Supplementary Exercises*: Topological Groups: Problem 3 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $H$ be a subspace of $G$ . Show that if $H$ is also a subgroup of $G$ , then both $H$ and $\overline{H}$ are topological groups.

Let $f:G\times G\rightarrow G$ be given by $f(x,y)=x\cdot y^{-1}$ . $\overline{H}$ is a subgroup iff $f(\overline{H}\times\overline{H})\subset\overline{H}$ , which is true as $f(\overline{H}\times\overline{H})=f(\overline{H\times H})\subset\overline{f(H\times H)}\subset\overline{H}$ ( $f$ is continuous, see also Theorem 19.5). Now, if $H$ (or $\overline{H}$ ) is a subgroup, then the restriction $f|_{H}:H\times H\rightarrow H$ (or $f|_{\overline{H}}$ ) is continuous (see Exercise 1), moreover, both $H$ and $\overline{H}$ are $T_{1}$ -spaces.

Subpages

Supplementary Exercises*: Topological Groups: Problem 3 Solution

Supplementary Exercises*: Topological Groups: Problem 3 Solution