Compactness implies closedness

Question

I was reading the post:

How to prove that a compact set in a Hausdorff topological space is closed?

Quoting the accepted answer:

Fix $x\in\mathbb{X}\setminus K$. Since $\mathbb{X}$ is Hausdorff, for each $y\in K$ there are disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. $\{V_y:y\in K\}$ is an open cover of $K$, so it has a finite subcover, say $\{V_y:y\in F\}$, where $F$ is some finite subset of $K$. Let $$U=\bigcap_{x\in F}U_x\;;$$ clearly $U$ is an open nbhd of $x$ disjoint from $K$. Since $x$ was an arbitrary point of $\mathbb{X}\setminus K$, $K$ must be closed.

Why is $\{V_y \mid y \in K\}$ an open cover of $K$? I mean, to use compactness in the argument, we need that the set $V_y \subseteq K$, otherwise I don't see how it can be part of an open cover in the subspace $K$.

Where does $A$ come from? – Angina Seng Mar 23 '18 at 19:03 — Angina Seng, Mar 23 '18 at 19:03
Look at the $U_x\cap K$ if you insist. – Angina Seng Mar 23 '18 at 19:06 — Angina Seng, Mar 23 '18 at 19:06

score 2 · Answer 1 · answered Mar 23 '18 at 19:07

2

Replace $V_y$ with $V_y\cap K$ and all will work:

each $V_y\cap K$ is a subset of $K$;
each $V_k\cap K$ is open in $K$;
$\left\{V_y\cap K\,\middle|\,y\in K\right\}$ is an open cover of $K$.

answered Mar 23 '18 at 19:07

José Carlos Santos

427,504

Yes, that was what I was thinking too. Does this mean that the proof given is wrong? – Mar 23 '18 at 19:08
@Math_QED Not at all. It uses a diiferent way of defining open cover when $K$ is a subset of a topological space $X$: an open cover is a set $\mathcal A$ of open subsets of $X$ such that $K\subset\bigcup_{A\in\mathcal A}A$. But this definition of open cover ledas to the same concept of compactness. – José Carlos Santos Mar 23 '18 at 19:11

score 2 · Accepted Answer · answered Mar 23 '18 at 19:13

2

There's two different notions of compactness you're dealing with: a compact subset of a topological space, and a compact space. But indeed when there's overlap, there is no distinction. That is,

Let $X$ be a topological space, and let $K\subset X$ be given. Then $K$ is compact (in the subspace topology) if and only if whenever $\{U_\alpha\}$ is a collection of open subsets of $X$ such that $K\subset\cup_\alpha K_\alpha$, there is a finite subset $\{U_1,\ldots,U_n\}$ of $\{U_\alpha\}$ such that $K\subset U_1\cup\cdots\cup U_n$.

The proof of this is quite straightforward, but a useful exercise if you're new to topology.

answered Mar 23 '18 at 19:13

Aweygan

23,232

Thanks a lot. This explains everything! – Mar 23 '18 at 19:16
You're welcome. Glad to help! – Aweygan Mar 23 '18 at 19:17
The proof indeed was quite straightforward. Thanks. Now, I can sleep without worrying :) – Mar 23 '18 at 19:52

score 0 · Answer 3 · answered Mar 23 '18 at 19:09

0

\begin{align} \{y\} &\subseteq V_y \\ K = \bigcup_{y \in K} y &\subseteq \bigcup_{y \in K} V_y \end{align}

answered Mar 23 '18 at 19:09

GNUSupporter 8964民主女神地下教會

17,627

I don't think you understand the question. $V_y$ must be open in $K$, which means that they have to be a subset of $A$ in the first place – Mar 23 '18 at 19:10
@Math_QED What is $A$? It never appeared in the question. Since each $V_y$ is open, the arbitrary union is also open. – GNUSupporter 8964民主女神地下教會 Mar 23 '18 at 19:14
@Math_QED Your question is "Why $\left{V_y\cap K,\middle|,y\in K\right}$ is an open cover of $K$", so I showed how $\bigcup_{y \in K} V_y$ contains $K$, while assuming that you know that it's open. In the definition for a Hausdorff space, $V_y$ is open in $\Bbb X$. – GNUSupporter 8964民主女神地下教會 Mar 23 '18 at 19:45
1

The accepted answer contains the cause of confusion. I upvoted your post btw. Thanks for the help. – Mar 23 '18 at 19:48

Compactness implies closedness

3 Answers3