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I saw the following three Theorems in my Topology class, that were left as a exercise:

Theorem 1: Let $(E,d)$ be a compact metric space and $(K_n)_{n \in \mathbb{N}}$ a decreasing sequence of non empty closed sets, then $\bigcap_{n \in \mathbb{N}} K_n$ $\neq \emptyset$.

Theorem 2: Let $(E,\mathcal{T})$ be a compact Hausdorff space and $(K_n)_{n \in \mathbb{N}}$ a decreasing sequence of compact non empty closed sets, then $ \bigcap_{n \in \mathbb{N}} K_n$ $\neq \emptyset$.

Theorem 3: Let $(E,\mathcal{T})$ be a compact Hausdorff space and $(K_n)_{n \in \mathbb{N}}$ a decreasing sequence of compact non empty closed sets, then $K_{\infty} := \bigcap_{n \in \mathbb{N}} K_n$ $\neq \emptyset$ and $K_{\infty}$ is compact.

Why are they true?

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    $\begingroup$ The first one. One can reformulate compactness such that a metric space is compact iff every family $(F_j, j \in J)$ of closed sets with $\bigcap_{j \in J} F_j = \emptyset$ there is a finite set $I \subseteq J$ such that $\bigcap_{i \in I} F_i = \emptyset$. So then if $ \bigcap_{n \in \mathbb{N}} K_n = \emptyset$ then there exists $I \subseteq J$ such that $\bigcap_{i \in I} K_i = \emptyset$ and since the sequence is decreasing there exists a $K_j = \emptyset$ and that is a contradiction. For the second and third I have yet two find an idea to prove it. $\endgroup$ – vaoy Jun 29 '17 at 13:36
  • $\begingroup$ math.stackexchange.com/questions/1709401/… $\endgroup$ – Guy Fsone Jun 29 '17 at 13:55
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Let say $\bigcap_{n \in \mathbb{N}} K_n=\emptyset$ . Then $(\bigcap_{n \in \mathbb{N}} K_n)^c=\emptyset^c$. It means $\bigcup_{n \in \mathbb{N}} K_n^c=E$, so the familly $\{K_n^c:n\in \mathbb{N} \}$ is open cover of $E$. Since $E$ is compact we have a finite subcover of the familly. It means there exists a finite subset $S$ of $ \mathbb{N}$ such that $\bigcup_{n \in S} K_n^c=E$. So $(\bigcup_{n \in S} K_n^c)^c=E^c$, it means $\bigcap_{n \in S} K_n=\emptyset$. Buy using decreasing property of the sequence, we have the smallest one of the familly $\{K_n:n\in S\}$. Say it is $K_{n_0}$ and so $K_{n_0}=\bigcap_{n \in S} K_n=\emptyset$. But all the $K_n$'s are non-empty. This is contradiction.

For the third one. $K_{\infty}$ is compact since it is closed and subset of a compact set say $K_n$

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A topological space $X$ is compact iff every collection of closed subsets of $X$ having the finite intersection property has non-empty intersection.

Since the sequence $(K_n)$ is decreasing it follows that for any finite subset $\Lambda$ of $\mathbb{N}$, $\bigcap_{n\in\Lambda}K_n\neq\emptyset$. Thus by the above characterization of compactness $\bigcap K_n\neq\emptyset$. So 1 and 2 hold.

Since a countable intersection of closed sets is closed and a closed subset of a compact space is compact 3 holds.

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