# User:TakuyaMurata/Topological groups

## Topological spaces

Given an arbitrary set ${\displaystyle X}$, a subset of the power set of ${\displaystyle X}$ is called a topology ${\displaystyle t}$ for ${\displaystyle X}$ if it includes

• (i) the empty set and ${\displaystyle X}$
• (ii) the union of any subset of ${\displaystyle t}$, and
• (iii) the intersection of any two members of ${\displaystyle t}$.

The members of ${\displaystyle t}$ are said to be open in ${\displaystyle X}$. Immediate examples of topologies are ${\displaystyle \{\varnothing ,X\}}$ and the power set of ${\displaystyle X}$. They are called trivial and discrete topologies, respectively. In the discrete topology, every set is open. On the other hand, ${\displaystyle \varnothing }$ and ${\displaystyle X}$ are the only subsets that are open in the trivial topology.

In application, we are usually given some set ${\displaystyle X}$ (which often carries an intrinsic structure on its own) and then induce a topology to ${\displaystyle X}$ by defining a ${\displaystyle \tau }$ according to how we want to do analysis on ${\displaystyle X}$. The key insight here is that saying some set is open and another closed is merely the matter of labeling. One way to induce a topology is to combine existing ones. For that, we use:

1 Lemma The intersection of topologies for the same set is again a topology for the set.
Proof: Let ${\displaystyle {\mathcal {F}}}$ be a family of topologies for the same set. If ${\displaystyle \tau \subset \cap {\mathcal {F}}}$, then ${\displaystyle \tau \subset }$ every member of ${\displaystyle {\mathcal {F}}}$, which is closed under unions. Thus, ${\displaystyle \cup \tau }$ is in every member of ${\displaystyle {\mathcal {F}}}$. The other properties can be verified in the same manner. ${\displaystyle \square }$

Another common method of inducing a topology is to define topology from convergence of filters. A filter ${\displaystyle s}$ is said to converge to ${\displaystyle x}$ if ${\displaystyle s}$ contains every neighborhood of ${\displaystyle x}$. The convergence depends on which sets are considered neighborhoods of ${\displaystyle x}$; in other words, it depends on a topology. We could go in the other way: given a collection of filters ${\displaystyle s\to x}$, we declare sets to be open so that ${\displaystyle s\to x}$. We only have to check this in fact defines a topology.

Let ${\displaystyle f}$ be a function to some topological space. Then the set of ${\displaystyle f^{-1}(G)}$ for every open set ${\displaystyle G}$ is a the weakest among topologies that make ${\displaystyle f}$ continuous. If ${\displaystyle {\mathcal {F}}}$ is a family of functions defined on the same set, then a weak topology generated by ${\displaystyle {\mathcal {F}}}$ is the intersection of the weakest topology that makes each member of ${\displaystyle {\mathcal {F}}}$ continuous. A weak topology is indeed a topology since the intersection of topologies for the same set is again a topology by Lemma.

Two important types of topological spaces are constructed by this method: product and quotient spaces.

A subset ${\displaystyle E\subset X}$ is said closed in ${\displaystyle X}$ when ${\displaystyle E\backslash X}$ is open. The intersection of all closed set containing ${\displaystyle E}$ is called the closure of ${\displaystyle E}$ and denoted by ${\displaystyle {\overline {E}}}$. By (iii) in the definition, the closure of ${\displaystyle E}$ is closed. Moreover, a set is closed if and only if ${\displaystyle E={\overline {E}}}$.

1 Lemma

• (i) ${\displaystyle \bigcup _{i\in I}{\overline {E_{i}}}\subset {\overline {\cup E_{i}}}}$ where the inequality holds if ${\displaystyle I}$ is finite, and
• (ii) ${\displaystyle {\overline {\bigcap _{i\in I}E_{i}}}\subset \bigcap _{i\in I}{\overline {E_{i}}}}$.

Proof: (i) Since ${\displaystyle {\overline {E_{j}}}\subset {\overline {\bigcup _{i}E_{i}}}}$ for all ${\displaystyle j\in I}$ the desired inequality holds, and since the finite union of closed sets is closed the equality holds if ${\displaystyle I}$ is finite. A similar argument shows (ii).${\displaystyle \square }$

A neighborhood of a point ${\displaystyle x\in X}$ is an "open" subset of ${\displaystyle X}$ containing ${\displaystyle x}$. A point ${\displaystyle x}$ is in the closure of ${\displaystyle E}$ if and only if every neighborhood of ${\displaystyle x}$ intersects ${\displaystyle E}$. Paraphrasing in the language of filter, ${\displaystyle x}$ is in the closure of ${\displaystyle E}$ if and only if a filter containing ${\displaystyle E}$ converges to ${\displaystyle x}$.

