A topological group is a group that is also a topological space satisfying the axiom, such that the maps sending to and sending to are continuous (note this means is a homeomorphism). We will use to denote an arbitrary topological group. The goal of these notes is to show that is completely regular ().
We take as given the fact that for any , the maps and are homeomorphisms of and the left coset quotient space for any arbitrary subgroup . Let denote the identity of .
Lemma) We call a neighborhood of symmetric if . For any neighborhood of , there exists a symmetric neighborhood of such that .
Proof) Consider the map defined as . As this is continuous, is a neighborhood of in the product topology. Let be a basis element containing , where and are open in . Taking , we can see that . Taking gives us our desired symmetric neighborhood.
Lemma 2) If we let be the collection of neighborhoods of , the following identity holds for any subset $S \subset G:
\overline{S} = \bigcap_{N \in \mathcal{N}_e} S \cdot N $$ *Proof)* First, assume that $x \in \overline{S}$. Then, for any open set $N \in \mathcal{N}_e$, $N^{-1}$ is open and $x \cdot N^{-1}$ is a neighborhood of $x$. Thus, this intersects $S$, implying $x \cdot N^{-1} \cap S \neq \emptyset \Rightarrow x \cap S \cdot N \neq \emptyset$. Therefore, $x \in \bigcap S \cdot N$ and $\overline{S} \subset \bigcap S \cdot N$. For the other direction, assume there is some $x \in G$ such that for any neighborhood $N$ of $e$, $x \in S \cdot N$. Let $V$ be an arbitrary neighborhood of $x$; $(x^{-1}V)^{-1}$ is a neighborhood of $e$, and thus $x \in S \cdot (x^{-1}V)^{-1} \Rightarrow V \cap S \neq \emptyset$. Thus, $x \in \overline{S}$ and $\bigcap S \cdot N \subset \overline{S}$. **Theorem)** $G$ is completely regular. *Proof)* Our goal is to show that any arbitrary element and disjoint closed set can be separated by a continuous function. Using the fact that $f_\alpha$ is a homeomorphism, we can assume this arbitrary element to always be $e$. Thus, our goal is as follows: given some closed set $A$ not containing $e$, show there exists a continuous function $f: G \to [0, 1]$ such that $f(e) = 0$ and $f(A) = 1$. We hope to use a Urysohn-like construction which works with the group structure alone without normality. Note Lemma gives us a good starting family of open sets: let $V_0$ be a neighborhood of $e$ disjoint from $A$, and for any $n$ let $V_{n+1}$ be a neighborhood of $e$ such that $V_{n+1} \cdot V_{n+1} \subset V_n$. The binary nature of this splitting operation motivates us to use the binary rationals (rationals of the form $k/2^l$, which we denote as $B$) as our countable index. We construct a family of neighborhoods $(U_b)_{b \in B}$ as follows: for any $b \leq 0$, let $U_b = \emptyset$ and for any $b > 1$, let $U_b = G$ with $U_1 =.V_0$. If $0 < b < 1$, let the binary expansion of $b$ be $b = 2^{-n_1} + 2^{-n_2} + \ldots + 2^{-n_k}$ ($0 < n_1 < \ldots < n_k$). We then let $U_b = V_{n_1} \cdot V_{n_2} \cdot \ldots \cdot V_{n_k}$. Note that this is always a subset of $V_0$ as $V_{n_1} \cdot \ldots V_{n_k} \subset V_1 \cdot V_2 \cdot \ldots \cdot V_{n_k} \subset V_1 \cdot V_1 \subset V_0$. To proceed with the Urysohn argument, we have to show whenever $\alpha < \beta$, $\overline{U_\alpha} \subset U_\beta$. If $\alpha \leq 0$ or $\beta > 1$, this is trivial; we thus only have to consider when $0 < \alpha < \beta \leq 1$. In this case, we can see that for any $k$ and $l$ such that $0 < k/2^l < (k+1)/2^l \leq 1$, $U_{k/2^l} \subset U_{k/2^l} \cdot V_{l} \subset U_{(k+1)/2^l}$, which implies $\overline{U_{k/2^l}} \subset U_{(k+1)/2^l}$. For any $0 < \alpha < \beta \leq 1$, we can find a $k$ and $l$ such that $\alpha \leq k/2^l < (k+1)/2^l \leq \beta$, and thus $\overline{U_\alpha} \subset U_\beta$. We now have the proper family of open sets to complete the proof. Define $f: G \to [0,1]$ as: $$ f(g) = \inf \{r \in B | g \in U_r\}ff(e) = 0f(A) = 1g \in G0 \leq f(g) \leq 1f[0,\alpha)(\beta,1]Gf^{-1}([0,\alpha))\bigcup_{b < \alpha} U_bf(g) \in (\beta,1] \iff\gamma > \betag \notin U_\gamma\iff\gamma > \betag \notin \overline{U_\gamma}\overline{U_p} \subset U_qp < qf^{-1}((\beta,1]) = \bigcup_{\gamma > \beta} (\overline{U_\gamma})^C$, which is clearly open. We are done.