- #1

TeethWhitener

Science Advisor

Gold Member

- 1,987

- 1,421

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?