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I'm trying to prove it using the following definition:

X is called compact if each of its open covers has a finite subcover.

I've tried the following

Assume that {p} is compact (Here is the problem, I don't know how to explain it formally.)

So {p} has a finite subcover $\bigcup_{i=1}^n Z_i$ and let $\bigcup_{i=1}^n V_i$ a finite subcover of K.

Then

$$\bigcup_{i=1}^n Z_i \times \bigcup_{i=1}^n V_i = \bigcup_{i=1}^n U_i \times V_i $$

It's a finite subcover of $ \{p\} \times K$.

Also I'd like to know, Can I choose the finite number of subcovers of compact sets?

Because I chose the same number of subcover for {p} and K, and I don't know if it's right.

Thank you for your attention.

Kutz
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  • I may be wrong but it seems to me that you are using the definition of compactness incorrectly. You have to start with a (not necessarily finite) cover of ${p} \times K$ and show that you can find a finite subcover of this cover. To get a better understanding of the definition you might want to show that ${p}$ is compact (in fact, this is true for any finite set). Another helpful thing would be to show that images of compact sets under continuous maps are compact, too. – Matthias Klupsch Mar 16 '18 at 15:30
  • Any singleton $;{p}\subset\Bbb R^m;$ is compact as any open cover of $;{p};$ trivially has a finite subcover (take any single set in the cover that contains $;p;$ ...and that's all!), but the beginning of your "proof" is wrong: you must prove that any cover of $;{p}\times K;$ in $;\Bbb R^{m+n};$ has a finite subcover...and not begin with some finite subcover that nobody knows where it appeared from... – DonAntonio Mar 16 '18 at 15:32

3 Answers3

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Singletons $\{p\}$ are compact because if $\bigcup_{i\in I}U_i\ni p$ is a cover by open sets, then there is some $i_0i\in I$ such that $U_{i_0}\ni p$. then $\bigcup_{i\in\{i_0\}}U_i\ni p$ is a finite sub-cover; it has only one open set.

Assume that $f:\mathbb{R}^n\to \mathbb{R}^n\times\mathbb{R}^m$ is the embedding $f(x)=(x,p)$, which is continuous. Let $$\bigcup_{i\in I}U_i\supset K\times \{p\}$$ be an open cover.

Then $\bigcup_{i\in I}f^{-1}(U_i)\supset K$ is an open cover of $K$. Therefore there is a finite subset $J\subset I$ such that $\bigcup_{i\in J}f^{-1}(U_i)\supset K$. Hence $\bigcup_{i\in J}U_i\supset K\times \{p\}$ is a finite cover of $K\times\{p\}$.

SphericalTriangle
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  • It seems reasonably easy to show $;f;$ is open, yet I think one needs to show (and at least remark) that $;f;$ is continuous, which is what's needed in order to have that the inverse image of an open set is open... – DonAntonio Mar 16 '18 at 15:36
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Let $K$ be an arbitrary subset of $\mathbb R^n$ and $p$ a point in $\mathbb R^m$. The map $y \mapsto (p, y)$ is a homeomorphism from $K$ onto $\{p\} \times K$. Therefore $K$ is compact if and only if $\{p\} \times K$ is compact.

fred goodman
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Since both $K$ and $\{p\}$ are compact, $K \times \{p\}$ is compact by Tychonoff's theorem.

David Lui
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