# intersections and compact sets

• Oct 26th 2009, 07:20 PM
mathgirlie
intersections and compact sets
Show that if K is compact and F is closed, then K intersect F is compact.

Alright this is what I have so far.

Suppose K is compact, which implies that K is closed and bounded. Since F is closed and K is closed, by theorem that means that K intersect F is closed as well. Since K intersect F is closed, that means that every cauchy sequence in K int. F has a limit that is an element of K int. F.

This is where I get stuck, I know that the definition of a compact set is:
A set K in R is compact if every sequence in K has a sub sequence that converges to a limit that is also in K.

I feel like that is what I concluded in my proof, so I could just say that means that K int. F is compact, however, it really isn't saying anything about sub sequences. Any help would be great!!!
• Oct 26th 2009, 08:57 PM
hatsoff
Quote:

Originally Posted by mathgirlie
Show that if K is compact and F is closed, then K intersect F is compact.

Alright this is what I have so far.

Suppose K is compact, which implies that K is closed and bounded. Since F is closed and K is closed, by theorem that means that K intersect F is closed as well. Since K intersect F is closed, that means that every cauchy sequence in K int. F has a limit that is an element of K int. F.

This is where I get stuck, I know that the definition of a compact set is:
A set K in R is compact if every sequence in K has a sub sequence that converges to a limit that is also in K.

I feel like that is what I concluded in my proof, so I could just say that means that K int. F is compact, however, it really isn't saying anything about sub sequences. Any help would be great!!!

If you're working in $\mathbb{R}^n$, then we can use the Bolzano-Weierstrass theorem which states that $S\subset\mathbb{R}^n$ is sequentially compact if and only if it is closed and bounded.

A proof for this theorem may be found here: Bolzano?Weierstrass theorem - Wikipedia, the free encyclopedia

Since $F,K$ are closed with $K$ bounded, then $F\cap K$ is closed and bounded, and thus by Bolzano-Weierstrass is sequentially compact. $\blacksquare$
• Oct 27th 2009, 01:28 AM
Enrique2
The result is true in general. This question appeared two days ago in other thread, Here is the argument.

If $F\subset K$, $F$ closed and $F=\cup_{i\in I}(V_i\cap F)$,
$V_i$ open in $K$, then $K=\cup_{i\in I}V_i\cup K\setminus F$. If $K$ is compact then there exists $J\subset I$ finite such that $K=\cup_{i\in J}V_i\cup K\setminus F$. Hence $F=\cup_{i\in J}V_i\cap F$, i. e. $F$ is also compact
• Oct 27th 2009, 04:53 AM
HallsofIvy
Any compact set is closed so the "intersection of K and F" is closed.

That is, $F\cap K$ is a closed subset of a compact set and so is compact. "Bounded" or, in fact, even being in a metric space, is not necessary.

Added: As Tonio pointed out to me in another thread, my cavalier statements about compact sets being closed and a closed subset of a compact set being compact are only true in a Hausdorff space.
• Oct 27th 2009, 07:34 AM
mathgirlie
thanks for the help everyone!