# Proof that Every Compact Set is Bounded

1. Jan 6, 2012

### gwsinger

I came across this proof and have a question about the bolded portion:

Consider the following objection to the bolded: In order for $\mathcal{G}$ to be an open cover of $K$ its sets must contain all of the points of $K$. The sets of $\mathcal{G}$ are $B_r(p)$ for some fixed $p$, and so as $r$ gets larger and larger, it will allow some $B_r(p)$ to consume more and more points of $K$. But how do we know that there is some finite $r$ that is large enough to consume ALL the points of $K$? Isn't this essentially what we're trying to prove here to begin with?

2. Jan 6, 2012

### lugita15

In the bolded portion, we are not assuming that there is some r that is large enough to consume all of K (that's going to be proved a few lines later). All that is being said in the bolded portion is that for any point x in K, there exists an r such that $x\in B_{r}(p)$. In other words, any point x in K is an element of some set in G (although conceivably different points x may belong to different sets in G). Since this is true of all points x in K, G is a cover of K.

After the bolded portion, the rest is pretty straightforward. Since G is an open cover of K and K is compact, G has a finite subcover of K, and it's easy to encompass all the points within that finite subcover using a single r.

3. Jan 6, 2012

### wisvuze

The main idea in this proof is that the compactness condition ( that there is a finite refinement ) applies to *all* open covers. Thus, we only need to pick a "clever" choice of a covering to get the point across ( that K is bounded ).
Imagine a point in a space. You can contain it in a ball, then contain it in a bigger ball, and so on. ( you can prove that all these balls will actually cover K, it is fairly trivial, since we are dealing with a metric space )
Now, by hypothesis, we can find a finite version of this collection of balls, and the result follows ( by taking the maximum of the ball sizes ).

4. Jan 6, 2012

### gwsinger

But how do we know that? Why can we just assert that all of the $x$'s in $K$ will belong to some $B_{r}(p)$? This seems to me to be the meat of the proof.

As wisvuze points out, of course if this is true then it will follow that there will be a finite number of such balls, so that we can then take the radius of the largest ball, choose an $M$ greater than this radius, and then declare our set to be bounded.

5. Jan 6, 2012

### wisvuze

Just from the beginning of the proof ( that is, without assuming the result ), you cannot just assert that all x in K will be contained some ball, otherwise we'd be done. The point is that for every x in K , there is SOME ball with SOME radius r so that B_r ( p ) contains x. This is how the open cover works, we collect all these balls and it is our cover. Note that we might have to collect an infinite amount of these balls, however.. the compactness hypothesis will come to the rescue ( to save us, in this particular case )

[ as an illustration, our cover will look something like this: ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( p ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) x ) ) ) ) ) ) ) ) ] each ( ) is an open ball of some radius, and x denotes an arbitrary x in K

6. Jan 6, 2012

### micromass

Staff Emeritus
Because that might not be true. For example, $\mathbb{R}$ is not contained in any ball. But we CAN say that $\mathbb{R}\subseteq \bigcup_{r>0} B_r(p)$ for some p.

All that means is that for each x, there exists an r>0 such that $x\in B_r(p)$. But the r is potentially different for each x!! There is no unique r that satisfies this.

7. Jan 6, 2012

### lugita15

I think you forgot what I said in the parentheses:
The assertion is not that all of the x's in K will belong to one set $B_{r}(p)$. The assertion is that each individual x will belong to some set $B_{r}(p)$, even though different points x1 and x2 may belong to different sets $B_{r1}(p)$ and $B_{r2}(p)$. So each point x will belong to some set in G (and let me emphasize, not necessarily all the points x will belong to the same set in G). Thus G is a cover of K.

8. Jan 6, 2012

### lugita15

Wow, micromass, wisvuze, and myself all made the same point in the same minute :rofl:

9. Jan 7, 2012

### imurme8

It might clarify things to notice that this is exactly the problem of switching the order of two quantifiers:

$\mathcal{G}\subseteq \bigcup\limits_{r>0} B_{r}(p)$ means $(\forall x\in \mathcal{G})(\exists r\in \mathbb{R}) (x \in B_r(p))$

$\mathcal{G}$ bounded means $(\exists r\in \mathbb{R})(\forall x\in \mathcal{G}) (x\in B_r(p))$

The classic example of this is that the first statement is like saying "everybody loves somebody", while the second is like saying "there is somebody who loves everybody".

10. Jan 7, 2012

### imurme8

Or even better, "Everybody is bounded by some number" versus "There is some number that bounds everybody".

11. Jan 7, 2012

### spamiam

Maybe this is also tripping you up: any two points in a metric space are a finite distance from each other. This is implicit in the definition of a metric since $d : X \times X \to \mathbb{R}^+ \cup \{0\}$ and $\mathbb{R}$ does not have "infinity" as an element.

A more explicit proof might go like this:

Fix $p \in X$. For any $x \in X$, $d(x,p) < \infty$. Thus each x is contained in some open ball $B(p;r)$ with $r \in \mathbb{R}^+$: take $r = d(x,p) + 1$, for instance. Therefore $\mathcal{G}=\{B_r(p) \, | \, r\in\mathbb{R}^+\}$ covers X, so it also covers K. K is compact, so this open cover contains a finite subcover and then, like you said, just take the ball with maximum radius from this subcover to show K is bounded.

Last edited: Jan 7, 2012