# the intersection of a collection of compact sets is compact

• Feb 28th 2010, 01:43 PM
inthequestofproofs
the intersection of a collection of compact sets is compact
Hello

I have to prove that the intersection of a collection of compact sets is compact

This is what I have so far:
Each set in the collection is compact, thus each set is closed and bounded.
Each set is bounded if it is bounded above and below (i.e. there exists a B in R such that x <= B for every x in the set. There is an L in R such that x >= L for every x in the set.

Let Bj be the upper bound for each set j =1,...,n. Choose max (B1,B2,... Bj) =b. Thus, this b is the least upper bound of the collection of compact sets. Let Lj be the lower bound for each set j =1,...., n. Choose min (L1, L2,....Lj) =l. Then, l is the greatest lower bound of the collection of compact sets. Since the collection is bounded above and below, it is bounded.

Since each set in the collection is compact, each set is closed. Thus, the intersection of the collection of sets must be closed as well.

Since the intersection of the collection of compact sets is both closed and bounded, then the intersection is compact

what do you think?
• Feb 28th 2010, 02:01 PM
southprkfan1
Everything you said is true only if you are dealing with the space R^m, where m is finite.

If you are dealing in infinite dimensional spaces (R^inf) or other spaces entirely (eg the set of continuous functions) then the closed and bounded condition is no longer sufficient.
• Feb 28th 2010, 02:58 PM
Plato
Quote:

Originally Posted by inthequestofproofs
I have to prove that the intersection of a collection of compact sets is compact
Since each set in the collection is compact, each set is closed. Thus, the intersection of the collection of sets must be closed as well.

As has been pointed out, you must be careful about the nature of the space.
But the intersection of closed subsets is closed. Call it ${M}$.
Say that $\mathcal{O}$ is a collection of open sets that cover $M$.
Because $M$ is closed then its complement $M^c$ is open.
Now define $\mathcal{O}^*=\mathcal{O}\cup \{M^c\}$.
Is it true that $\mathcal{O}^*$ an open cover of any compact set in the original collection?
Now can you finish?