Topological groups are completely regular

I am studying topological groups, and I have been able to do quite a lot on my own by proving the propositions in this link on my own, but when I read up wikipedia that topological groups are all completely regular, I wasn't able to either find a proof or do it myself, which got me concerned.

The question is ; can anyone confirm me that topological groups are indeed completely regular, and if so, where could I find a proof, or do you know it?

One proof I would be interested in is that accordingly to this Wikipedia page concerning uniform spaces, a topological group can be equipped with the structure of a uniform space in a canonical way, and uniform spaces are completely regular. I don't know how to prove this one either.

I would also be interested in a more direct approach if possible for the purposes of giving a talk on topological spaces ; if a direct proof is shorter than the one going through uniform spaces, it would give me more time in my talk to mention other things.

Solution 1:

Have a look at this:

http://www.math.wm.edu/~vinroot/PadicGroups/519probset1.pdf"

Problem 2 is what you want.

Solution 2:

Every uniform space $(X,\mathcal D)$ is completely regular.

sketch of a proof: Suppose $F$ is closed in $(X,\mathcal D)$ and $p\in F^c$. There's a pseudometric uniformity $P$ on $X$, such that: $$\mathcal D=\mathcal D_P=\bigcup_{d\in P}\mathcal D_d$$ Where $\mathcal D_d$ is the usual uniformity by the pseudometric $d:X^2\to [0,\infty)$.

For each $d\in P$, define $$f_d:X\to\Bbb R$$ $$f_d(x)=\inf_{c\in F}d(c,x)$$ and $$g_d:X\to \Bbb R$$ $$g_d(x)=d(p,x)$$ $f_d$ and $g_d$ are continuous. It's not hard to prove there's some $d_0\in P$ with $$(\forall a\in X)(f_{d_0}(a)\ne 0\text{ or } g_{d_0}(a)\ne 0)$$ This may help. Define $$h:X\to [0,1]$$ $$h(x)=\frac{g_{d_0}(x)}{g_{d_0}(x)+f_{d_0}(x)}$$ $h$ is continuous and $$h(p)=0,\quad h(F)=\{1\}$$

Edit:

linked