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Consider a sequence of closed sets $A_n\subset \mathbb{R}$ such that $A_{n+1}\subset A_n$. Define $$ A = \bigcap\limits_n A_n. $$ If $A_1$ is bounded then $A_n$ is compact for all $n$. Hence if $A_n$ are not empty for all $n$ then $A$ is not empty.

But given a metric space $X$, can we make it always compact with say $$ \rho'(x,y) = \tan^{-1}\rho(x,y) $$ where $\rho$ is an original metric and $\rho'$ is a new one. Also this procedure will preserve closeness of sets.

This mean that regardless of compactness of $A_1$, if $A_n$ are not empty for all $n$ then $A$ is not empty?

Edited: I would like to formulate the question more direct: if $A_1$ is no bounded but all $A_n$ are closed and non empty, does it mean that $A$ is not empty?

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No, because the new metric will induce the same topology on $X$, and the compactness is an invariant of the topology, not the metric. (The problem is, you won't get a complete metric space in general.) –  Akhil Mathew Jun 17 '11 at 15:25
    
@Akhil, thanks - but I slightly changed a question. Any comments now? –  Ilya Jun 17 '11 at 15:27
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Bounded is not the same as compact. Compactness is a topological property, so it can't be affected by changing the metric if the same topology is induced. –  Qiaochu Yuan Jun 17 '11 at 15:31
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@PEV: "closed and bounded" is not equivalent to compact in general metric spaces; you need "complete and totally bounded" instead. –  Arturo Magidin Jun 17 '11 at 15:50
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@PEV: In the original setting, yes, but not once you switch the metric to $\rho(x,y)=\arctan|x-y|$. –  Arturo Magidin Jun 17 '11 at 15:55

4 Answers 4

up vote 7 down vote accepted

In $\mathbb{R}^n$ with the usual metric, "compact" is equivalent to "bounded and closed" (Heine-Borel-Lebesgue). However, the equivalence does not hold in arbitrary metric spaces.

The correct statement for arbitrary metric spaces is:

Theorem. Let $(X,d)$ be a metric space. A subset $S$ of $X$ is compact if and only if it is complete and totally bounded.

Recall that a set $S$ is "complete" if every Cauchy sequence in $S$ converges; and is "totally bounded" if for every $\epsilon\gt 0$ there exists a finite number of open balls of radius $\epsilon$ that cover $S$.

In $\mathbb{R}^n$ with the usual topology, totally bounded and bounded are equivalent, as are "complete" and "closed". But, for example, if you change the metric to $d(x,y)=\arctan|x-y|$, then the space is bounded but not totally bounded.

Added. To see that $\mathbb{R}$ is bounded but not totally bounded under $d$, note that $0\leq\arctan|x-y|\lt\frac{\pi}{2}$, so the space is bounded, contained in the ball of radius $\frac{\pi}{2}$ centered at $0$. To see that $\mathbb{R}$ is not totally bounded under $d$, pick $\epsilon\gt 0$ that is very small. Then $y\in B_d(x,\epsilon)$ if and only if $\arctan|x-y|\lt\epsilon$, if and only if $|x-y|\lt\tan(\epsilon)$. For $\epsilon$ very close to $0$, we have $\tan(\epsilon)\approx\epsilon$, so these balls are "essentially" the same size they would be in the Euclidean metric; for very small $\epsilon\gt 0$, then, you cannot cover $\mathbb{R}$ with finitely many $\epsilon$-$d$-open balls, so $\mathbb{R}$ is not totally bounded under $d$.

So you cannot make the closed set $A_1$ compact merely by replacing the metric with an equivalent-but-bounded metric, because this exchange does not respect the property of being totally bounded.

For an example in the usual metric where $A_k$ is and nonempty and closed for all $k$, but the intersection is empty, just take $A_n=[n,\infty)$ in $\mathbb{R}$. Easily generalized to $\mathbb{R}^n$ with the Euclidean metric.

