Show that one point compactification of $\Bbb Q$ is not Hausdorff .

Question

Show that one point compactification of $\Bbb Q$ is not Hausdorff .

Suppose to the contrary $\Bbb Q$ is Hausdorff, then we take two points $x,\infty\in\Bbb Q$, $x\ne\infty$, and produce disjoint open sets $U,V$ such that $x\in U$ and $\infty\in V$. By definition, we know that $\Bbb Q\setminus V$ is a closed and compact space such that $x\in U\subseteq\Bbb Q\setminus V$.

Please help me to complete the proof.

Answer 1

Hint: What do the compact subsets of $\Bbb Q$ look like? What does an open neighbourhood of $x$ look like?

Answer 2

Suppose $0 \in \Bbb Q \subseteq \alpha(\Bbb Q)$ and $\infty \in \alpha(\Bbb Q)$ ahve disjoint open neighbourhoods $U$ and $V$ in $\alpha(\Bbb Q)$.

Then $U$ is open in $\Bbb Q$ (in its original topology) and $V=\Bbb Q\setminus K$, where is $K \subseteq \Bbb Q$ is compact. By disjointness of $U$ and $V$ we must have that $0 \in U \subseteq K$. But then let $m$ be any irrational in $U$ and $q_n$ be a rational sequence in $U$ converging to $m$ (all in the usual topology on $\Bbb R$, using that $U$ contains some open interval around $0$). As this sequence is sequence in $K$ without convergent subsequence in $\Bbb Q$ (so not in $K$ either), this contradicts (sequential) compactness of $K$; contradiction.

We really just use that in $\Bbb Q$, no compact subset (like $K$) can have non-empty interior.