Given: $X$ Hausdorff, $K \subset X$ compact.

Show: If $p \in X \setminus K $, then there are disjoint $O_1, O_2$ such that, $p \in O_1$ and $K \subset O_2$ open.

My answer to this problem doesn't use the property of $K$ being compact, which makes me uncertain, whether my proof is correct.

Thus I would be happy, if someone could quickly proofread it.

My answer: Using $X$ Haussdorf: $\forall k \in K, \exists O_p , O_k \subset X$ open, such that $p \in O_p$ and $k \in O_k$ (1)

Thus $K \subset \bigcup\limits_{k \in K}^{} O_k = O_2$ union of open sets. Since $O_p$ and $O_k$ disjoint for any k, $O_p \cap O_2 = \emptyset \quad \square $

I guess my proof goes awry in (1), as I create a set $O_p$ for every $O_k$. Let's get four specific sets: $O_p$ for a $O_k$ and $O_{p'}$ for $O_{k'}$.

While $p \in O_p, O_{p'} $ and $O_p \cap O_k = \emptyset$, $O_{p'} \cap O_{k'}= \emptyset$, this does not imply $O_p \cap \cap O_{k'} = \emptyset = O_{k} \cap O_{p'} $ and I guess my proof breaks. However, what would be an example to further illustrate, that $K$ being compact is actually needed?


For sure there is a flaw in your proof. The issue is that $O_p$ depends on $k \in K$ in (1).

$\mathbb R^2$ equipped with the Euclidean norm is Haussdorf. Take for $K$ the open disk $D$ centered on $(1,0)$ of radius $1$ and for $p$ the origin. It is impossible to find $O_1, O_2$ as described in your question.


It's a bit easy to see why this breaks for open sets as per a previous answer. The harder part is finding how it breaks for closed, but not compact, sets. The guiding principle should be that you want a point that on the surface seems like an accumulation point but isn't and you want to search for a topology that fits that bill.

Take $\Bbb R$ with the topology generated by the usual open sets without countably many points. This means that the set $\left\{\frac{1}{n}: n\in \Bbb N\right\} $ is closed. Are there disjoint open sets separating this from $0$?

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    $\begingroup$ I'm very confused as to what topology you want to put on $\mathbb{R}$. $\endgroup$ – MartianInvader Jun 19 '18 at 23:30
  • $\begingroup$ Take the usual open sets generated by open intervals, then take out countably many points from those. It's the cocountable topology. I didn't want to use that term in case OP wasn't familiar. $\endgroup$ – Cameron Williams Jun 20 '18 at 1:00
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    $\begingroup$ @MartianInvader A very nice topology! $\endgroup$ – mathcounterexamples.net Jun 20 '18 at 19:58
  • $\begingroup$ Although formally nothing is wrong with this, may I suggest you replace the "open" in your first sentence with "non-closed"? They way it is written now, it could reenforce the common "closed = opposite of open" misconception to beginners. $\endgroup$ – Torsten Schoeneberg Jun 23 '18 at 21:32

The main flaw is that the neighbourhood of $p$ also depends on $k \in K$:

For each $k \in K$, we have that $p \neq k$ (as $p \notin K$) and so Hausdorffness of $X$ gives us open sets $U_k$ and $O_k$ such that

$$p \in U_k, k \in O_k ,U_k \cap O_k = \emptyset$$

Now, of course $K \subseteq \bigcup_{k \in K} O_k$ so we have an open cover of $K$ that, by compactness of $K$, we can reduce to finitely many $O_{k_1},\ldots, O_{k_n}$ such that $$K \subseteq O:=\bigcup_{i=1}^n O_{k_i}$$

Now define $$O_p = \bigcap_{i=1}^n U_{k_i}$$

which is an open neighbourhood of $p$ as a finite intersection of neighbourhoods of $p$. And if $x \in O$ it means that $x \in O_{k_i}$ for some $i \in \{1,\ldots,n\}$ but then $x \notin U_{k_i} \supseteq O_p$, so $x \notin O_p$ and so $O_p \cap O = \emptyset$.

It's clear that we cannot have this separation for any $K$ and $p \notin K$: any time $p \in \overline{K}\setminus K$ we will have that we cannot separate $p$ from $K$, almost by definition of closure.

Thanks to Hausdorffness $K$ is closed in $X$, as we saw above, and we can even extend Hausdorffness to points and disjojnt compact sets; even to compact sets and compact sets (when disjoint), beyond a point and a point.


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