Definition: one point compactification

Let $X$ be any topological space, and let $ \infty$ be any object which is not already an element of $X$. Put $ X^{*}=X\cup \{\infty \}$, and topologize $X^* $ by taking as open sets all the open subsets $U$ of $X$ together with all subsets $V$ which contain $\infty $ and such that $X\setminus V$ is closed and compact

Show that the one point compactification of $\mathbb{Q}$ which is $\mathbb{Q}^*$ is Not Hausdorff.

What to do? What...

My Attempt:

Suppose to the contrary $\mathbb{Q}^*$ is Hausdorff, then we take two points $x$, $\infty \in \mathbb{Q}^*$, $x \neq \infty$, and produce disjoint open sets $U,V$ such that $x \in U$ and $\infty \in V$

By definition, we know that $\mathbb{Q} \backslash V$ is a closed and compact space such that $x \in U \subseteq \mathbb{Q} \backslash V$

We wish to produce a contradiction such that $\mathbb{Q} \backslash V$ is not closed, or not compact. But we know that $\mathbb{Q} \backslash V$ has to be closed since $V$ is open, therefore we need to show $\mathbb{Q} \backslash V$ is not compact.

Let $\mathcal{U}$ be an open cover of $\mathbb{Q} \backslash V$. Since $\mathbb{Q} \backslash V$ is claimed to be compact, then $\mathcal{U}$ has a finite subcover $\{U_i|i \in F\}$, $F$ is finite in $\mathbb{Q} \backslash V$. Then for all $x \in \mathbb{Q}\backslash V$, $\exists i \in F$ s.t. $x \in U_i$ (...Ugh everything seems fine...)

Can someone provide me with some help as to how to go on with this proof? Thanks a bunch.


Given your open sets $U$ and $V$ in $\Bbb Q^*$, you can continue the argument as follows.

$U$ is an ordinary open nbhd of $x$ in $\Bbb Q$, so it contains an open nbhd of $x$ of the form $(a,b)\cap\Bbb Q$ for some irrational $a,b\in\Bbb R$. Let $W=(a,b)\cap\Bbb Q$; clearly $W\cap V=\varnothing$, so in particular $\infty\notin\operatorname{cl}_{\Bbb Q^*}W$. Thus,

$$\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q=W\;.$$

But this is clearly impossible, since $W$ is not compact.

If you want an explicit example of an open cover of $W$ with no finite subcover, let $\langle a_n:n\in\Bbb N\rangle$ and $\langle b_n:n\in\Bbb N\rangle$ be sequences in $(a,b)$ such that

  • $\langle a_n:n\in\Bbb N\rangle$ is strictly decreasing and converges to $a$,
  • $\langle b_n:n\in\Bbb N\rangle$ is strictly increasing and converges to $b$, and
  • $a_0<b_0$.

Let $U_n=(a_n,b_n)\cap\Bbb Q$ for $n\in\Bbb N$; then $\{U_n:n\in\Bbb N\}$ is the desired cover.

  • $\begingroup$ Hi, how do you reach the conclusion that: $\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q=W\;.$? I understand why $\operatorname{cl}_{\Bbb Q^*}W=\operatorname{cl}_{\Bbb Q}W=[a,b]\cap\Bbb Q$, but then how does that equal to $W$? And what do you mean by $W$ is not compact? Do you just mean that we defined $W = (a,b) \cap \Bbb{Q}$ but later reached the conclusion $W = [a,b] \cap \Bbb{Q}$...do you mean the latter $ [a,b] \cap \Bbb{Q}$ is compact? How? Thanks for ur help $\endgroup$ – Olórin Aug 9 '16 at 22:45
  • 1
    $\begingroup$ @John: $a$ and $b$ are irrational, so $[a,b]\cap\Bbb Q=(a,b)\cap\Bbb Q$. \\ I mean that $W$ is not compact. I even exhibited an open cover with no finite subcover. $\endgroup$ – Brian M. Scott Aug 10 '16 at 0:10
  • $\begingroup$ I agree, I just wish to understand your top proof. So we let $W = (a,b) \cap \mathbb{Q}$ which is open and not compact, then we found that $\text{cl}_\mathbb{Q}^* W = W$, which implies it is closed and compact... so we reach a contradiction. I think what I am not getting is why is $(a,b) \cap \mathbb{Q}$ not compact in $\mathbb{Q}^*$, and why if we show that $\text{cl}_\mathbb{Q}^* W = W$ is closed, then it is compact. After all we are not working in $\mathbb{R}$, so the intuition about closed and bounded = compact doesn't hold. $\endgroup$ – Olórin Aug 10 '16 at 0:25
  • $\begingroup$ @John: Every closed subset of a compact space is compact. Since $W$ is closed in $\Bbb Q^*$, it must be compact. But in fact it's just $(a,b)\cap\Bbb Q$ with its usual topology, which is not compact. This contradiction shows that $x$ and $\infty$ cannot in fact be separated by disjoint open sets. $\endgroup$ – Brian M. Scott Aug 10 '16 at 0:35

Let $C$ be a compact subset of $\mathbb Q$. Then because the inclusion map $\mathbb Q\to\mathbb R$ is continuous it follows then $C$ is a compact subset of $\mathbb R$, so it is bounded and has all its limit points.

Let $a,b\in\mathbb R$ with $a<b$ such that $(a,b)\cap\mathbb Q\subseteq C$, then the closure of $C$ in $\mathbb R$ includes $(a,b)$. So by contradiction $C$ has empty interior. Therefore every open set containing $\infty$ is dense.

Finally let $x\in \mathbb Q$ and let $U,V$ be open sets with $x\in U$ and $\infty\in V$, then $V$ is dense so $U\cap V\ne\emptyset$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.