User:TakuyaMurata/Topological groups

Topological spaces edit

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

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

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

In application, we are usually given some set $X$ (which often carries an intrinsic structure on its own) and then induce a topology to $X$ by defining a $\tau$ according to how we want to do analysis on $X$ . 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 ${\mathcal {F}}$ be a family of topologies for the same set. If $\tau \subset \cap {\mathcal {F}}$ , then $\tau \subset$ every member of ${\mathcal {F}}$ , which is closed under unions. Thus, $\cup \tau$ is in every member of ${\mathcal {F}}$ . The other properties can be verified in the same manner. $\square$

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

Let $f$ be a function to some topological space. Then the set of $f^{-1}(G)$ for every open set $G$ is a the weakest among topologies that make $f$ continuous. If ${\mathcal {F}}$ is a family of functions defined on the same set, then a weak topology generated by ${\mathcal {F}}$ is the intersection of the weakest topology that makes each member of ${\mathcal {F}}$ 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 $E\subset X$ is said closed in $X$ when $E\backslash X$ is open. The intersection of all closed set containing $E$ is called the closure of $E$ and denoted by ${\overline {E}}$ . By (iii) in the definition, the closure of $E$ is closed. Moreover, a set is closed if and only if $E={\overline {E}}$ .

1 Lemma

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

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

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

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 $A$ is in a filter then any set containing $A$ is in the filter (if $A$ is a neighborhood of a point , then any larger set should also be a neighborhood of that point.) In fact, given a point $x$ in some set $X$ , let $f=\{A\subset X:x\in A\}$ . Then $f$ is an ultrafilter on $X$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

f(X)=f({\overline {f^{-1}(V)}})\subset {\overline {f(f^{-1}(V))}}\subset {\overline {V}}

.

\square

Compact sets and Hausdorff spaces edit

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 $K$ be a compact set, and suppose $f[s]$ is an ultrafilter on K. Then s is an ultrafilter and so converges to, say, x. By continuity, f[s] converges to f(x). $\square$

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

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

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

x_{n}\to \infty

implies

f(x_{n})\to \infty .

.

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

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 $X\times Y\to Y$ is closed.

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

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

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 $f:X\to \mathbf {R}$ such that $f=1$ on F and $f(x)=0$ .

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 $X'$ and a continuous injection $i:X\to X'$ such that for any continuous map $f:X\to K$ (where $K$ is a compact space) there is a unique continuous map $F:X'\to K$ with $f=F\circ i$

Proof: See [1] (See also: [2])

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

1 Theorem A topological space is compact if and only if

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

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

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

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

x\in \cap {\tilde {s}}\subset \cap s

.

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

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

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

K\subset G_{1}\cup G_{2}\cdots \cup G_{n}\subset \bigcup G_{\alpha }

for some

n

.

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

\{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 $K_{j}$ be a sequence of compact sets, and suppose that its union is covered by some open cover $\Gamma$ . For each $j=1,2,...$ , since $\Gamma$ is an open cover of $K_{j}$ , it admits a finite subcover $\gamma _{j}$ of $K_{j}$ . Now, $\gamma _{1}\cup \gamma _{2}...$ is a countable subcover of $\Gamma$ . $\square$ .

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

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

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

,

contradicting that $E$ is infinite. $\square$ .

The points x and y, separated by their respective neighbourhoods U and V.

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

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

.

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

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

.

\square

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

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

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

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

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

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

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

Topological groups edit

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

1 Theorem

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

Proof: Left to the reader. $\square$

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

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

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

|f(y)-f(x)|<\epsilon \qquad

for all

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 $W^{2}\subset V$ .) By compactness, $K$ is contained in the union of some $x_{1}V_{x_{1}},x_{2}V_{x_{2}},...,x_{n}V_{x_{n}}$ . Let $V=\cap _{k}V_{x_{k}}$ . It follows that: