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A subset $U\subset \mathbf{R}^n$ is open if and only if it can be covered by open balls. (An open ball is a subset of the form $\{x\in \mathbf{R}^n: d(x,y) < r\}$, where $y$ is the origin of the open ball, $r$ is the radius, and $d(\cdot, \cdot)$ is the standard metric on $\mathbf{R}^n$.)

Is it true that a subset $X\subset\mathbf{R}^n$ is closed if and only if it can be covered by a finite number of closed balls?

If $X$ is closed, its complement can be covered by open balls. That's all I got.

just answered my own question. It's not possible. Take the complement of the open ball around the origin of radius $1$.

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Think about unbounded sets. –  Weltschmerz Oct 19 '11 at 19:40
    
Or take a closed line segment in $\mathbb R^2$. –  Grumpy Parsnip Oct 19 '11 at 19:41

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The answer is no. For example, $\mathbb Z$ in $\mathbb R$ is closed (since it is discrete, and every sequence converges if and only if it is eventually constant) but a finite union of closed balls has a finite diameter.

The diameter of a set $A$ is defined as: $\operatorname{diam}(A)=\sup\{|x-y|:x,y\in A\}$, the supremum of distances between the elements of $A$.

However, if we consider compact subsets, then the answer is almost yes, that is a finite union of closed balls is compact, and a compact set is a subset of such finite union of closed balls. This comes from the fact that a subset of $\mathbb R^n$ is compact if and only if it is closed and bounded. Since it is compact it is covered by a finite union of open balls. Take the closed balls with the same radius and centers and you have it.

Added. Several things to note:

  1. A finite union of closed balls is always closed, and its diameter is always finite.

  2. Every bounded set can be covered by a closed ball, regardless of being open or closed or anything else.

  3. Not every set is the union of closed balls, it is not the intersection of finite union of closed balls either. Closed sets can get very complicated, and the simplest and best way to describe them is ultimately as "complement of open".

  4. Closed sets which contain no open subset (for example a singleton) cannot be union of closed balls, since closed balls have positive radii and thus contain an open set.

This list can get very long, so I'll stop now.

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Asaf, the OP asked whether a set is closed iff it can be covered by a finite set of closed balls. The definition of "covered" is ambiguous. But if you take it to mean that the set is included in a union of finitely many closed balls, then that does not imply closed or compact. If you take it to mean that the set is equal to a union of closed balls, then you can't get sets without interior. –  Grumpy Parsnip Oct 19 '11 at 19:57
    
In other words, the statement "A subset of $\mathbb R^n$ is compact iff it is covered by finitely many closed balls." is false no matter what definition of "covered" is used. –  Grumpy Parsnip Oct 19 '11 at 20:02
    
@Jim: I added some notes and clarifications. –  Asaf Karagila Oct 19 '11 at 20:18
    
Asaf, thanks. I still take issue with your statement "if we consider compact subsets, then the answer is yes." One direction of the "iff" holds, but not the other. –  Grumpy Parsnip Oct 20 '11 at 0:08
    
@Jim: Right. I have added another sentence about that. –  Asaf Karagila Oct 20 '11 at 0:10

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