Compactness of point and compact set product

  • #1
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 [itex] B \subset R^m [/itex] and [itex] x \in R^n [/itex] then [itex]\{x\}\times B \subset R^{n+M}[/itex] 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.
 

Answers and Replies

  • #2
22,129
3,297
Hi SrEstroncio! :smile:

Just use that

[tex]\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)[/tex]

is a homeomorphism. Thus B is compact if and only if [itex]\{x\}\times B[/itex] is.
 
  • #3
Hi SrEstroncio! :smile:

Just use that

[tex]\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)[/tex]

is a homeomorphism. Thus B is compact if and only if [itex]\{x\}\times B[/itex] 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 [itex] \{x\}\times B [/itex], and I want to prove there is a finite subcollection of sets in A such that said subcollection covers [itex] \{x\}\times B [/itex].
 
Last edited:
  • #4
22,129
3,297
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 [itex] \{x\}\times B [/itex], and I want to prove there is a finite subcollection of sets in A such that said subcollection covers [itex] \{x\}\times B [/itex].

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

[tex]\varphi:B\rightarrow \{x\}\times B:b\rightarrow (x,b)[/tex]

and

[tex]\psi:\{x\}\times B\rightarrow B:(x,b)\rightarrow b[/tex]

These functions are eachothers inverse.

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

[tex]\{\varphi^{-1}(G_i)~\vert~i\in I\}[/tex]

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

[tex]\{\varphi^{-1}(G_i)~\vert~i\in F\}[/tex]

Thus

[tex]\{\psi^{-1}(\varphi^{-1}(G_i))~\vert~i\in F\}[/tex]

is a finite cover of [itex]\{x\}\times B[/itex]. The proof follows since

[tex]\psi^{-1}(\varphi^{-1}(G_i))=G_i[/tex]

Is that more clear?
 
  • #5
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 [itex] \phi [/itex] and [itex] \psi [/itex], which is to say, how would you do it "by foot"?

sorry for the hassle, thanks in advance
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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.)
 

Related Threads on Compactness of point and compact set product

  • Last Post
Replies
1
Views
5K
Replies
8
Views
6K
  • Last Post
Replies
10
Views
3K
Replies
8
Views
8K
Replies
6
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
14
Views
6K
Replies
4
Views
15K
Top