Cantor Intersection Theorem

**Slazenger3** · March 26th 2010, 06:07 AM

Suppose that H is a closed bounded set of real numbers and that (U sub n) is an expanding sequence of open sets.

(a) Explain why the sequence of sets H \ (U sub n) is a contracting sequence of closed bounded sets.

(b) Use the Cantor intersection theorem to deduce that if H \ (U sub n) does not equal an empty set for every n, then the intersection from n=1 to infinity, of (H \ (U sub n)) does not equal an empty set.

**Drexel28** · March 26th 2010, 10:02 PM

Originally Posted by Slazenger3

Suppose that $H$ is a closed bounded set of real numbers and that $U_n$ is an expanding sequence of open sets.

(a) Explain why the sequence of sets $H-U_n$ is a contracting sequence of closed bounded sets.

Note that $U_n\subseteq U_{n+1}\implies H-U_{n}\supseteq H- U_{n+1}$ . So, they're clearly contracting. Also, We know that $H$ is closed and $U_n$ open, thus $H-U_n=H\cap \left(U_n\right)'$ is the intersection of two closed sets, thus closed. Lastly, note that $H\supseteq H-U_1\supseteq H-U_2\supseteq\cdots$ and so $\text{diam }H\geqslant \text{diam }H-U_1\geqslant \text{diam }H-U_2\geqslant\cdots$

(b) Use the Cantor intersection theorem to deduce that if $H-U_n$ does not equal an empty set for every n, $\bigcap_{n=1}^{\infty}\left\{H-U_n\right\}\ne\varnothing$

Which Cantor intersection theorem? The one that deals with complete spaces or the one that deals with compact spaces? Noting the insistence on bounded subspaces I assume the latter.

Merely note from the above that given a finite subclass $H-U_{n_1},\cdots,H-U_{n_j}$ of $\left\{H-U_n\right\}_{n\in\mathbb{N}}$ that $\left(H-U_{n_1}\right)\cap\cdots\cap \left(H-U_{n_j}\right)=H-U_\alpha$ where $\alpha=\max\{n_1,\cdots,n_j\}$ and since by assumption this is non-empty we see that $\left\{H-U_n\right\}_{n\in\mathbb{N}}$ is a class of closed subsets of the compact space $H$ with the FIP. It follows that $\bigcap_{n=1}^{\infty}\left\{H-U_n\right\}\ne\varnothing$

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