In a sense, a filter is a generalization of the notion of "neighborhoods of a point." This explains why a filter, by definition, doesn't contain the empty set (the empty set is a neighborhood of no point as well as the property ${\displaystyle A}$ is in a filter then any set containing ${\displaystyle A}$ is in the filter (if ${\displaystyle A}$ is a neighborhood of a point , then any larger set should also be a neighborhood of that point.) In fact, given a point ${\displaystyle x}$ in some set ${\displaystyle X}$, let ${\displaystyle f=\{A\subset X:x\in A\}}$. Then ${\displaystyle f}$ is an ultrafilter on ${\displaystyle X}$.

Let ${\displaystyle f:A\to B}$. We say ${\displaystyle f}$ is continuous when its pre-image of every open set in ${\displaystyle B}$ is open.

1.4 Theorem Let ${\displaystyle f:X\to Y}$. The following are equivalent:

• (i) ${\displaystyle f}$ is continuous.
• (ii) If ${\displaystyle V\subset Y}$ is an open subset and ${\displaystyle f(x)\in V}$, then there exists a neighborhood ${\displaystyle U}$ of ${\displaystyle x}$ such that ${\displaystyle f[U]\subset V}$.
• (iii) If a filter ${\displaystyle s\to x}$, then ${\displaystyle f[s]\to f(x)}$.

Proof: (i) ${\displaystyle \Rightarrow }$ (ii) ${\displaystyle f(x)\in V}$ means that ${\displaystyle x\in f^{-1}[V]}$. Thus, ${\displaystyle f^{-1}[V]}$ is a neighborhood of ${\displaystyle x}$ and ${\displaystyle f[f^{-1}[V]]\subset V}$. Take ${\displaystyle U=f^{-1}[V]}$. (ii) ${\displaystyle \Rightarrow }$ (iii): If ${\displaystyle V}$ is a neighborhood of ${\displaystyle f(x)}$, then there is ${\displaystyle U}$ as in (ii). Since ${\displaystyle U}$ is a neighborhood of ${\displaystyle x}$, by convergence, ${\displaystyle U\in s}$; thus, ${\displaystyle f[U]\in f[s]}$ and so ${\displaystyle V\in f[s]}$. (iii) ${\displaystyle \Rightarrow }$ (i): We claim:

${\displaystyle f[{\overline {E}}]\subset {\overline {f[E]}}}$

for any subset ${\displaystyle E\subset Y}$. Suppose a filter ${\displaystyle s}$ containing ${\displaystyle E}$ converges to some point ${\displaystyle x\in X}$. By (iii), ${\displaystyle f[s]\to f(x)}$. This proves the claim. (By the way, the claim is then another equivalent definition of continuity.) Now, given a closed subset ${\displaystyle F\subset X}$, applying the claim we get:

${\displaystyle f[{\overline {f^{-1}[F]}}]\subset {\overline {f[f^{-1}[F]]}}\subset {\overline {F}}=F}$

Thus, ${\displaystyle {\overline {f^{-1}[F]}}\subset f^{-1}[F]}$ and so ${\displaystyle f^{-1}[F]}$ is closed. Since ${\displaystyle Y\backslash f^{-1}(U)=f^{-1}(X\backslash U)}$, ${\displaystyle f^{-1}}$ is an open mapping then. ${\displaystyle \square }$

In particular, (iii) means that if ${\displaystyle X}$ is a discrete space, then every function on ${\displaystyle X}$ is continuous.

1 Theorem The composite of continuous functions is again continuous.
Proof: If a filter ${\displaystyle s\to x}$, then ${\displaystyle f[s]\to f(x)}$ and so ${\displaystyle (g\circ f)[s]\to (g\circ f)(x)}$. ${\displaystyle \square }$

A subset of a topological space is locally closed if it can be written as the intersection of a closed set and open set. (Note: sets here are allowed to be empty; so, closed sets and open sets are locally closed.) Equivalently, a set is locally closed if and only if it is open in its closure.

