Prove -- The product of two compact spaces is compact

Gold Member

Main Question or Discussion Point

I'm attempting to prove that the product of two compact topological spaces is compact. My attempt at a proof runs something like this:

Let ##Q## and ##R## be compact, and ##Q \times R = S##. From the product topology, any open set of ##S## has to have the form ##S_{AB} = Q_A \times R_B## (where ##Q_A## and ##R_B## are open in their respective topologies), so any open cover of ##S## can be written
$$\bigcup_{(i,j) \in I \times J} S_{ij} =\bigcup_{(i,j) \in I \times J} (Q_i \times R_j)$$
where ##I## and ##J## are possibly infinite.

Now we show that ##\bigcup_i Q_i## covers ##Q## and ##\bigcup_j R_j## covers ##R## if and only if ##\bigcup_{i,j} (Q_i \times R_j)## covers ##S##.

1) "Only if" direction: Let ##\bigcup_i Q_i## cover ##Q## and ##\bigcup_j R_j## cover ##R##. Then
$$\bigcup_i Q_i \times \bigcup_j R_j = \bigcup_{(i,j)} (Q_i \times R_j)$$
since the Cartesian product distributes across unions. I claim this set covers ##S##. Assume it doesn't. Then ##\exists (q,r) \in S## such that ##(q,r) \notin \bigcup_{(i,j)} (Q_i \times R_j)##; in other words, ##q \notin \bigcup_i Q_i ## or ##r \notin \bigcup_j R_j ##. But this can't happen because ##\bigcup_i Q_i## and ##\bigcup_j R_j ## cover ##Q## and ##R## respectively.
Therefore, if ##\bigcup_i Q_i## covers ##Q## and ##\bigcup_j R_j## covers ##R##, then ##\bigcup_{(i,j)} (Q_i \times R_j)## covers ##S##.
2) "If" direction (I prove the contrapositive): Suppose ##\bigcup_j R_j## does not cover ##R##. Then ##\exists r \in R## such that ##r \notin \bigcup_j R_j##. This means that ##\exists (q,r) \in S## such that ##(q,r) \notin \bigcup_i Q_i \times \bigcup_j R_j = \bigcup_{(i,j)} (Q_i \times R_j)##. So ##\bigcup_i Q_i \times \bigcup_j R_j = \bigcup_{(i,j)} (Q_i \times R_j)## doesn't cover ##S##.

Since ##Q## and ##R## are compact, there exist finite subcovers of ##\bigcup_i Q_i## and ##\bigcup_j R_j## (call them ##\bigcup_A Q_A## and ##\bigcup_B R_B##). We've just proven that the open covers of ##S## are precisely the Cartesian product of covers of ##Q## and ##R##. So for any cover of ##S## (given by ##\bigcup_{(i,j)} (Q_i \times R_j)##), there exists a finite subcover given by ##\bigcup_{(A,B)} (Q_A \times R_B)##. Therefore, ##S## is compact.

Is this a valid proof?

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Erland
From the product topology, any open set of ##S## has to have the form ##S_{AB} = Q_A \times R_B## (where ##Q_A## and ##R_B## are open in their respective topologies)
No, an open set in ##S## is a union of sets of this kind. This union can be taken from an arbitrary large family of sets of this kind.

PeroK
Homework Helper
Gold Member
Let ##Q## and ##R## be compact, and ##Q \times R = S##. From the product topology, any open set of ##S## has to have the form ##S_{AB} = Q_A \times R_B## (where ##Q_A## and ##R_B## are open in their respective topologies), so any open cover of ##S## can be written
$$\bigcup_{(i,j) \in I \times J} S_{ij} =\bigcup_{(i,j) \in I \times J} (Q_i \times R_j)$$
where ##I## and ##J## are possibly infinite.
This is not true. For example, if you have an open cover, you could add a further set ##A \times B## where these sets do not appear in your open cover. This is still an open cover, but it does not include ##A \times R_1## etc.

Gold Member
Ok, I think what's tripping me up is that a subcover is a subset of the collection of open sets that makes up the cover, rather than a subset of the union of those sets. I'll think a little harder and get back to you.

Svein
This is the Tychonoff theorem. Google it (the proof is not trivial).

S.G. Janssens
This is the Tychonoff theorem.
For two (or finitely many) spaces Tikhonov's theorem (equivalent to AC) is not necessary and, indeed, overkill in my opinion. I would recommend that the OP tries to continue his attempts to give his own proof. It is a good exercise.

Cruz Martinez
Gold Member

If ##Q\times R = S##, then as @Erland said, any open set in ##S## has the form ##S_{ab} = \bigcup_{a,b}(Q_{a}\times R_{b})##, where ##a\in A## and ##b\in B##. Then an open cover of ##S## is a collection of sets ##S_i## such that
$$S \subseteq \bigcup_i S_i=\bigcup_i \bigcup_{a,b}(Q_{a}\times R_{b})$$
where ##i\in I##. I think I can push the indexes in the separate unions together like so:
$$\bigcup_i \bigcup_{a,b}(Q_{a}\times R_{b}) = \bigcup_{(a,b,i)} (Q_{(a,i)} \times R_{(b,i)})$$
where ##(a,b,i)\in A\times B \times I##, or something along these lines.
From my first post, I know that in order for ##\bigcup_i S_i## to cover ##S##, ##\bigcup_{(a,i)}Q_{(a,i)}## has to cover ##Q## and ##\bigcup_{(b,i)}R_{(b,i)}## has to cover ##R##. Since ##Q## and ##R## are compact, a finite subset of ##(a,i) \in A\times I## covers ##Q## (call it ##(\alpha, j) \in A' \times J##) and likewise for ##R## (call it ##(\beta,j) \in B' \times J##). In order for ##A' \times J## to be finite, both ##A'## and ##J## have to be finite. This means that there is a subcover of
$$\bigcup_{(a,b,i)} (Q_{(a,i)} \times R_{(b,i)})$$
indexed by ##J \subseteq I## such that ##J## is finite.

What it boils down to is that I'm not quite confident about how to handle the union of unions while preserving the granularity of the individual sets in the cover of ##S##. Is the idea of the proof above even on the right track?