1. Apr 8, 2012

### cragar

1. The problem statement, all variables and given/known data
Decide whether the following propositions are true or false. If the claim is valid supply a short proof, and if the claim is false provide a counterexample.
a) An arbitrary intersection of compact sets is compact.
b)A countable set is always compact.
3. The attempt at a solution
a) If I took an infinite amount of intersections of closed intervals of the real line, I could get a set that is not bounded, And by the Heine-Borel theorem a set is compact if and only if it is closed and bounded.
b) The set of naturals is countable but not bounded so again by the Heine-Borel theorem this is not true.

2. Apr 8, 2012

### Fredrik

Staff Emeritus
You may have confused intersection with union in 3a. An intersection of intervals is smaller than (in the sense of being a subset of) any of the intervals. 3b is OK, if you're only supposed to prove it for subsets of ℝ.

Are those statements about subsets of ℝ, subsets of an arbitrary metric space, subsets of an arbitrary Hausdorff space, or subsets of an arbitrary topological space?

3. Apr 8, 2012

### cragar

ok so for 3a) if we had the intersection of [0,1] and [1,2] the intersection would just be the point 1. I guess this could be the constant sequence 1, so maybe the statement is true. Since the sequences are bounded they would have convergent sub sequences.
We are proving these things for subsets of the real line. And we are studying the topology of the real line.

4. Apr 8, 2012

### Fredrik

Staff Emeritus
OK, in that case, you just need to prove that every intersection of compact sets is closed and bounded. This is very easy. (Use what you know about intersections of closed sets).

5. Apr 10, 2012

### cragar

So I need to prove that intersections of closed sets are closed.
So i basically need to prove that any subset of a closed set is closed.
We know that these sets are bounded. So we know because they are bounded that they contain a convergent sub sequence. So I guess I need to prove that the limit point is part of the set.

6. Apr 10, 2012

### Fredrik

Staff Emeritus
Yes.

No. This isn't true. ℝ is closed, but the set of positive real numbers is not. It's not hard to come up with lots of other examples.

On the other hand, if you prove that every subset of a bounded set is bounded, that would be useful.

No need to talk about subsequences if your goal is to prove that intersections of closed sets are closed.