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Could someone provide me with an example of a metric space having a nested decreasing sequence of bounded closed sets with empty intersection? I first thought of Cantor set but the intersection is not empty!

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    $\begingroup$ Can you think of sets which are bounded and closed, but not compact? Because if you ever include a compact set in your sequence, Cantor's intersection theorem will imply there is a non-empty intersection. (As is noted in an answer, we can't do this in $\mathbb R$ or $\mathbb R^n$ due to the Heine-Borel theorem) $\endgroup$ – Milo Brandt Aug 15 '15 at 3:49
  • $\begingroup$ @MiloBrandt: Your comment was indeed what I was hinting at in my answer. $\endgroup$ – user21820 Aug 15 '15 at 11:02
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Let $\mathbb N$ be endowed with the discrete metric. In this metric space, every subset is bounded (although not necessarily totally bounded) and closed. Moreover, the subsets \begin{align*} A_1\equiv&\,\{1,2,3,4,\ldots\},\\ A_2\equiv&\,\{\phantom{1,\,}2,3,4,\ldots\},\\ A_3\equiv&\,\{\phantom{1,2,\,}3,4,\ldots\},\\ \vdots&\, \end{align*} are nested, and their intersection is empty.


However, if you stay within the realm of $\mathbb R$ endowed with the usual Euclidean metric, than you can't have a situation like the one above:

Claim: Suppose that $$A_1\supseteq A_2\supseteq A_3\supseteq\ldots$$ is a countable family of non-empty, closed, bounded subsets of $\mathbb R$. Then, $\bigcap_{n=1}^{\infty} A_n\neq\varnothing$.

Proof: By the Heine–Borel theorem, $A_n$ is compact for each $n\in\mathbb N$. For the sake of contradiction, suppose that $\bigcap_{n=1}^{\infty} A_n=\varnothing$. This is equivalent to $\bigcup_{n=1}^{\infty} A_n^{\mathsf c}=\mathbb R$. In particular, $$A_1\subseteq\bigcup_{n=1}^{\infty} A_n^{\mathsf c}.$$ Since $A_1$ is compact and the sets $(A_n^{\mathsf c})_{n=1}^{\infty}$ form an open cover of it, there must exist a finite subcover. That is, there exists some $m\in\mathbb N$ such that $$A_1\subseteq\bigcup_{n=1}^m A_n^{\mathsf c}=A_m^{\mathsf c},$$ where the second equality follows from the fact that $$A_1^{\mathsf c}\subseteq A_2^{\mathsf c}\subseteq A_3^{\mathsf c}\subseteq\ldots.$$ Now, $A_1\subseteq A_m^{\mathsf c}$ means that if a point is in $A_1$, then it must not be in $A_m$, so that $A_1\cap A_m=\varnothing$. But $A_m\subseteq A_1$ (given that the sets are nested), so that $A_1\cap A_m= A_m=\varnothing$, which contradicts the assumption that $A_m$ is not empty. This contradiction reveals that the intersection $\bigcap_{n=1}^{\infty} A_n$ must not be empty. $\quad\blacksquare$

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  • $\begingroup$ Are these sets bounded? What I see is they are bounded from below not from above. Could You please explain why they are bounded? $\endgroup$ – Fabian Aug 15 '15 at 3:57
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    $\begingroup$ @Fabian You're thinking in terms of the Euclidean metric, in which case each set is unbounded, indeed. However, consider the discrete metric on $\mathbb N$. This is defined as $d(m,n)=1$ if $m\neq n$ and $d(m,n)=0$ if $m=n$. It is not difficult to show that this is a legitimate metric. Also, the distance between any two distinct points is 1. This implies that if you take a “ball” of “diameter” 2 around the point $n=1$, then this ball will contain the whole space! That is, $$\{m\in\mathbb N\,|\,d(m,n)<2\}=\mathbb N,$$ and the whole space fits into a ball, which implies boundedness. $\endgroup$ – triple_sec Aug 15 '15 at 4:05
  • $\begingroup$ Got it! Great explanation! Thank you triple_sec . $\endgroup$ – Fabian Aug 15 '15 at 4:09
  • $\begingroup$ @Fabian I added a proof that the intersection may never be empty if you stay in $\mathbb R$. $\endgroup$ – triple_sec Aug 15 '15 at 4:17
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Another simple example is to look at the "punctured line": $(-\infty, 0) \cup (0, \infty)$, which is just the real numbers with $0$ removed. The sets $A_n = \{ x \ \in \Bbb R \,|\, |x| \le 1/n \text{ and } x \ne 0 \}$ are closed and bounded in the punctured line, but their intersection is empty.

