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jjou

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**[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 [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?

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