(Proof Checking) Prove that if $C,C'$ are compact subsets of a hausdorff space $X$ then $C\cap C'$ is compact in $X$.

Question

Prove that if $C,C'$ are compact subsets of a hausdorff space $X$ then $C\cap C'$ is compact in $X$.

I am tempted to use the following argument. Let $U = \{U_i|i\in I\}$ be some open cover of $C\cap C'$ where $I$ is a possibly infinite indexing set. Also, let $V= \{ V_j | j\in J \}$ be an open cover for $C$ for some indexing set $J$. Now let $W$ be the sum of $U$ and $V$, that is, $W$ contains all open sets in $U$ and $V$. Then $W$ is an open cover for $C$. Because $C$ is compact we may find a finite subcover for $C$, call it $W_f$. Then we remove from $W_f$ all sets that were not originally in $U$. We are left with an open cover for $C\cap C'$ that is finite, and is a subcover of $U$. Therefore $C\cap C'$ is compact.

I feel I made a mistake somewhere because I did not use the Hausdorff condition. Can anyone point out the mistake and possible provide a hint to a proof that is correct? Thanks advance!

Answer 1

Since $C$, $C'$ are compact and $X$ is a hausdorff space, they are closed. Therefore $C\cap C'$ is closed. Now, closed subsets of compact sets are compact.

Answer 2

In Hausdorff spaces compact sets are closed and in compact spaces closed sets are compact.

$C$ and $C'$ are compact subsets of Hausdorff space $X$ hence are closed. Then $C$ equipped with the subspace topology is a compact space. $C'\cap C$ is closed in this compact space, hence is compact.

Answer 3

Here is an example of two compact subsets in a non-Hausdorff space whose intersections fails to be compact:

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Let $ℚ$ be the equipped with the topology where the only open sets in $ℚ$ are either $ℚ$ itself, or subsets of $ℤ$.

Clearly, $ℚ$ and any proper superset of $ℤ$ is compact, because any open cover of such a set must contain $ℚ$ itself.

But $ℤ/3 ∩ ℤ/2 = ℤ$ is not compact (being infinite, discrete).

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So, yeah, simply use that

• compact subsets of Hausdorff spaces are closed, and
• closed subsets of compact spaces are compact,

and that’s it.

Answer 4

For your proof: what guarantees that taking the finite subcovering of W you have remaining elements of $U$?

Second here is what I think is a proof:

1. Take a covering of $C' \cap C$ call it $U$.

2. Take $V=U \cup \{X-C\cap C'\}$ which is a open covering of $C$ (that is a consequence of Hausdorff because the fact that $C$ and $C'$ are compact implies closed and then their intersection is closed and then their complement is open).

3. For $C$ take the finite subcovering of $V$, $V_1,...V_n$, if $X-C\cap C'$ is in the subcovering remove it and you have a subcovering of $C' \cap C$