User:TakuyaMurata/Topological groups

Topological spaces
Given an arbitrary set $$X$$, a subset of the power set of $$X$$ is called a topology $$t$$ for $$X$$ if it includes 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.
 * (i) the empty set and $$X$$
 * (ii) the union of any subset of $$t$$, and
 * (iii) the intersection of any two members of $$t$$.

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 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).$$\square$$
 * (i) $$\bigcup_{i \in I} \overline{E_i} \subset \overline{\cup E_i}$$ where the inequality holds if $$I$$ is finite, and
 * (ii) $$\overline{\bigcap_{i \in I} E_i} \subset \bigcap_{i \in I} \overline{E_i}$$.

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:'' 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:
 * (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 $$f[s] \to f(x)$$.
 * $$f[\overline{E}] \subset \overline{f[E]}$$

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. 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. $$\square$$
 * (i) X is irreducible.
 * (ii) Every nonempty open subset of X is dense.
 * (iii) Every open subset of X (in particular X) is connected.

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
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: 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). $$\square$$
 * (i) Every product space of compact spaces is compact.
 * (ii) Axiom of Choice.

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 $$f = F \circ i$$
 * [[Image:Stone–Cech compactification.png|120px]]

Proof: See (See also: )

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 Proof: ($$\Rightarrow$$) (ii) is immediate. For (i), we first remark that the following are equivalent: 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, $$x$$,
 * (i) Every open cover of $$X$$ contains a subcover that is a finite set, and 
 * (ii) X is Hausdorff.
 * (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).
 * $$x \in \cap \tilde{s} \subset \cap s$$.

($$\Leftarrow$$) 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: 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. $$\square$$
 * (i) Axiom of Choice.
 * (ii) Every product topology of compact topologies is compact. (Tychonoff's product theorem)
 * (iii) Something has empty intersection.

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. $$\square$$.

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 $$x \ne y$$, there is some $$f \in \mathcal{F}$$ such that $$f(x) \ne f(y)$$.

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, $$f(x) \ne f(y)$$. 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 $$f(x) \ne y$$ 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
Let G be a topological group, which we don't assume to be Hausdorff.

1 Theorem Proof: Left to the reader. $$\square$$
 * (i) G is Hausdorff if and only if $$\{e\}$$ is closed.
 * (ii) If G is Hausdorff, then its discrete subgroup is closed.

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 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:
 * $$|f(y) - f(x)| < \epsilon$$    for all $$y \in x V$$.

Indeed, if $$x \in K$$, then $$x \in x_k V_{x_k}$$ for some $$k$$ and so:
 * $$|f(y) - f(x)| \le |f(y) - f(x_k)| + |f(x_k) - f(x)| < 2\epsilon $$   for all $$y \in x V$$.

for any $$y \in x V \subset x_k {V_{x_k}}^2$$.

1. Corollay Every function in $$C_c(G)$$ is uniformly continuous. (Actually, true for C_0(G)?)