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Question: Let $\{F_n\mid n \in \mathbb{N}\}$ be a family of non empty closed subsets of the Euclidean space $\mathbb{R}^p$ with $F_1$ bounded and $F_{n+1} \subset F_n$ for every $n\in\mathbb{N}$. Show that $\bigcap_{n\in\mathbb{N}}F_n\neq\emptyset$.

Source: It was asked in an exam in the year 2005.

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    $\begingroup$ Think of the Heine-Borel theorem. $\endgroup$ – Michael Greinecker Jun 22 '13 at 20:47
  • $\begingroup$ thank you so much @AlexBecker I'll keep this in mind in future. $\endgroup$ – Mathy Jun 22 '13 at 21:05
  • $\begingroup$ Out of curiosity, how did you get to the meta.math.SE site to begin with? We've had a few users posting math questions there, and were curious as to how this is happening. $\endgroup$ – Alex Becker Jun 22 '13 at 21:07
  • $\begingroup$ I was looking at MathJax basic tutorial at meta.math.SE and I might have hit "Ask Question" there only. $\endgroup$ – Mathy Jun 22 '13 at 21:21
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    $\begingroup$ @Mathy : You had the curly braces in $\{F_n\mid n\in\mathbb N\}$ outside the math environment. I put the inside. Just use backslashes on them. $\endgroup$ – Michael Hardy Jun 22 '13 at 21:36
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Let's rephrase this question as:

Let $F_1$ be a closed, bounded subset of $\mathbb R^p$ and let $F_n$ be a sequence of closed nonempty subsets of $F_1$ such that $F_{n+1}\subset F_n$ for all $n\in\mathbb N$. Show that $\cap_{n\in\mathbb N} F_n\ne \emptyset$.

This is the same as saying that $F_1$ satisfies what's called the finite-intersection property. The crucial insight for this problem is that the finite-intersection property is equivalent to compactness. I'll prove the direction you need.

Theorem: Let $F$ be a compact space and $F_n$ a sequence of closed nonempty subsets such that $F_{n+1}\subset F_n$ for all $n\in\mathbb N$. Then $\cap_{n\in\mathbb N}F_n\ne\emptyset$.

Proof: Suppose otherwise. Then $\{F\setminus F_n\}_{n\in\mathbb N}$ is an open cover, so we have some $N$ such that $\{F\setminus F_n\}_{n\leq N}$ is a subcover. But then $F\setminus F_N=F$, so $F_N=\emptyset$, a contradiction.

Now simply observe that $F_1$ is compact by Heine-Borel.

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Here's a proof which uses the Bolzano-Weierstrass theorem. Choose a point $x_i \in F_i$. The sequence $\{x_i\}_{i = 1}^\infty$ lies in the sequentially compact set $F_1$, and therefore has convergent subsequence $\{x_{i_k}\}_{k=1}^\infty$ which converges to some $x \in F_1$. For any given $n$, the subsequence lies in the closed set $F_n$ when $i_k \ge n$, and therefore $x \in \bigcap_{n \in \mathbb{N}} F_n$

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