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I must prove that

$M\times N$ compact $\implies$ $M$ compact and $N$ compact

using the definition that, if a metric space $M$ is compact, then every cover has an open finite sub cover.

$$M=\cup A_{\lambda}\implies M = A_{\lambda1}\cup\cdots\cup A_{\lambda_n}$$ where $A_\lambda$ is an open cover.

So, I must prove that if I have an open cover in $M\times N$, with a finite open subcover, I must have an open cover in $M$ and $N$ with finite open subcovers. However, I don't have any idea on how to prove that.

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    $\begingroup$ Do you know that continuous images of compact sets are compact? $\endgroup$ – Hmm. Jun 25 '16 at 14:33
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    $\begingroup$ Do you consider the empty space to be compact? $\endgroup$ – GEdgar Jun 25 '16 at 14:35
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    $\begingroup$ "So, I must prove that if I have an open cover in $M\times N$..." No, that's not what you need to prove! If you want to prove that $M$ is compact you must prove that if you have an open cover of $M$ then it has a finite subcover. (Hint: If $(A_\lambda)$ is an open cover of $M$ then ... is an open cover of $M\times N$.) $\endgroup$ – David C. Ullrich Jun 25 '16 at 14:37
  • $\begingroup$ @Hmm. yes i do. $\endgroup$ – Guerlando OCs Jun 25 '16 at 14:38
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    $\begingroup$ To expand on the hint given by @DavidC.Ullrich, remember that the whole space is open by the definition of topology. $\endgroup$ – parsiad Jun 25 '16 at 14:38
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So, I must prove that if I have an open cover in $M\times N$, with a finite open subcover, I must have an open cover in $M$ and $N$ with finite open subcovers. However, I don't have any idea on how to prove that.

No, that is not what you must prove.

What you must prove is that assuming that $M\times N$ is compact, if you have an open cover of $M$, then it has a finite subcover. (And then similarly for $N$, but that goes the same way by symmetry).

Assuming that $M\times N$ is compact is not the same as assuming that you have a particular open cover of $M\times N$. It is, by definition, the same as knowing that you are allowed to construct an open cover of $M\times N$ in any way you like, and then there's an oracle that will pick out a finite subset of that which covers all of $M\times N$. You will need the freedom to choose a cover of $M\times N$ in a deliberate way in order to make your proof go through.

In other words instead of assuming that somebody gives you a cover of $M\times N$, what you're given is an open cover of $M$. And once you see that, you can decide on your own which cover (or covers) of $M\times N$ to apply the definition of compactness to.

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You're confused about the definitions here, and what is given vs. what is to be proved.

You are assuming that $M \times N$ is compact. This means (by definition) that given any open cover $\mathscr{U} = \{U_\lambda: \lambda \in \Lambda\}$ of $M \times N$, we can find a finite subover of $\mathscr{U}$.

From this we want to show that $M$ is compact and $N$ is compact.

So we have to show that $M$ is compact. So take any open cover $\{O_i : i \in I \}$ of $M$. Then for every $i$ define $U_i = O_i \times N$. As $U_i$ is open in $M$, $U_i$ is open in $M \times N$, as a product of two open sets.

Also, $\{U_i : i \in I \}$ is a cover: if $(x,y) \in M \times N$, then $x \in O_j$ for some $j \in I$ (as the $O_i$ cover $M$), and then $(x,y) \in U_j$ for that $j$.

So we have an open cover of $M \times N$, and we can apply the assumption that $M \times N$ is compact. So we can find finitely many $i_1,\ldots,i_N \in I$ such that $M \times N$ is covered by the $U_{i_1},\ldots,U_{i_N}$. But then, if $x \in M$, pick any $p \in N$ (we do need that $N$ is non-empty!) and note that $(x,p) \in M \times N$, so is in $U_{i_j} = O_{i_j} \times N$ for some $j \in \{1,\ldots,N\}$. This means (definition of Cartesian product, essentially!) that $x \in O_{i_j}$ for that $j$ and so $\{O_{i_1},\ldots,O_{i_N}\}$ form a finite subcover of the $O_i$.

As the starting open cover was arbitrary, $M$ is compact.

Now you also have to show that $N$ is compact, in a similar way.

Of course, $M = \pi_1[M \times N]$ where $\pi_1$ is the continuous projection from $M \times N$ onto $M$. And continuous maps preserve compactness, and this is another way to see this. The above proof is really a special case for the proof that: $X$ compact, $f: X \rightarrow Y$ continuous, then $f[X]$ is compact too, for the projection. Similarly $N = \pi_2[M \times N]$ for the projection onto the second coordinate.

That both spaces must be non-empty is clear, otherwise $\emptyset \times N = \emptyset$ is compact but we cannot prove anything about $N$, e.g.

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