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Let $\{X_n\}$ be a sequence of closed and bounded subsets of a complete metric space such that $X_n\supset X_{n+1}$ for every positive integer $n$ and $\lim_{n\rightarrow\infty}(\text{diam }X_n)=0$. Prove that $\bigcap_{n=1}^{\infty}X_n$ contains exactly one point.

Every closed subset of a complete metric is complete, so $X_1,X_2,\ldots$ are complete. Also, the intersection of closed subsets is closed, so $\bigcap_{n=1}^{\infty}X_n$ is closed. Suppose there are two points $a,b$ in $\bigcap_{n=1}^{\infty}X_n$. Then $\lim_{n\rightarrow\infty}(\text{diam }X_n)\geq d(a,b)$, a contradiction. Now suppose there is no point in $\bigcap_{n=1}^{\infty}X_n$. How is this a contradiction?

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    $\begingroup$ Note that we also need the assumption: $X_n$ is non-empty for all $n$. $\endgroup$ – AD. Jun 20 '13 at 5:33
  • $\begingroup$ @AD. True, thanks. The problem statement that I copied doesn't have it either. $\endgroup$ – PJ Miller Jun 20 '13 at 5:34
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HINT: For each $n\in\Bbb Z^+$ let $x_n\in X_n$. Show that $\langle x_n:n\in\Bbb Z^+\rangle$ is a Cauchy sequence. Where must its limit be?

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  • $\begingroup$ Fix $\epsilon$. Since $\lim_{n\rightarrow\infty}(\text{diam }X_n)=0$, there exists $N$ such that if $n>N$ and $a,b\in X_n$, then $d(a,b)<\epsilon$. Let $i,j>N$. Since $x_i,x_j\in X_N$, we have $d(x_i,x_j)<\epsilon$, implying that $(x_n)$ is a Cauchy sequence. Since the metric space is complete, this sequence is convergent to a point $x$ in the metric space. For each $n$, the set $X_n$ contains $x_n,x_{n+1},\ldots$, and since it's closed, the limit point $x$ must belong to $X_n$ as well. So $x\in\bigcap_{n=1}^{\infty}X_n$, as desired. $\endgroup$ – PJ Miller Jun 20 '13 at 5:31
  • $\begingroup$ @PJMiller: Looks good! $\endgroup$ – Brian M. Scott Jun 20 '13 at 5:32
  • $\begingroup$ Wait, is the boundedness condition only needed in order for $\text{diam}$ to be defined? $\endgroup$ – PJ Miller Jun 20 '13 at 5:33
  • $\begingroup$ It looks like if we define $\text{diam}$ to be $\infty$ when the set is unbounded, then the problem holds without the boundedness condition. $\endgroup$ – PJ Miller Jun 20 '13 at 5:34
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    $\begingroup$ @PJMiller: Yes, but since you need the diameters to go to $0$, you might as well just throw away the unbounded sets: there can be only finitely many of them. $\endgroup$ – Brian M. Scott Jun 20 '13 at 5:43
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Hint

Pick $x_n\in X_n$ (which must exist, see the comment above), show that $(x_n)_n=1^\infty$ is Cauchy in $\bigcap X_k$. Finish off with the Hint of Biran M. Scott.

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