Let $K \subset R^{n} $ be a compact set and $p \in R^m$ then $ \{p \} \times K$ is compact in $R^{n +m}$

Question

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.

Answer 1

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 $K\times\{p\}$.

Answer 2

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.

Answer 3

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