Compactness of point and compact set product

[SrEstroncio]

I was reading Spivak's Calculus on Manifolds and in chapter 1, section 2, dealing with compactness of sets he mentions that it is "easy to see" that if $B \subset R^m$ and $x \in R^n$ then $\{x\}\times B \subset R^{n+M}$ is compact. While it is certainly plausible, I can't quite get how to handle set products when dealing with covers.
I was wonering if anyone could sketch a proof of this for me, I've been stuck on that page for days now.

note: this is not homework, just doing some self study.

 

[micromass]

Hi SrEstroncio!

Just use that

$$\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)$$

is a homeomorphism. Thus B is compact if and only if $\{x\}\times B$ is.

 

[SrEstroncio]

Hi SrEstroncio!

Just use that

$$\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)$$

is a homeomorphism. Thus B is compact if and only if $\{x\}\times B$ is.

While I don't doubt there's nothing wrong with your argument, I am not familiar with homeomorphisms and your proof seems a little out of my grasp right now. I am trying to prove it by means of covers, I suppose A is a cover of $\{x\}\times B$, and I want to prove there is a finite subcollection of sets in A such that said subcollection covers $\{x\}\times B$.

 

[micromass]

While I don't doubt there's nothing wrong with your argument, I am not familiar with homeomorphisms and your proof seems a little out of my grasp right now. I am trying to prove it by means of covers, I suppose A is a cover of $\{x\}\times B$, and I want to prove there is a finite subcollection of sets in A such that said subcollection covers $\{x\}\times B$.

OK, let me translate it for you then. We have the following continuous functions:

$$\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)$$

and

$$\psi:\{x\}\times B\rightarrow B:(x,b)\rightarrow b$$

These functions are each others inverse.

Now, take $\{G_i~\vert~i\in I\}$ an open cover of $\{x\}\times B$. Then

$$\{\varphi^{-1}(G_i)~\vert~i\in I\}$$

forms an open cover of B. Because B is compact, it has a finite subcover

$$\{\varphi^{-1}(G_i)~\vert~i\in F\}$$

Thus

$$\{\psi^{-1}(\varphi^{-1}(G_i))~\vert~i\in F\}$$

is a finite cover of $\{x\}\times B$. The proof follows since

$$\psi^{-1}(\varphi^{-1}(G_i))=G_i$$

Is that more clear?

 

[SrEstroncio]

I get it now, thanks.

Now, at the risk of seeming kind of stubborn, imagine you've just been given the definition of compact sets and you were immediately asked to prove this (which is the case with Spivak's book), how would you do it without constructing the functions $\phi$ and $\psi$, which is to say, how would you do it "by foot"?

sorry for the hassle, thanks in advance

 

[HallsofIvy]

Every point in {x} X B is of the form (x, b) where b is in B.

Let F:{x} X B-> B be defined by F(x, b)= b.

Further, it has the obvious inverse F^{-1}(b)= (x, b).

If {U} is an open cover for {x}X B then {F(U)} is an open cover for B.

Since B is compact, there is a finite subcover, {U_n}. Then F^{-1}({U_n}) is finite subcover of the original cover.

(Of course, you need to show all of those statements.)