A topological space is said to be connected if it is not a disjoint union of nonempty open sets. Equivalently, a topological space is connected if it has no proper open closed subsets.

A topological space is said to be irreducible if it cannot be written as a union of two proper closed subsets.

1. Theorem Let X be a topological space. Then the following are equivalent.

• (i) X is irreducible.
• (ii) Every nonempty open subset of X is dense.
• (iii) Every open subset of X (in particular X) is connected.

Proof: (i) ${\displaystyle \Rightarrow }$ (ii): Let ${\displaystyle U\subset X}$ be a nonempty open subset. X is then the union of ${\displaystyle {\overline {U}}}$ and the complement of U. By (i), ${\displaystyle {\overline {U}}}$ cannot be a proper subset. (ii) ${\displaystyle \Rightarrow }$ (iii): Let A and B be open subsets. By (ii), ${\displaystyle A}$ is dense and so intersects B. (iii) ${\displaystyle \Rightarrow }$ (i): If X is not irreducible, then X is the union of proper closed subsets A and B. Then ${\displaystyle (X\backslash A)\cup (X\backslash B)}$ is the disjoint union of nonempty open subsets; thus, not connected. ${\displaystyle \square }$

If X is irreducible, then ${\displaystyle \operatorname {C} (X,\mathbf {C} )}$ is a domain. (Consider the zero set of the product ${\displaystyle fg}$)

1. Corollary A continuous image of an irreducible space X is again irreducible.
Proof: Let ${\displaystyle f:X\to f(X)}$ be a continuous map. Let ${\displaystyle V\subset f(X)}$ be an open subset. Then ${\displaystyle f^{-1}(V)}$ is open. By (ii) in the theorem and continuity, we have:

${\displaystyle f(X)=f({\overline {f^{-1}(V)}})\subset {\overline {f(f^{-1}(V))}}\subset {\overline {V}}}$. ${\displaystyle \square }$

## Compact sets and Hausdorff spaces

A subset of a topological space is said to be compact if every ultrafilter in it converges to exactly one point. (Note: This definition, due to Bourbaki, is slightly different but much simpler than one in literature.) In particular, a compact set is closed, and a closed subset of a compact space is compact.

1 Theorem A continuous function sends compact sets to compact sets.
Proof: Let ${\displaystyle K}$ be a compact set, and suppose ${\displaystyle f[s]}$ is an ultrafilter on K. Then s is an ultrafilter and so converges to, say, x. By continuity, f[s] converges to f(x).${\displaystyle \square }$

1 Corollary A function with compact graph is continuous.
Proof: Let ${\displaystyle f:X\to Y}$ be a function that has compact graph. Let ${\displaystyle \pi :\operatorname {gra} (f)\to Y}$ and ${\displaystyle i:X\to \operatorname {gra} (f)}$ be canonical surjection and injection, respectively. Then ${\displaystyle f=\pi ^{-1}\circ i}$. By hypothesis, ${\displaystyle \pi }$ is a closed map; its inverse is thus continuous. Hence, ${\displaystyle f}$ is continuous. ${\displaystyle \square }$

A function is said to be proper if the pre-image of a compact set is compact.

1 Theorem A function ${\displaystyle f:X\to Y}$ is proper if and only if

${\displaystyle x_{n}\to \infty }$ implies ${\displaystyle f(x_{n})\to \infty .}$.

Here ${\displaystyle x_{n}\to \infty }$ means that every compact set contains only finite many ${\displaystyle x_{n}}$.
Proof: (${\displaystyle \Rightarrow }$) is obvious. For the converse, suppose f is not proper. Then there exists a compact subset ${\displaystyle K\subset X}$ such that ${\displaystyle f^{-1}(K)}$ is not compact. The non-compactness allows us to find a strictly increasing sequence ${\displaystyle U_{n}}$ of open subsets such that ${\displaystyle f^{-1}(K)\not \subset U_{n}}$ for every ${\displaystyle n}$ but the union ${\displaystyle \bigcup U_{n}}$ contains ${\displaystyle K}$. Inductively, we can then find a sequence ${\displaystyle x_{n}\in f^{-1}(K)}$ such that ${\displaystyle x_{n}\in U_{n}\backslash U_{n-1}}$. Now, ${\displaystyle f(x_{n})\not \to \infty }$ since ${\displaystyle f(x_{n})\in K}$ for all ${\displaystyle n}$. On the other hand, ${\displaystyle x_{n}\to \infty }$. Indeed, suppose ${\displaystyle L\subset X}$ is a compact subset. Since ${\displaystyle \{x_{1},x_{2},...\}}$ consists of isolated points; thus, closed, the set ${\displaystyle \{x_{1},x_{2},...\}\cap L}$ is a compact subset, of which ${\displaystyle U_{n}}$ is an open cover. Thus, there is some ${\displaystyle n_{0}}$ such that ${\displaystyle U_{n_{0}}\supset \{x_{1},x_{2},...\}\cap L}$. Thus, ${\displaystyle x_{n}\not \in L}$ for all ${\displaystyle n>n_{0}}$. ${\displaystyle \square }$

