Introduction to Topological manifolds Problem 4-15

Question

I am doing problem 4-15 from Lee's book (2nd Edition). The problem has following 4 parts.

${\bf a.}$ If $U$ is open subgroup of a topological group $G$ then $U$ is closed.

This is trivial. Because $U^c=\bigcup\limits_{g:gU\neq U}gU$ is open.

${\bf b.}$ If $U$ is any (open) neighborhood of $1,$ then the subgroup $<U>$ generated by $U$ is both open and closed.

I have a solution for this. I replace $U$ by $U\cup U^{-1}$ and then observe that $\langle U\rangle=\bigcup\limits_{m=1}^{\infty}U^m.$ It suffices therefore to show that $U^m$ is open for each $m,$ which follows from a simple induction argument and the fact that $UV$ is open if $U, V$ are open.

${\bf \text{My problems here is that I am not using that } U\text{ is a neighborhood of }1.}$

$\bf\text{Is it correct that the subgroup generated by any open subset of }G\text{ is open?}$

${\bf c.}$ If $U$ is connected neighborhood of $1$ then $\langle U\rangle$ is connected.

Here it is important that $1\in U.$ I argue that $1\in U^m$ for each $m.$ And then since $U^m$ is connected (continuous image of Cartesian products of $U$), the union over all $U^m$ is connected and we are done.

${\bf d.}$ If $G$ is connected then any connected neighborhood of $1$ generated the whole group.

This follows from ${\bf b}.$ Because $\langle U\rangle$ is both open and closed, and since is non-empty it must be the whole $G$ (since $G$ is connected.)

${\bf Question}$ My doubt arises mainly from this last part. Since I am only using part ${\bf b}$ to conclude that $\langle U \rangle$ is whole $G.$ If everything here is correct, then ${\bf \text{Every open subset of }G\text{ generates whole }G}.$ This, of course, seems like something too good to be true. I am wondering if I am missing something?

Answer 1

You're right -- it's not necessary for the open set to contain the identity, but it must at least be nonempty. So a stronger statement of (b) and (d) would be:

• (b) If $U$ is any nonempty open subset of $G$, then the subgroup $\langle U\rangle$ generated by $U$ is both open and closed.
• (d) If $G$ is connected, then every nonempty open subset of $G$ generates $G$.

I hadn't noticed that -- thanks for pointing it out. Most references I've seen state these results in terms of a neighborhood of the identity, and that's what's normally used; but it's interesting to note that the stronger results are also true.

Answer 2

Well, $⟨U⟩=\cup_{1=m}^{\infty}U^m\cup U^ {−m}$ is only true if $1\in U$ (since $1$ needs to be in $⟨U⟩$, so it needs to be an element of your union).