I am having trouble understanding the answers here.

I am trying to prove that a compact subset of a Hausdorff space is closed.

Following the proof is difficult, perhaps because Brian reused letters for different things(although I get they are arbitrary, I can't follow it.) The second answer uses nets and filters, which I don't know of, so would like to avoid for now.

Now trying to follow Brians proof, I get something like the following:

Let $S$ be our Hausdorff space, and $C\subset S$ be a compact subspace of $S$. (easy to remember, think $S$ for space, $C$ for compact)

To show that this subspace is closed, we can show the complement of the subspace is open. I.e. $S/C$ is open. Now following the proof of Brian, we take some element $x\in S/C$ and since $S$ is Hausdorff(which means $C$ is Hausdorff), we have for each $y\in C$ that there are disjoint open sets $U_y,V_y$ such that $x\in U_y$ and $y\in V_y$ now here is a little confusing, either he has reused $x$, or for some reason $x\in C$, which shouldn't be possible, since $x\in S/C$. Let's assume this is reusing the letter. Then the collection of all of these disjoint open sets, lets call it $\{\alpha_i,i\in C\}$ is an open cover of $C$, and since $C$ is compact, it has a finite subcover, $\{\alpha_i|u\in F\}$ where $F$ is a finite subset of $C$. Let $$U=\bigcap_{x\in F}V_y=\emptyset$$ Right? So then $U$ is open, since all empty sets are open... Now the proof doesn't even seem to follow.

I have no idea what's going on!

  • $\begingroup$ No, I didn’t reuse letters for different things. For each $y\in C$ there is a pair of open sets, $U_y$ and $V_y$, such that $y\in V_y$, $x\in U_y$ and $U_y\cap V_y=\varnothing$. The nbhd $U_y$ of $x$ is labelled with a point of $C$ because it depends on that point: for each $y\in C$ you may get a different nbhd $U_y$ of $x$. Eventually you’re going to intersect finitely many of those nbhds of $x$ to get a nbhd of $x$ disjoint from $C$. $\endgroup$ – Brian M. Scott Jun 18 '15 at 4:54
  • $\begingroup$ @BrianM.Scott So for each point in our subspace, we have two open sets for it, one which it is within, and another that is around some other point, such that they don't intersect, sure, I am not really sure why we are giving them subscripts of $y$ though. So this is just the Hausdorff condition so far. Then we say that the subscript is because $x\in U_y$ is a neighborhood that is centered on a point in $C$(which to me says that $x\in C$). For each point in $C$ we have a different neighborhood around (?)$x$(?). Eventually $x$ is outside of $C$, hmmm. How can I visualize this? I think perhaps... $\endgroup$ – Hausdorff Jun 18 '15 at 5:19
  • $\begingroup$ ... I am thinking about this all wrong, maybe because I am still in a metric space mindset. I should be thinking about this as elements of $\tau$. $\endgroup$ – Hausdorff Jun 18 '15 at 5:19
  • $\begingroup$ Give me a few minutes, and I’ll make a rough sketch. $\endgroup$ – Brian M. Scott Jun 18 '15 at 5:21
  • $\begingroup$ @BrianM.Scott Thank you so much $\endgroup$ – Hausdorff Jun 18 '15 at 5:21

Here’s a sketch that may help a little.

enter image description here

The large black oval represents the compact set $C$, $x$ is the point outside $C$, and $y$ and $z$ are two points of $C$ that I’ll use to illustrate the choice of open nbhds. The space is Hausdorff, so $x$ and $y$ have disjoint open nbhds; in the picture they’re the smaller black ovals, $V_y$ being the nbhd of $y$, and $U_y$ the corresponding nbhd of $x$. Similarly, $x$ and $z$ have disjoint open nbhds; I’ve drawn them in blue, $V_z$ being the nbhd of $z$, and $U_z$ being the corresponding nbhd of $x$.

The idea is to do this for each point of $C$. Thus, corresponding to each point $p\in C$ we get two open sets, which I’ve called $U_p$ and $V_p$: $V_p$ is an open nbhd of $p$, $U_p$ is an open nbhd of $x$, and the sets $U_p$ and $V_p$ are disjoint. I can do this because the space is Hausdorff.

The sets $V_p$ with $p\in C$ form an open cover $C$: each of them is open, and for any $p\in C$ I know that $p\in V_p$, so they cover $C$. $C$ is compact, so only finitely many of them are needed in order to cover $C$. This means that there is a finite subset $\{p_1,\ldots,p_n\}$ of $C$ such that $$C\subseteq V_{p_1}\cup\ldots\cup V_{p_n}\;.$$

Now I look at the corresponding open nbhds of $x$, $U_{p_1},\ldots,U_{p_n}$. Specifically, I let

$$U=U_{p_1}\cap\ldots\cap U_{p_n}\;;$$

since $U$ is an intersection of only finitely many open sets, $U$ is also open. I claim that $U\cap C=\varnothing$. Suppose, on the contrary, that there is a point $y\in U\cap C$. We saw above that

$$C\subseteq V_{p_1}\cup\ldots\cup V_{p_n}\;,$$

so there is a $k\in\{1,\ldots,n\}$ such that $y\in V_{p_k}$. But $U\subseteq U_{p_k}$, and $y\in U$, so $y\in U_{p_k}$, which is impossible: we chose $U_{p_k}$ and $V_{p_k}$ to be disjoint (like $U_y$ and $V_y$ in the sketch, or $U_z$ and $V_z$). Thus, no such point $y$ exists, and $U$ is therefore an open nbhd of $x$ disjoint from $C$.

Since we can do this for any point $x$ not in $C$, $C$ must be closed.

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  • $\begingroup$ I successfully followed it, after a few more readings it will feel natural. Thank you so much! $\endgroup$ – Hausdorff Jun 18 '15 at 5:58
  • $\begingroup$ @Hausdorff: Excellent! You’re very welcome. $\endgroup$ – Brian M. Scott Jun 18 '15 at 5:58

Alternative write up to Brian's(as interpreted from Brian's write up):

Let $A$ be a compact subset of the Hausdorff space $X$. Take $x\in A$ and $y\in X/A$. Then $x\ne y$ and by the definition of Hausdorff spaces, there are two disjoint open sets $G_{x,y}$ and $H_{x,y}$ of $X$ such that $x\in G_{x,y}$ and $y\in H_{x,y}$ with $G_{x,y}\cap H_{x,y}=\emptyset$

Since $\{G_{x,y}|x\in A\}$ for each $y\in X/A$ is an open cover of $A$, i.e. we have $A\subset \cup_{x\in A} G_{x,y}$. Since we have compactness for $A$, there is a finite subcover of $A$ consisting of the elements $x_1,\cdots,x_n$ of $A$, such that $$A\subset \cup_{i=1}^n G_{x_i,y}$$

Also for each $i=1,\cdots,n$ there is an open set $H_{x_i,y}$ such that $y\in H_{x_i,y}$. Let $H_y = \cap_{i=1}^n H_{x_i,y}$ and note that $y\in H_y$ and $H_y$ is an open set of $X$. Furthermore, $H_y\cap(\cup_{i=1}^n G_{x_i,y})=\emptyset$

Thus $H_y\subset X/A$ and therefore $X/A = \cup_{y\in {X/A}} H_y$ is an open set and so $A$ is closed.

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  • $\begingroup$ The last line is the magic line. $\endgroup$ – rainman Sep 7 '19 at 6:01

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