# Topology: Nested, Compact, Connected Sets

1. Mar 1, 2008

### jjou

[SOLVED] Topology: Nested, Compact, Connected Sets

1. Assumptions: X is a Hausdorff space. {K_n} is a family of nested, compact, nonempty, connected sets. Two parts: Show the intersection of all K_n is nonempty and connected.

That the intersection is nonempty: I modeled my proof after the widely known analysis proof. I took a sequence (x_n) such that $$x_n\in K_n$$ for all n. Assuming x_n has a limit point x (AM I ALLOWED TO ASSUME THE SEQUENCE HAS A LIMIT POINT?), then x is in the sequential closure of K_n, which is contained in the closure of K_n, which is equal to K_n: $$x \in SCl(K_n) \subset Cl(K_n) = K_n$$ (since X is Hausdorff, all compact sets are closed). Thus $$x\in K_n$$ for all n, so it is in the intersection. Therefore the intersection is non-empty. This all hinges on the fact that I assumed there was a limit point ... am I talking in circles, or is this okay?

That the intersection is connected: I'm guessing I should be using contradiction. So, suppose the intersection $$K=\bigcap^{\infty}K_n$$ is not connected, then there exists open sets U, V such that $$U\cap V=\emptyset$$, $$U\cap K\neq\emptyset$$, $$V\cap K\neq\emptyset$$, and $$K\subset U\cup V$$. I also know then that $$U\cap K_n\neq\emptyset$$ for any n and likewise for V. But I don't know that there is any n for which $$K_n\subset U\cup V$$ - which would be the contradiction I am looking for, since every K_n is connected. Or is this not the right method at all?

Last edited: Mar 2, 2008
2. Mar 1, 2008

### morphism

What do you mean by "limit point"?

For connectedness, we require our set to be in the union of U and V, not their intersection (which is empty!).

3. Mar 2, 2008

### jjou

Eek, you're right. That was a horrible typo.

By "limit point," I mean any point such that any neighborhood of that point contains points of the sequence... I think.

4. Mar 2, 2008

### jjou

For showing that the intersection is connected, two ideas - can somebody check them?

METHOD 1
$$K_1 \cap K_2 = K_2$$ is connected. Then, for any $$n\in\mathbb{N}$$, we have $$\bigcap_{i=1}^{n} K_i = K_n$$ is connected. Then let $$n\rightarrow\infty$$..? Or is that oversimplifying the problem?

METHOD 2
Suppose not connected. There exists open sets U, V such that (all assumptions from above). Then consider $$U\cup V$$. There must exist n such that $$K_n\subset U\cup V$$ since {K_n} is a decreasing sequence of nested subsets. In other words, I can view the intersection as the "limit" of the sequence of intersections $$I_n=\bigcap_{i=1}^n K_i$$. Thus, any neighborhood containing K must also contain an element of the sequence. So I take $$U\cup V$$ as my neighborhood containing K and then get my contradiction..?

Please check these for me. Still having trouble with showing K is nonempty. Can someone please offer a hint? Thank you! :)

5. Mar 2, 2008

### jjou

I got that K is nonempty using the finite intersection property, so part 1 is done.

Still wondering about part 2 (connectedness). Can someone please check the ideas I posted previously?

6. Mar 2, 2008

### jjou

Nevermind, got it. :)