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Why it's not totally bounded? The new metric allows us to use balls in fact of any radius in old metric still of the radius smaller then $\epsilon$ in the new metric. –  Ilya Jun 17 '11 at 16:03
    
@Gortaur: under the metric $d(x,y)=\arctan|x-y|$, $B(x,\epsilon)$ consists of all $y$ such that $\arctan|x-y|\lt\epsilon$, i.e., $|x-y|\lt \tan(\epsilon)$. For small $\epsilon$, $\tan\epsilon\approx \epsilon$, so you cannot cover $\mathbb{R}$ with finitely many balls of $d$-radius $\epsilon$ with small $\epsilon$, even though $\mathbb{R}$ has finite $d$-radius. So "the space" ($\mathbb{R}$) is not totally bounded in $d$. When you said you could "make $X$ compact" by switching to a finite metric, you seem to assume that bounded implies totally bounded, but it does not. –  Arturo Magidin Jun 17 '11 at 16:27
    
@Arturo: Ok, I explain how I see it. To cover $\mathbb{R}$ in the new metrics with balls of radius $\epsilon$ is the same as to cover $(-\pi/2,\pi/2)$ with the very same balls in the usual Euclidean metric and then find their images under $\tan$. –  Ilya Jun 17 '11 at 16:37
    
@Gortaur: The reason this doesn't work is that $\arctan(x-y)\neq \arctan(x)-\arctan(y)$; the image under $\tan$ of the interval $(\pi/4,\pi,3)$ is not a ball of diameter $\frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}$: it goes from $1$ to $\sqrt{3}$, and the diameter is $\sqrt{3}-1\approx 0.73\gt 0.26\approx \frac{\pi}{12}$. Those small intervals further away from $0$ on $(-\pi/2,\pi/2)$ become very large intervals in the usual Euclidean metric after you apply tangent. –  Arturo Magidin Jun 17 '11 at 16:43
    
@Arturo: what if we take $\rho'(x,y) = |\arctan(x) - \arctan(y)|$. Will we make a space compact? –  Ilya Jun 20 '11 at 8:25

I'm adding my comment as answer:

A counterexample in $\mathbb R$ is $A_n=\langle n,\infty)$.

However, nested closed set have non-empty intersection in a compact space. The same is true in a complete metric space if you add the assumption that the diameter of $A_n$ tends to 0.

The result for complete metric spaces is often called Cantor's intersection theorem. See wikipedia or wikibooks.

In compact spaces, every system of closed sets with finite intersection property has non-empty intersection. In fact, this condition is equivalent to compactness.

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There are several properties in play here: Compactness, boundedness and being closed. Compactness and being closed are topological properties. This means that if two metrics give the same collection of open sets (that is, the metrics are "topologically equivalent"), then they will also give the same collection of closed sets (obvious) and the same collection of compact sets. However, boundedness is not a topological property, for instance by the example you gave.

In every metric space, it is true that a compact set is bounded and closed. This implies, for instance, that if a set is compact w.r.t. one metric, then it is bounded w.r.t. any metric which is topologically equivalent to it. However, if a set if bounded w.r.t. to one metric, this says nothing about compactness of the set w.r.t. any metric. Indeed, one can always define a new equivalent metric which is bounded, in the way you did.

In $\mathbb{R}$ and more generally, in $\mathbb{R}^n$ (with the usual metric), the above relation also works backwords, that is: a closed and bounded set is compact. The problem here is that for most metrics, and for most metric spaces, this is no longer true. A metric space for which this is true is said to have the Heine-Borel property. In some sense, such metrics are particularly good, since it is often easier to tell if a set is closed and/or bounded than if it is compact. Thus one would often like to do the opposite of what you did: to switch from a metric without the HB property to an equivalent metric with the HB property. Unfortunately, this is not always possible: for some metric spaces (which don't have the HB property), there doesn't exist any other metric (topologically equivalent to the original metric) which has the HB property.

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The property of metric spaces you mention (or very similar one) is called "nice closed balls" in this book books.google.com/… I did not find a book where "Heine-Borel" property is used in this sense - usually it's equivalent to compactness google.com/… –  Martin Sleziak Jun 17 '11 at 16:52

Let $A$ be any infinite set, and define a "distance" on $A$ by putting $d(x,y)=1$ if $x \ne y$, and $d(x,x)=0$. Then every subset of $A$ is closed and bounded.

However, $A$ has no infinite compact subset. And if $X$ is any infinite subset of $A$, one can produce an infinite nested sequence $(X_i)$ of non-empty subsets of $X$ whose intersection is empty.

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