- #1

Eclair_de_XII

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- TL;DR Summary
- non-trivial intersection. Denote the sequence as ##K_n## with the property that for any ##n##, ##K_{n+1}\subset K_n##. Show that there is a point in ##\cap_{n\in \mathbb{N}} K_n##.

Definition: A set is sequentially compact if all sequences contained in the set contain a subsequence that converges to a point in the set.

Let ##N\in \mathbb{N}## and suppose that ##m\geq N##. Let ##x\in K_m##. Since ##K_m\subset K_{m-1}\subset \ldots \subset K_N##, it follows that ##x## is an element of all ##K_n## for ##n\leq N## and all ##n\in [N,m]##.

I'm very sure I'm misunderstanding something very crucial here, since I didn't need to invoke the fact that these sets are sequentially compact.

Let ##N\in \mathbb{N}## and suppose that ##m\geq N##. Let ##x\in K_m##. Since ##K_m\subset K_{m-1}\subset \ldots \subset K_N##, it follows that ##x## is an element of all ##K_n## for ##n\leq N## and all ##n\in [N,m]##.

I'm very sure I'm misunderstanding something very crucial here, since I didn't need to invoke the fact that these sets are sequentially compact.