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https://en.wikipedia.org/wiki/Alexandroff_extension

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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $U\cap V\ne\emptyset$.

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