# One point compactification of the rationals is not second countable

As a counter example to show that second countable is a property which is not preserved under a continuous function, I consider the rationals under the discrete topology, which is second countable, and the existence of any other countable space that is not second countable is sufficient.

I chose the one point compactification of the rationals as such a space because several sources say this is true, but I am very lost on how to go about proving this. I have studied other questions on this site about the compact subsets of Q, and aim to form an open set in Q* that couldn't possibly be formed by a countable union of open sets, but I do not see how this is possible.

I appreciate any insight. Thanks

• Why the rationals with the discrete topology and not the integers with the discrete topology (which is also the induced subspace topology in the reals)? – Dan Rust Dec 19 '13 at 18:38
• I don't think this is true. If you are putting the discrete topology on $\mathbb{Q}$ you might as well be talking about $\mathbb{N}$. In this case the one point compactification is homeomorphic to the set $\{0\}\cup\{\frac{1}{n}:n\in\mathbb{N}\}$. – Owen Sizemore Dec 19 '13 at 18:38
• @OwenSizemore I think the OP took the standard topology of $\mathbb{Q}$ for the one-point-quasicompactification. That should be clarified, of course. – Daniel Fischer Dec 19 '13 at 18:42
• Ah, yes I mean the standard topology for the compactification. I picked rationals with discrete instead of natural numbers for simplicity in choosing the identity map instead of some bijection, but this seems like a trivial point. – user116719 Dec 20 '13 at 5:21

It is true that the one-point compactification of the rationals with their usual topology is not second countable: it is not first countable at the new point. Unfortunately, neither of the proofs that come to mind is completely elementary. I’ll sketch the simpler one. Let $\Bbb Q^*$ be the one-point compactification of $\Bbb Q$ with the usual topology, and let $p$ be the point at infinity.
I’ll show first that $\Bbb Q^*$ is a $US$-space, meaning that no sequence in $\Bbb Q^*$ converges to more than one point. Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb Q$ that converges in $\Bbb Q^*$. If $\sigma$ converges to some $x\in\Bbb Q$, the set $K=\{x\}\cup\{x_n:n\in\Bbb N\}$ is a compact subset of $\Bbb Q$, $\Bbb Q^*\setminus K$ is therefore an open nbhd of $p$ disjoint from $K$, and $\sigma$ does not converge to $p$. The subspace $\Bbb Q$ is Hausdorff, and all Hausdorff spaces are $US$-spaces, so $\sigma$ does not converge to any point of $\Bbb Q$ other than $x$, and $x$ is therefore the unique limit of $\sigma$. If $\sigma$ does not converge to any point of $\Bbb Q$, then of course its unique limit must be $p$. Thus, $\Bbb Q^*$ is a $US$-space.
$\Bbb Q^*$ is not Hausdorff: if $x\in\Bbb Q$, $x$ and $p$ cannot be separated by disjoint open sets. For suppose that $U$ is an open nbhd of $p$. Then $K=\Bbb Q^*\setminus U$ is a compact subset of $\Bbb Q$ and therefore nowhere dense in $\Bbb Q$, so $K$ cannot contain an open nbhd of $x$.
Finally, every first countable $US$-space is Hausdorff, so $\Bbb Q^*$ cannot be first countable. To see this, suppose that $X$ is a first countable $US$-space that is not Hausdorff, and let $x,y\in X$ be such that $U\cap V\ne\varnothing$ whenever $U$ and $V$ are open nbhds of $x$ and $y$, respectively. Suppose that $\{B_n:n\in\Bbb N\}$ is a countable open base at $x$, and $\{W_n:n\in\Bbb N\}$ is a countable open base at $y$. For each $n\in\Bbb N$ let $x_n\in B_n\cap W_n$; then $\langle x_n:n\in\Bbb N\rangle$ converges to both $x$ and $y$, contradicting the hypothesis that $X$ is a $US$-space. Thus, $X$ must be Hausdorff.
There are much easier examples of countable spaces that are not second countable. One is the Arens-Fort space; given a countable family $\mathscr{B}$ of open nbhds of the one non-isolated point of the space, it’s not hard to construct an open nbhd of that point that doesn’t contain any member of $\mathscr{B}$. If you know a little about ultrafilters, another fairly simple example is to let $p$ be a free ultrafilter on $\Bbb N$ and let $X=\{p\}\cup\Bbb N$. $\Bbb N$ has the discrete topology, and a local base at $p$ is $\big\{\{p\}\cup U:U\in p\big\}$. This is a subspace of the Katětov extension of $\Bbb N$, and I proved in this answer that it’s not first countable at $p$.