# Prove that there exists an open set $V$ with compact closure such that $K⊆V⊆\overline V⊆ U$

Suppose that $$U$$ is open in a locally compact Hausdorff space $$X$$ and $$K\subseteq U$$ is a compact set. Then there exists an open set $$V$$ with compact closure such that $$K\subseteq V\subseteq\overline V\subseteq U$$. Theorem 2.5

Suppose $$X$$ is a Hausdorff space, $$K\subset X$$ compact and $$p\in K^c.$$ Then there exists $$U,V\in\tau$$ such that $$p\in U,K\subset V$$ and $$V\cap U=\varnothing$$

Theorem 2.6

Let $$\{K_\alpha\}$$ be a collection of compact sets of a Hausdorff space. If $$\displaystyle\bigcap_\alpha K_\alpha=\emptyset,$$ then there exists $$\alpha_1,\dots,\alpha_n$$ such that $$\displaystyle\bigcap_{k=1}^n K_{\alpha_k}=\emptyset.$$

I am having confusions with the proof, in the first paragraph we have $$K\subseteq\ U_x\subseteq\overline U_x,\forall x\in K$$, I think.

But $$K$$ is compact, so $$K\subseteq\bigcup_{i=1}^n U_{x_i}\subseteq\bigcup_{i=1}^n\overline U_{x_i}.$$This last set is compact. If $$G=\bigcup_{i=1}^n U_{x_i},$$ then $$K$$ lies in an open set with compact closure?

How will I know that $$\overline G=\overline{\bigcup_{i=1}^n U_{x_i}}$$ is compact?

I only know that $$\bigcup_{i=1}^n\overline U_{x_i}$$ is compact.

2nd qstn. where it says $$K\subset W_p$$ and $$p\not\in\overline W_p$$ is that because $$K\subset W_p\subset\overline W_p$$ and thus $$p$$ can't be in $$\overline W_p$$ ?

3rd qstn. This is a collection of compacts $$\{C\cap\overline G\cap\overline W_p\}$$ and is empty. All are compacts because $$C$$ and $$\overline W_p$$ are closed thus intersected with compact $$\overline G$$ will be compact, correct?

And is also empty because, suppose it's not. Then we would get a contradiction with $$p\in C$$ and $$p\not\in\overline W_p$$, right?

4th qstn. This contention $$\overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}\subset U$$ (not explicitly mentioned in the proof) is because if it weren't truth i.e. $$p\in\overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}$$ and $$p\not\in U$$, then $$p\in U^c=C!$$ with $$C\cap \overline G\cap\overline W_{p_1}\cap...\cap\overline W_{p_n}=\emptyset$$

Thank you.

• – user486983
Feb 4, 2020 at 5:33

First question: note that $$\bigcup_{i=1}^n \overline{U}_{x_i}$$ is a compact, and hence closed set (note: Hausdorff) containing $$G$$. The closure of $$G$$ is therefore contained in this set. It is a closed subset of a compact set, and hence compact.

Second question: According to the statement of Theorem 2.5, there exist disjoint open $$U_p$$ and $$W_p$$ such that $$p \in U_p$$ and $$K \subseteq W_p$$. Since $$p \in U_p \subseteq X \setminus W_p$$, we have that $$p$$ is in the interior of $$X \setminus W_p$$, which is to say, $$p \notin \overline{W}_p$$.

Third question: I don't really follow your reasoning here. The reason why I saw that the intersection was empty was because it was contained in $$C = X \setminus U$$, and every $$p \in C$$ fails to be in $$\overline{W}_p$$. That is, every point in $$C$$ is excluded from the intersection, hence no points can be in the intersection.

Fourth question: It's not really a question, but I agree!

• Thank you Theo. My third question wasn't well redacted.. I edit again
– user486983
Jun 2, 2019 at 4:05
• In the second question did you mean $K\subseteq V$ instead of $W_p\subseteq V$? $\$ Btw did my argument had a flaw?
– user486983
Jun 2, 2019 at 4:27
• @Isa Yes, I made a bit of an error. I've now fixed it. I'm not 100% sure what your argument is saying, but I think it doesn't work. Just knowing that compact $K$ lies in a larger set $W_p$, and that $p$ is separated from $K$, does not imply $p \notin W_p$. Jun 2, 2019 at 15:28
• I see. $p$ could be in the larger set $\overline W_p$ that contains $K$ and $p\not\in K$
– user486983
Jun 2, 2019 at 16:57
• Why did you mention $C = X \setminus U$ first?, my argument skipt this and directly goes (by contradiction) to $p\in C$ and $p\not\in \overline W_p$, not sure if it's right.
– user486983
Jun 2, 2019 at 17:29

The closure of that union is in that finite union of the closures which is known compact and a closed subset of a compact set is compact .