Topology: Nested, Compact, Connected Sets

[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?

 

[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!).

 

[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.

 

[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! :)

 

[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?

 

[jjou]

Nevermind, got it. :)