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Let X be a topological space and A, B $ \subset $ X compact.

a) Show that A $ \cup $ B is compact.

b) Show that if X is Hausdorff, then A $\cap$ B is compact.

c) Show that the hypothesis that X is Hausdorff cannot be omitted in the previous item.


My attempt

a) Let X be a topological space and A, B $\cup$ where A and B are compact. We will claim A $\cup$ B is compact.

Let $\{O_{\alpha}\}_{\alpha \in I}$ be an open for A $\cup$ B, so A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\alpha}$.

But A $\subset$ A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\alpha}$.

So since A is compact, A $\subset$ $\bigcup_{i=1}^n O_{\alpha_i}$ wherer $\alpha_i \in I$

Likewise,

B $\subset$ A $\cup$ B $\subset$ $\bigcup_{\alpha \in I} O_{\beta}$.

So since B is compact, B $\subset$ $\bigcup_{j=1}^m O_{\beta_j}$, wherer $\beta_i \in I$.

therefore, A $\cup$ B $ \subset \bigcup_{\alpha \in I}^n \cup \bigcup_{\beta \in I}^m$, ando so, A $\cup$ B is compact.

b) Since X is $ T_2 $ and A, B $ \subset $ X compact, we have that A and B are closed.

So, let $ U_{\alpha} $ be an open cover of A $ \cap $ B

So, we have A $ \cap $ B $ \subset \bigcup_{\alpha} U_{\alpha} $

So, B $ \subset \bigcup_{\alpha} U_{\alpha} \cup$ (X\A)

Since A is closed (X∖A) is open

Therefore,

$ U_{\alpha} \cup $ X∖A is an open cover of A.

Now we have that B is compact, so it follows by definition that a finite subcover, {$U_{\alpha_1}, U_{\alpha_2},…, U_{\alpha_n }$, X∖A} of A exists.

Therefore,

A $\cap$ B $\subset \{U_{\alpha_1}, U_{\alpha_2},…, U_{\alpha_n}\}$ and therefore A $\cap B$ is compact.

c) If it is not $ T_2 $ we cannot assume A, B closed and in this case the demonstration does not take place.

Thanks in advance for any comments.

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    $\begingroup$ I think for c) you are expected to give an example non Hausdorff space where this fails. $\endgroup$
    – copper.hat
    Dec 7 '20 at 17:22
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    $\begingroup$ For point (b): $A,B \subset X$ are compact in a $T_2$ space so they are closed. This means that also $A \cap B$ is closed. In conclusion, since $A \cap B \subset A$ (or in $B$, it's the same) which is compact, you know that also $A \cap B$ is compact. $\endgroup$
    – Vajra
    Dec 7 '20 at 17:22
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    $\begingroup$ Here, a concrete counterexample can be found. $\endgroup$ Dec 7 '20 at 17:22
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Your arguments for (a) and (b) are basically correct, but in places you have badly misused notation. For example, in (b) you write:

So, let $U_\alpha$ be an open cover of $A\cap B$.

So, we have $A\cap B=\bigcup_\alpha U_\alpha$.

If $U_\alpha$ is an open cover of $A\cap B$, as you say in the first sentence, it is a family of open sets. In the second sentence, however, you make it clear that $U_\alpha$ is a single member of some indexed family of open sets. Thus, your first sentence should have read something like this:

So, let $\{U_\alpha:\alpha\in I\}$ be an open cover of $A\cap B$.

Later you write:

$U_\alpha\cup X\setminus A$ is an open cover of $A$

This says that a single set $U_\alpha\cup X\setminus A$ covers $A$. There are two problems here. First, the expression is ambiguous: should it be read $(U_\alpha\cup X)\setminus A$, or should it be read $U_\alpha\cup(X\setminus A)$? Use parentheses when there’s any real possibility of ambiguity. The real problem, though, is that you actually meant something completely different: you meant that $\{U_\alpha:\alpha\in I\}\cup\{X\setminus A\}$ is an open cover of $A$.

By the way, you can avoid indexing altogether. For (a), for instance, let $\mathscr{U}$ be an open cover of $A\cup B$. $A$ and $B$ are compact, and $\mathscr{U}$ is clearly an open cover of each of them, so there are a finite $\mathscr{U}_A\subseteq\mathscr{U}$ that covers $A$ and a finite $\mathscr{U}_B\subseteq\mathscr{U}$ that covers $B$. But then $\mathscr{U}_A\cup\mathscr{U}_B$ is a finite subset of $\mathscr{U}$ that covers $A\cup B$, so $A\cup B$ is compact. Similarly, in (b) you can start with an open cover $\mathscr{U}$ of $A\cap B$. Then $\mathscr{U}\cup\{X\setminus A\}$ is an open cover of the compact set $B$, there is a finite $\mathscr{U}_B\subseteq\mathscr{U}$ such that $\mathscr{U}_B\cup\{X\setminus A\}$ covers $B$. $(A\cap B)\cap(X\setminus A)=\varnothing$, so $\mathscr{U}_B$ covers $A\cap B$, which is therefore compact.

Your answer for (c) is insufficient: you’ve shown that if $X$ is not Hausdorff, the argument that you gave for (b) cannot be carried out, but that does not rule out the possibility that there is some other argument that does not require $X$ to be Hausdorff. To answer (c), you must come up with an actual counterexample. One of the nicest is a convergent sequence with two limits:

Example. Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are any two points not in $\Bbb N$. Points of $\Bbb N$ are isolated. For each finite $F\subseteq\Bbb N$ let $$B(p,F)=\{p\}\cup(\Bbb N\setminus F)$$ and $$B(q,F)=\{q\}\cup(\Bbb N\setminus F)\,;$$ the sets $B(p,F)$ form a local open nbhd base at $p$, and the sets $B(q,F)$ form a local open nbhd base at $q$. Thus, a set $U\subseteq X$ is open if and only if either $U\subseteq\Bbb N$, or $X\setminus U$ is finite. $X$ is a $T_1$ space and is almost Hausdorff: the only two points that do not have disjoint open nbhds are $p$ and $q$.

Now let $A=X\setminus\{q\}$ and $B=X\setminus\{p\}$; I leave the easy verification that $A$ and $B$ are compact to you. $A\cap B=\Bbb N$, and clear $\big\{\{n\}:n\in\Bbb N\big\}$ is an open cover of $A\cap B$ with no finite subcover.

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