Compactness and Nested Sequences: A Proof Dilemma

[qinglong.1397]

Homework Statement

Prove that a subset $$A$$ of $$R^{n}$$ is compact if and only if every nested sequence $$\{A_n\}_{n=1}^{\infty}$$ of relatively closed , non-empty subsets of $$A$$ has non-empty intersection

The Attempt at a Solution

I can prove $$\rightarrow$$, but not $$\leftarrow$$. Would you please give me a hint? Thank you very much!

 

[micromass]

Take an open cover $$(G_i)_{i\in I}$$ which does not have an open subcover (so indeed, we are proving things by contradiction).
Take $$G_0$$ in this open cover such that this set is not entire A.
Take $$G_1$$ in this open cover such that $$\{G_0,G_1\}$$ does not cover A.
...

We end up with a sequence of sets $$(G_n)_n$$.

For the rest of the proof, I'll give a hint. Take a look at the complements of the sequence $$G_0,G_0\cup G_1, G_0\cup G_1\cup G_2, ...$$

 

[qinglong.1397]

Take an open cover $$(G_i)_{i\in I}$$ which does not have an open subcover (so indeed, we are proving things by contradiction).
Take $$G_0$$ in this open cover such that this set is not entire A.
Take $$G_1$$ in this open cover such that $$\{G_0,G_1\}$$ does not cover A.
...

We end up with a sequence of sets $$(G_n)_n$$.

For the rest of the proof, I'll give a hint. Take a look at the complements of the sequence $$G_0,G_0\cup G_1, G_0\cup G_1\cup G_2, ...$$

You mean $$(G_i)_{i\in I}$$ does not have a finite subcover, don't you? I am sorry that I do not agree with you. I cannot see any contradiction from your hint. Since the union of any family of finitely many elements of $$(G_i)_{i\in I}$$ does not equal $$A$$, then the intersection of the collection of the complements of the sequence $$G_0,G_0\cup G_1, G_0\cup G_1\cup G_2, ...$$ is not empty, naturally.

 

[micromass]

Yes, but this was only a sketch of the proof. The point is that we can take the sequence $$(G_n)_n$$ such that its union does cover A.
I agree that if you take the sequence like I said, thaen there is no contradiction. But you have to the sequence kinda special...

 

[qinglong.1397]

Yes, but this was only a sketch of the proof. The point is that we can take the sequence $$(G_n)_n$$ such that its union does cover A.
I agree that if you take the sequence like I said, thaen there is no contradiction. But you have to the sequence kinda special...

Yeah, maybe there is such kind of sequence, but how could you prove that? I think, unless A has some special structure, it does not have the sequence we want. Remember that A is an abstract subset of $$R^n$$. I really do not know how to construct the sequence.

 

[micromass]

Well, the construction of the sequence is the hardest part of the problem. Let $$(G_i)_i$$ be our open cover. Then we know that

$$A\subseteq \bigcup_{i\in I}{G_i}$$

For every rational number q in $$\bigcup_{i\in I}{G_i}$$, we take an element $$G_q$$ in our open cover. If we do that for every rational, then we have obtained a sequence $$(G_q)_q$$ which still covers A.

 

[qinglong.1397]

In my opinion, it makes no difference. I am so stupid. Maybe you can give me more definite hint. Or, complete proof!

 

[micromass]

Take $$(G_i)_{i\in I}$$ be an open cover without finite subcover. Like in my post 6, I can assume that I is countable. Thus without loss of generality, I take it that I=N.

Then $$G_0\cup...\cup G_n$$ does not cover A, although $$\bigcup_{n\in \mathbb{N}}{G_n}$$ does cover A.

With complementation, this yields a contradiction.

 

[qinglong.1397]

I see, finally. Thank you very much!