1 Theorem Every proper map into a locally compact space is a closed map. (Recall that we assume every locally compact space is Hausdorff.)

1 Theorem Every continuous bijection from a compact space to a Hausdorff space is a homeomorphism. 1 Theorem A topological space X is compact if and only if, for all topological spaces Y, the projection ${\displaystyle X\times Y\to Y}$ is closed.

A Hausdorff space X is said to be compactly generated if closed subsets A of X are exactly subsets such that ${\displaystyle A\cap K}$ is closed in K for all compact subsets ${\displaystyle K\subset X}$.

1 Theorem (Tychonoff's product theorem) The following are equivanelt:

• (i) Every product space of compact spaces is compact.
• (ii) Axiom of Choice.

Proof: (i) ${\displaystyle \Rightarrow }$ (ii). Let ${\displaystyle \{X_{\alpha }\}_{\alpha \in A}}$ be a collection of compact spaces and ${\displaystyle \pi _{\alpha }}$ be a projection from ${\displaystyle \prod X_{\alpha }\to X_{\alpha }}$. Let ${\displaystyle {\mathcal {F}}}$ be an ultrafilter on ${\displaystyle \prod X_{\alpha }}$. For each ${\displaystyle \alpha }$, since ${\displaystyle \pi _{\alpha }({\mathcal {F}})}$ is again an ultrafilter and ${\displaystyle X_{\alpha }}$ is compact, ${\displaystyle \pi _{\alpha }({\mathcal {F}})}$ must converge. Then it follows that ${\displaystyle {\mathcal {F}}}$ converges. Axiom of Choice implies that the product space is compact. Conversely, let ${\displaystyle \{X_{\alpha }\}_{\alpha \in A}}$ be a nonempty collection of nonempty sets. Let ${\displaystyle p}$ be a point such that ${\displaystyle p\in X_{\alpha }}$ for all ${\displaystyle \alpha }$. Such a ${\displaystyle p}$ must exist; if not, the intersection of ${\displaystyle X_{\alpha }}$ is the universal set, contradicting that it is a proper class. For each ${\displaystyle \alpha }$, let ${\displaystyle Y_{\alpha }=X_{\alpha }\cup \{p\}}$ and ${\displaystyle \tau _{\alpha }=\{0,p,X_{\alpha },Y_{\alpha }\}}$. Then ${\displaystyle \tau _{\alpha }}$ is a topology for ${\displaystyle Y_{\alpha }}$ and since finiteness is compact. Using (i) ${\displaystyle \prod Y_{\alpha }}$ is compact and thus ${\displaystyle \{X_{\alpha }\}}$ has the finite intersection property and this implies the statement equivalent to (ii). ${\displaystyle \square }$

A topological space X is called a Tychonoff space if it is Hausdorff and, given a closed set F and a point x outside F, there exists a continuous function ${\displaystyle f:X\to \mathbf {R} }$ such that ${\displaystyle f=1}$ on F and ${\displaystyle f(x)=0}$.

1 Theorem Every locally compact space is Tychonoff.

1 Theorem (Stone-Čech compactification) Let X be a Tychonoff space. Then there exists a compact space ${\displaystyle X'}$ and a continuous injection ${\displaystyle i:X\to X'}$ such that for any continuous map ${\displaystyle f:X\to K}$ (where ${\displaystyle K}$ is a compact space) there is a unique continuous map ${\displaystyle F:X'\to K}$ with ${\displaystyle f=F\circ i}$

A cover of the set E is a collection of sets {${\displaystyle G_{\alpha }}$} such that ${\displaystyle E\subset \bigcup G_{\alpha }}$. An open cover of ${\displaystyle E}$ is a cover of ${\displaystyle E}$ consisting of open sets, or equivalently, a subset of the topology whose union contains ${\displaystyle E}$. A subset of cover of ${\displaystyle E}$ is called a subcover if it is again a cover of ${\displaystyle E}$.