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Do you know a theorem about nested bounded closed sets having non-empty intersection? If you do then you would need to find a metric space that does not satisfy the conditions of that theorem. $\mathbb{R}$ satisfies that theorem and hence you're not going to find a counter-example there. But there is a smaller metric space sitting inside it, namely $\mathbb{Q}$, that will give you a counter-example.

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  • $\begingroup$ I do not understand the downvote. I feel that my hint (without explicitly stating $\mathbb{Q}$) was just right, rather than just giving a complete solution like the other answerers that leave nothing for the asker to try. $\endgroup$ – user21820 Aug 15 '15 at 11:01
  • $\begingroup$ Hello user21820, I upvoted your answer, and sorry to see users downvote for no reason. Thank you for your hint. $\endgroup$ – MATH Aug 15 '15 at 17:45
  • $\begingroup$ @MATH: Thanks. Anyway the downvoter removed his downvote already, so perhaps he is happier with the explicit example of $\mathbb{Q}$. =) $\endgroup$ – user21820 Aug 16 '15 at 2:21
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I initially missed the fact that bounded sets were desired.

Let the metric space be $(0,1)$, the set of all real numbers between $0$ and $1$, not including the endpoints, with the usual metric $d(x,y)=|x-y|$.

Then the example can be that the $n$th set is $(0,\ 1/n]$. This is closed within this space; it contain all of its limit points in the space.

The first thing I thought of was $\displaystyle \bigcap_{n=0}^\infty [n,+\infty)$.

If the object called $+\infty$ were included in the space, with the appropriate topology so that $n\to\infty$, then these sets would not be closed, but would become closed if one added $+\infty$ to them as a new member, and then the intersection would not be empty because $+\infty$ would be a member of it.

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  • $\begingroup$ The sets should be bounded. $\endgroup$ – Carl Mummert Dec 3 '15 at 21:30
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If $(X,d)$ is a metric space then $$d'(x,y)=\min\{d(x,y),1\}$$ is a metric on $d$, as well. Moreover, the metrics $d$ and $d'$ generate the same topology (hence the same subsets of $X$ are closed in $(X,d)$ and in $(X,d')$).

Every subset is bounded in $(X,d')$.

See also: Proof that every metric space is homeomorphic to a bounded metric space


So it suffices to find a metric space which contains some nested closed subsets with empty intersection. Then you can look at the same system of subsets in the modified metric, and you will have an example of closed bounded sets.

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Possibly a simpler example is the usual metric on the rationals, and the closed sets $$C_k=\left[{\lfloor \sqrt{2}k\rfloor\over k}, {\lceil \sqrt{2}k\rceil\over k }\right].$$ Intuitively, the intersection of the $C_k$s is $\{\sqrt 2\}$, but that's not rational, so in $\mathbb{Q}$ their intersection is empty.

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  • $\begingroup$ Note that what's really being used here is that $\mathbb{Q}$ is incomplete - I'm essentially taking a non-convergent Cauchy sequence (one "converging" to $\sqrt{2}$) and turning it into a decreasing sequence of bounded closed sets with empty intersection. As e.g. Martin Sleziak's answer shows, this is not necessary - there are complete metric spaces in which nonetheless we have decreasing sequences of bounded closed sets with empty intersection. $\endgroup$ – Noah Schweber Dec 4 '15 at 0:07

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