Intuition for theorem about compact subsets of topological groups Let $G$ be a topological group, let $K$ be a compact subset of $G$, and let
$U$ be an open subset of $G$ such that $K \subset U$. Then, there is an open set $V$ containing the identity such that $KV \subset U$.
I understand the statement of this theorem but it seems artificial - I don't really understand the intuition behind it. If someone could provide me some insight that would be great. 
 A: A compact subset of a topological group $G$ is "small" with a "sharp boundary" (if you draw it on paper, it should have a solid line denoting its boundary, not a dotted line).  In contrast, an open subset of $G$ is "fuzzy".
Given an element $g\in G$ and a subset $S\subset G$, observe that $Sg=\{sg:s\in S\}$ is the "translate" of $S$ by $g$. (For intuition, consider the case $G=\mathbb{R}$ with the operation $+$, so that $Sg$ would instead be written $S+g$.)
The set $V$ is going to be like a neighborhood $(-\epsilon,\epsilon)$ of $0$ in $\mathbb{R}$, where $\epsilon$ is "sufficiently small". The set $V$ will consist of all the "vectors" along which we will "translate" $K$.
Given a compact $K$ and open $U$ containing $K$, the theorem says you can find a "small enough" open neighborhood $V$ of $0$ such that "translating" $K$ by any element of $V$ still lies inside $U$.
$$KV=\bigcup_{v\in V}Kv \subset U$$
In other words, if $K$ is a small, sharp object contained inside a fuzzy object $U$, I can find a sufficiently small amount of fuzziness $V$ to add to $K$ such that the fuzzy-fied object $KV$ still lies entirely in $U$.