1 Theorem A topological space is compact if and only if

• (i) Every open cover of ${\displaystyle X}$ contains a subcover that is a finite set, and
• (ii) X is Hausdorff.

Proof: (${\displaystyle \Rightarrow }$) (ii) is immediate. For (i), we first remark that the following are equivalent:

• (a) Every nonempty collection of closed subsets of ${\displaystyle X}$ with the finite intersection property is nonempty.
• (b) Every nonempty collection of closed subsets of ${\displaystyle X}$ with empty intersection contains a finite subcollection with empty intersection.
• (i).

To see the equivalence of (i) and (b), consider the collection consisting of ${\displaystyle U_{j}^{c}}$, given an open cover ${\displaystyle U_{j}}$. We shall prove (a). Let ${\displaystyle s}$ be a nonempty collection of closed sets with the finite intersection property, and ${\displaystyle {\tilde {s}}}$ be an ultrafilter containing ${\displaystyle s}$. Since ${\displaystyle {\tilde {s}}}$ converges to, say, ${\displaystyle x}$,

${\displaystyle x\in \cap {\tilde {s}}\subset \cap s}$.

(${\displaystyle \Leftarrow }$) Because of (ii), we only have to show that every ultrafilter converges to some point. Let ${\displaystyle s}$ be an ultrafilter. Since ${\displaystyle s}$ has the finite intersection property, by an equivalent form of (i), ${\displaystyle \cap s}$ has a point, say, ${\displaystyle x}$. For every neighborhood ${\displaystyle U}$ of ${\displaystyle x}$, we have either${\displaystyle U\in s}$ or ${\displaystyle U^{c}\in s}$. The latter not being the case, we have ${\displaystyle U\in s}$. In other words, ${\displaystyle s\to x}$ ${\displaystyle \square }$

1 Corollary If ${\displaystyle K_{n}\supset K_{n_{1}}\supset ...}$ is a sequence of compact sets, then ${\displaystyle \cap _{n=1}^{\infty }K_{n}}$ is nonempty.

A set ${\displaystyle K}$ is compact if every open cover {${\displaystyle G_{\alpha }}$} of it has a finite subcover; i.e.,

${\displaystyle K\subset G_{1}\cup G_{2}\cdots \cup G_{n}\subset \bigcup G_{\alpha }}$ for some ${\displaystyle n}$.

For example, let ${\displaystyle \{G_{\alpha }\}}$ be open cover of the finite set ${\displaystyle \{a_{1},a_{2},...a_{n}\}}$. Then for each ${\displaystyle a_{k}}$, we can find some ${\displaystyle G(a_{k})}$. It thus follows that a finite set (e.g., the empty set) is compact since

${\displaystyle \{a_{1},a_{2},...a_{n}\}\subset G(a_{1})\cup G(a_{2})\cup ...G(a_{n})\subset \bigcup G_{\alpha }}$

Conversely, every compact subset of a discrete space is finite.

1 Theorem If a topological space is the union of countably many compact sets, then any of its open cover admits a countable subcover.
Proof: Let ${\displaystyle K_{j}}$ be a sequence of compact sets, and suppose that its union is covered by some open cover ${\displaystyle \Gamma }$. For each ${\displaystyle j=1,2,...}$, since ${\displaystyle \Gamma }$ is an open cover of ${\displaystyle K_{j}}$, it admits a finite subcover ${\displaystyle \gamma _{j}}$ of ${\displaystyle K_{j}}$. Now, ${\displaystyle \gamma _{1}\cup \gamma _{2}...}$ is a countable subcover of ${\displaystyle \Gamma }$. ${\displaystyle \square }$.

1 Theorem The following are equivalent:

• (i) Axiom of Choice.
• (ii) Every product topology of compact topologies is compact. (Tychonoff's product theorem)
• (iii) Something has empty intersection.

