Convergence
\begin{theorem}
Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Let the infinite series:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous at $x$.
\end{theorem}
\begin{proof}
Let $\sequence {s_n}$ be sequence of real functions $S \to \R$ such that:
:$\ds \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_n} x$
for each $n \in \N$ and $x \in S$.
By Sum Rule for Continuous Real Functions:
:$s_n$ is continuous at $x$ for all $n \in \N$.
Since additionally $s_n \to f$ uniformly, we have by Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function:
:$f$ is continuous at $x$.
{{qed}}
\end{proof}
|
22972
|
\section{Uniformly Convergent Series of Continuous Functions Converges to Continuous Function/Corollary}
Tags: Continuous Functions, Uniform Convergence
\begin{theorem}
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous for all $n \in \N$.
Let the infinite series:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous.
\end{theorem}
\begin{proof}
Let $x \in S$.
Then $f_n$ is continuous at $x$ for all $n \in \N$.
Since:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
converges uniformly to $f$, we have by Uniformly Convergent Series of Continuous Functions Converges to Continuous Function:
:$f$ is continuous at $x$.
As $x \in S$ was arbitrary, we have that:
:$f$ is continuous.
{{qed}}
Category:Uniform Convergence
Category:Continuous Functions
\end{proof}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.