[SOLVED] Topology: Nested, Compact, Connected Sets

jjou

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 [tex]x_n\in K_n[/tex] 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: [tex]x \in SCl(K_n) \subset Cl(K_n) = K_n[/tex] (since X is Hausdorff, all compact sets are closed). Thus [tex]x\in K_n[/tex] 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 [tex]K=\bigcap^{\infty}K_n[/tex] is not connected, then there exists open sets U, V such that [tex]U\cap V=\emptyset[/tex], [tex]U\cap K\neq\emptyset[/tex], [tex]V\cap K\neq\emptyset[/tex], and [tex]K\subset U\cup V[/tex]. I also know then that [tex]U\cap K_n\neq\emptyset[/tex] for any n and likewise for V. But I don't know that there is any n for which [tex]K_n\subset U\cup V[/tex] - 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
[tex]K_1 \cap K_2 = K_2[/tex] is connected. Then, for any [tex]n\in\mathbb{N}[/tex], we have [tex]\bigcap_{i=1}^{n} K_i = K_n[/tex] is connected. Then let [tex]n\rightarrow\infty[/tex]..? 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 [tex]U\cup V[/tex]. There must exist n such that [tex]K_n\subset U\cup V[/tex] 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 [tex]I_n=\bigcap_{i=1}^n K_i[/tex]. Thus, any neighborhood containing K must also contain an element of the sequence. So I take [tex]U\cup V[/tex] 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. :)