Proof: Let ${\displaystyle X}$ be a product topology and ${\displaystyle \{\pi _{\alpha }\}}$ be a collection of projections on ${\displaystyle X}$. Let ${\displaystyle {\mathcal {F}}}$ be an ultrafilter on ${\displaystyle X}$. For each ${\displaystyle \alpha }$, since ${\displaystyle \pi _{\alpha }({\mathcal {F}})}$ is again an ultrafilter and ${\displaystyle \pi _{\alpha }(X)}$ is compact, ${\displaystyle \pi _{\alpha }({\mathcal {F}})}$ converges. From the lemma 1.something it follows that ${\displaystyle {\mathcal {F}}}$ converges. If (i) is true, then the convergence implies that ${\displaystyle X}$ is compact. To show (ii) implies (iii), Let ${\displaystyle \{X_{\alpha }\}}$ be a nonempty collection of nonempty sets. Also, let ${\displaystyle a}$ be a element such that ${\displaystyle a\not \in \cap X_{\alpha }}$. Such an element must exist since the contrary means that ${\displaystyle \cap X_{\alpha }}$ is the universal set. For each ${\displaystyle \alpha }$ let ${\displaystyle Y_{\alpha }=X_{\alpha }\cup \{a\}}$ and induce the topology ${\displaystyle \tau _{\alpha }}$ by letting ${\displaystyle \tau _{\alpha }=\{0,\{a\},X_{\alpha },Y_{\alpha }\}}$. Then since its topology is finite, each ${\displaystyle Y_{\alpha }}$ is compact. That (iii) implies (i) is well known in set theory. ${\displaystyle \square }$

We say a point is isolated if the set ${\displaystyle {x}}$ is both open and closed, otherwise called an limit point. If a set has no limit point, then the set is said to be discrete.

1 Lemma Every infinite subset ${\displaystyle E}$ of a compact space ${\displaystyle K}$ has a limit point.
Proof: Let ${\displaystyle E\subset K}$ be infinite and discrete. Then ${\displaystyle E}$ is closed since it contains all of limit points of ${\displaystyle E}$. Since ${\displaystyle E}$ is a closed subset of a compact set, ${\displaystyle E}$ is compact. It now follows: for each ${\displaystyle x\in E}$, the singleton ${\displaystyle \{x\}}$ is open and thus the collection ${\displaystyle \{\{x\}:x\in E\}}$ is an open cover of ${\displaystyle E}$, which admits a finite subcover. Hence, we have:

${\displaystyle E\subset \{x_{1},x_{2},...x_{n}\}}$,

contradicting that ${\displaystyle E}$ is infinite. ${\displaystyle \square }$.

We say a topological space is Hausdorff if two distinct points are covered by disjoint open sets. This definition can be strengthened considerably.

1 Theorem A topological space ${\displaystyle X}$ is Hausdorff if and only if every pair of disjoint compact subsets of ${\displaystyle X}$ can be covered by two disjoint open sets.
Proof: Let ${\displaystyle K_{1},K_{2}\subset X}$ be compact and disjoint, and ${\displaystyle x\in K_{1}}$ be fixed. For each ${\displaystyle y\in K_{2}}$ we can find disjoint open sets ${\displaystyle A(y)}$ and ${\displaystyle B(y)}$ such that ${\displaystyle x\in A(y)}$ and ${\displaystyle y\in B(y)}$. Since ${\displaystyle K_{2}}$ is compact, there is a finite sequence ${\displaystyle y_{1},y_{2},...,y_{n}\in K_{2}}$ such that:

${\displaystyle K_{2}\subset B(y_{1})\cup B(y_{2})...B(y_{n})}$.

Let ${\displaystyle A(x)=A(y_{1})\cap A(y_{2})\cap ...A(y_{n})}$, and ${\displaystyle B=B(y_{1})\cup B(y_{2})...B(y_{n})}$. Since ${\displaystyle A(x)}$ is disjoint from any of ${\displaystyle B(y_{1})...B(y_{n})}$, ${\displaystyle A(x)}$ and ${\displaystyle B}$ are disjoint. ${\displaystyle A(x)}$ is open since it is a finite intersection of open sets. Since ${\displaystyle K_{1}}$ is compact, there is a finite sequence ${\displaystyle x_{1},...,x_{n}\in K_{1}}$ such that:

${\displaystyle K_{1}\subset A(x_{1})\cup A(x_{2})...A(x_{n})\subset B^{c}\subset {K_{2}}^{c}}$. ${\displaystyle \square }$

1 Theorem A topological space is Hausdorff if and only if a filter ${\displaystyle s}$ on it converges, if it ever does, to at most one point.
Proof: Suppose ${\displaystyle {\mathcal {F}}}$ converges to two distinct points ${\displaystyle x}$ and ${\displaystyle y}$. By the separation axioms, we can find disjoint subjects ${\displaystyle A}$ and ${\displaystyle B}$ such that ${\displaystyle x\in A}$ and ${\displaystyle y\in B}$. Since convergence, ${\displaystyle \{A,B\}\subset {\mathcal {F}}}$. But this then implies that ${\displaystyle A\cap B=0\in {\mathcal {F}}}$, a contradiction. ${\displaystyle \square }$

A Hausdorff space is said to be locally compact if every point in it has a compact neighborhood. A locally compact space is thus locally closed.

