Why, in this case, if G Hausdorff then G is discrete?

Question

Let $G$ a topological group. It is Hausdorff and locally compact. Why , if every point has a neighborhood that contains a finite number of points, then G is discrete?

Answer 1

This does not depend on the group structure at all.

Claim. Let $X$ be a Hausdorff topological space such that every point has a finite neighborhood. Then the topology is discrete.

Proof. For every $x\in X$ let $U_x=\{x, u_1,\ldots,u_n\}$ be a finite open set containing $x$. Apply the Haudorff property to each of the pairs $(x,u_i)$ in turn to obtain open sets separating $x$ from each of the $u_i$. Intersect $U_x$ with the sets separating $x$ from the $u_i$ to obtain the singleton set $\{x\}$, which is open since it is a finite intersection of open sets.

Answer 2

Let $x \in G$ and let $U$ be a finite open set such that $x \in U$. Since $G$ is Hausdorff each singleton set is closed. If you remove from $U$ all points except $x$ you will be left with $\{x\}$. This proves that each singleton set is open (since finite union of closed sets is closed). Hence all sets are open.

Answer 3

Your point $p$ has a finite neighbourhood $N$. Let the other points of $N$ be $p_1,\ldots,p_n$. By Hausdorffness, $p$ has a neighbourhood $U_j$ not containing $p_j$. Then $N\cap U_1\cap\cdots \cap U_n=\{p\}$ is a neighbourhood of $p$. Therefore $G$ is discrete.

Answer 4

Let $x$ be a point of $G$ and $U$ an open subset of $x$ whose element are $x_1=x,...,x_n$ since $G$ is Haussdorff there exists open subsets $U_i, V_i, i>1$ such that $x_i\in U_i, x\in V_i$ and $U_i\cap V_i$ is empty. Take $V=U\cap_i V_i$. $V$ is an open subset whose only element is $x$.