Compact Nested Sequences and Their Intersection

[JamesF]

Hi everyone. I feel like I'm really close to the answer on this one, but just out of reach :) I hope someone can give me some pointers

Homework Statement

Let $$A1 \supseteq A2 \supseteq A3 \supseteq \ldots$$ be a sequence of compact, nonempty subsets of a metric space $$(X, d)$$. Show that $$\bigcap A_n \neq \emptyset$$. (Hint: Let $$U_n = X-A_n$$)

The Attempt at a Solution

I tried showing by contradiction.

Suppose $$\bigcap A_n = \emptyset$$
Choose an open subcover $$U_n = X-A_n$$ (that's supposed to be set minus but I don't know how to do \ in tex). Then $$\bigcup U_n = (X-A_1) \cup (X - A_2) \ldots = X - (\bigcap A_n) = X$$

but where's the contradiction? So X is not compact, but that goes without saying and we can't infer much from that. What am I overlooking here? Or is this the wrong approach entirely?

Thank you for your assistance.

 

[Dick]

X may not be compact but A1 is. The U_n are a cover of A1 if you assume the intersection of all the Ai is empty. Hence there is a finite subcover. That seems headed for a contradiction.

 

[JamesF]

Thanks for your help, Dick. I was able to get the solution.


I have one more question on my current HW.
Is a set $$A_n = [n, \infty)$$ open or closed in $$\mathbb{R}$$? I would think so, but it's unbounded.

 

[Dick]

Thanks for your help, Dick. I was able to get the solution.


I have one more question on my current HW.
Is a set $$A_n = [n, \infty)$$ open or closed in $$\mathbb{R}$$? I would think so, but it's unbounded.

What do you think it would be?