We also give two other characterizations of Hausdorff spaces, which are sometimes useful in application.

1 Lemma A topological space ${\displaystyle T}$ is Hausdorff if and only if for every point ${\displaystyle p\in T}$ the set ${\displaystyle \{p\}}$ is the intersection of all of its closed neighborhoods.

1 Lemma Let ${\displaystyle {\mathcal {F}}}$ be a family of functions from ${\displaystyle X}$ to a Hausdorff space ${\displaystyle Y}$. Let ${\displaystyle \Gamma }$ be the union of ${\displaystyle f^{-1}(G)}$ taken all over open sets ${\displaystyle G}$ of ${\displaystyle Y}$ and ${\displaystyle f\in {\mathcal {F}}}$. Then ${\displaystyle \Gamma }$ satisfies the Hausdorff separation axiom if and only if for each ${\displaystyle x,y\in X}$ with ${\displaystyle x\neq y}$, there is some ${\displaystyle f\in {\mathcal {F}}}$ such that ${\displaystyle f(x)\neq f(y)}$.
Proof: First suppose the separation axiom. Then we can find two disjoint sets ${\displaystyle A,B\in \Gamma }$ such that ${\displaystyle x\in A}$ and ${\displaystyle y\in B}$. Then since by definition, there is some function ${\displaystyle f}$ and sets ${\displaystyle G_{1}}$ and ${\displaystyle G_{2}}$ such that ${\displaystyle A=f^{-1}(G_{1})}$ and ${\displaystyle B=f^{-1}(G_{2})}$. Thus, ${\displaystyle f(x)\neq f(y)}$. Conversely, suppose the family ${\displaystyle {\mathcal {F}}}$ separates points in ${\displaystyle X}$. Then by the separation axiom there are disjoint open open sets ${\displaystyle A}$ and ${\displaystyle B}$ such that ${\displaystyle x\in A}$ and ${\displaystyle y\in B}$. Then by the definition of ${\displaystyle \Gamma }$ ${\displaystyle f^{-1}(A)}$ and ${\displaystyle f^{-1}(B)}$ are disjoint and both open. ${\displaystyle \square }$

1 Theorem Let ${\displaystyle f:X\to Y}$ be continuous. If ${\displaystyle Y}$ is Hausdorff, then the graph of ${\displaystyle f}$ is closed.
Proof: Let ${\displaystyle E}$ be the complement of the graph of ${\displaystyle f}$. If ${\displaystyle (x,y)\in E}$, then, since ${\displaystyle f(x)\neq y}$ and ${\displaystyle Y}$ is Hausdorff, we can find in ${\displaystyle Y}$ disjoint neighborhoods ${\displaystyle U}$ and ${\displaystyle V}$ of ${\displaystyle f(x)}$ and ${\displaystyle y}$, respectively. It follows: ${\displaystyle (x,y)\in f^{-1}(U)\times V\subset E}$ since there is no point ${\displaystyle z}$ such that ${\displaystyle z\in f^{-1}(U)}$ and ${\displaystyle f(z)\in V}$. By continuity, ${\displaystyle f^{-1}(U)}$ is open; thus, ${\displaystyle E}$ is open. ${\displaystyle \square }$

1 Corollary Let ${\displaystyle f,g:A\to B}$ be continuous functions. Suppose ${\displaystyle B}$ is Hausdorff. If ${\displaystyle f=g}$ on some dense subset, then ${\displaystyle f=g}$ identically.
Proof: Let ${\displaystyle E}$ be the intersection of the graph of ${\displaystyle f}$ and the graph of ${\displaystyle g}$. ${\displaystyle E}$ is dense in the graph of ${\displaystyle f}$ by hypothesis and closed by the preceding theorem. ${\displaystyle \square }$

1 Theorem A dense subspace X of a compact Hausdorff space Y is locally compact if and only if X is an open subset of Y.

A graph of a function ${\displaystyle f}$ is the set consisting of ordered pairs ${\displaystyle (x,f(x))}$ for all ${\displaystyle x\in }$ the domain of ${\displaystyle f}$. In the set-theoretic view, of course this set is ${\displaystyle f}$. But since we usually do not see a function as a set, the notion is often handy to use.

