A topological group $G$ is compact whenever there is a compact $K \subset G$ such that $x K \cap K \neq \emptyset$ for all $x \in G$

Question

There is the following sufficient condition for a topological group to be compact:

Let $G$ be a topological group. Suppose there exists a compact set $K \subset G$ such that $x K \cap K \neq \emptyset$ for all $x \in G$. Then $G$ is compact.

I am curious whether there is a straightforward proof of this fact using net arguments, i.e., given a net $(g_i)_{i \in I} \subset G$ would it be possible to show that there exists a convergent subnet $(g_j)_{j \in J}$.

Any help or comment is appreciated.

Answer 1

We know that a topological space $X$ is compact iff for every net $(x_\alpha)$ in $X$ there is a subnet that converges to a point in $X$.

So let $(g_\alpha)$ be a net in $G$. Then there exists $(k_\alpha)\subset K$ such that $g_\alpha\cdot k_\alpha\in g_\alpha K\cap K$ for each $\alpha$. Since $g_\alpha\cdot k_\alpha\in K$ which is compact, there is a subnet (which for convenience we shall denote by $(g_\alpha\cdot k_\alpha)$ itself) which converges to an element $g\in K\subset G$. Also there is a subnet of $k_\alpha$ (which we will denote by $(k_\alpha)$) which converges to some $k\in K$. Thus $g_\alpha=(g_\alpha\cdot k_\alpha)\cdot k_\alpha^{-1}\to g\cdot k^{-1}$ and you are done.

Answer 2

Someone already gave a direct proof like you asked, so here's a proof without nets for those (like me) who prefer them:

The condition implies that for every $g\in G$ there are $x,y\in K$ such that $g=xy^{-1}$. Defining the map $\phi$ by $$G\times G\to G:(x,y)\mapsto x\cdot y^{-1},$$we see this means $\phi(K\times K)=G$. Since $\phi$ is continuous and $K\times K$ is compact, so is $G$.