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The proof is outlined as follows: (copied from Wikipedia but Apostol gives the same idea)

If a set is compact, then it is bounded: Consider the open balls centered upon a common point, with any radius. This can cover any set, because all points in the set are some distance away from that point. Any finite subcover of this cover must be bounded, because all balls in the subcover are contained in the largest open ball within that subcover. Therefore, any set covered by this subcover must also be bounded.

My concern is this: the proof seems to implicitly assume that the set is bounded by saying we can cover it with a countable number of concentric open balls. If the set is unbounded, we would not be able to do that since there will be a point in the set that is "infinitely" far from the center of the balls. Or, is my intuition breaking down somewhere?

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  • $\begingroup$ It follows from the definition of compactness: A set is compact iff every open cover has a finite subcover. $\endgroup$ – M47145 May 13 '16 at 19:11
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    $\begingroup$ Any two points have a finite distance in a metric space. If a set is unbounded, that means that for every $M$ there are points in the set which are farther away than $M$ from the given point. Consider $\mathbb{R}^n$. We have $\mathbb{R}^n = \bigcup_k B_k(0)$, since every point has a finite distance from the origin, so $x$ is contained in all balls $B_k(0)$ with $k > d(x,0)$. But since $\mathbb{R}^n$ is unbounded, any finite family of these balls leaves part of the space uncovered. $\endgroup$ – Daniel Fischer May 13 '16 at 19:11
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Your intuition is incorrect. Let $X$ be the space, and fix a point $p\in X$. For each $x\in X$, the distance $d(p,x)$ is a real number, so it’s finite. In particular, there is some positive integer $n$ such that $d(p,x)<n$. But then $x\in B(p,n)$, the open ball of radius $n$ centred at $p$, and it follows that

$$X=\bigcup_{n\in\Bbb Z^+}B(p,n)$$

and hence that $\{B(p,n):n\in\Bbb Z^+\}$ is an open cover of $X$.

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  • $\begingroup$ Ah, I see. I was missing the fact that although there are points "infinitely" far away (i.e. it is unbounded, I'm not implying that d(p,x) is actually infinite), there are also infinitely many balls in the union so they can account for all of the points in the original set. $\endgroup$ – dkv May 13 '16 at 19:22
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    $\begingroup$ @dtkv At least you're putting the word "infinitely" in quotes. There are no points infinitely far away. Every point is at finite distance. There are points arbitrarily far away. Today's new word. $\endgroup$ – David C. Ullrich May 13 '16 at 19:50

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