1 Lemma Let ${\displaystyle f:X\to Y}$ be a function. Suppose ${\displaystyle X}$ is locally compact. Then ${\displaystyle f}$ is continuous if and only if ${\displaystyle f}$ is continuos on every compact subset of ${\displaystyle X}$.
Proof: Suppose ${\displaystyle f(x)\in U}$ where ${\displaystyle U\subset X}$ is an open subset. By local compactness, ${\displaystyle f(x)}$ has a neighborhood ${\displaystyle V\subset K\subset U}$ where ${\displaystyle K}$ is a compact set. Since ${\displaystyle f|_{K}}$ is continuous by hypothesis, ${\displaystyle x}$ has a neighborhood ${\displaystyle W}$ such that: ${\displaystyle f(W)\subset V}$. ${\displaystyle \square }$

## Topological groups

Let G be a topological group, which we don't assume to be Hausdorff.

1 Theorem

• (i) G is Hausdorff if and only if ${\displaystyle \{e\}}$ is closed.
• (ii) If G is Hausdorff, then its discrete subgroup is closed.

Proof: Left to the reader. ${\displaystyle \square }$

1 Theorem (Baire) Let ${\displaystyle X}$ be a locally compact space. If ${\displaystyle X}$ is written as a union of countably many sets, then one of those sets contains a nonempty open subset.
Proof: Similar to one given in w:Baire category theorem.

1 Corollary If ${\displaystyle G}$ is countable and locally compact, then ${\displaystyle G}$ is discrete.
Proof: We write ${\displaystyle G=\{g_{1},g_{2},...\}}$. Then ${\displaystyle \{g_{j}\}}$ is then open for some ${\displaystyle j}$. It follows that ${\displaystyle \{g_{j}^{-1}\}}$ and so ${\displaystyle \{e\}}$ is open. (Recall that every finite subset of a Hausdorff space is closed.) Hence, every subset of ${\displaystyle G}$ is open and closed. ${\displaystyle \square }$

In particular, the only topology that makes ${\displaystyle \mathbf {Q} }$ a locally compact topological group is the discrete one.

1. Theorem Let ${\displaystyle K\subset G}$ be a compact subset. Then ${\displaystyle C(K)}$ consists of uniformly continuous functions.
Proof: We only show right uniform continuity, since the proof for the left uniform continuity is completely analogous. Let ${\displaystyle f\in C(K)}$, and ${\displaystyle \epsilon >0}$ be given. By continuity, for each ${\displaystyle x\in K}$, there is a neighborhood ${\displaystyle V_{x}}$ of e such that

${\displaystyle |f(y)-f(x)|<\epsilon \qquad }$    for all ${\displaystyle y\in x{V_{x}}^{2}}$.

(Note: by the continuity of translation, for any neighborhood V of e, one can always find another neighborhood W of e such that ${\displaystyle W^{2}\subset V}$.) By compactness, ${\displaystyle K}$ is contained in the union of some ${\displaystyle x_{1}V_{x_{1}},x_{2}V_{x_{2}},...,x_{n}V_{x_{n}}}$. Let ${\displaystyle V=\cap _{k}V_{x_{k}}}$. It follows that:

${\displaystyle |f(y)-f(x)|<\epsilon }$     for all ${\displaystyle y\in xV}$.

Indeed, if ${\displaystyle x\in K}$, then ${\displaystyle x\in x_{k}V_{x_{k}}}$ for some ${\displaystyle k}$ and so:

${\displaystyle |f(y)-f(x)|\leq |f(y)-f(x_{k})|+|f(x_{k})-f(x)|<2\epsilon }$    for all ${\displaystyle y\in xV}$.

for any ${\displaystyle y\in xV\subset x_{k}{V_{x_{k}}}^{2}}$.

1. Corollay Every function in ${\displaystyle C_{c}(G)}$ is uniformly continuous. (Actually, true for C_0(G)?)

## Notes

Most of materials in "Compact sets and Hausdorff space" section comes from [3]