# Showing A Compact Interval Is Closed

1. Oct 17, 2012

### Yagoda

1. The problem statement, all variables and given/known data
Basically, prove the Extreme Value Theorem.
"If f is a continuous function over the interval [a,b] then f reaches a max and a min on that interval."

2. Relevant equations
In this case they're more like definitions and things I have proved so far.
Intervals are connected.
The only connected subsets of ℝ are intervals.
A subset S of ℝ is compact iff every infinite subset of S has a limit point in S.
Closed intervals are compact
A subset of ℝ is closed iff it contains all its limit points

3. The attempt at a solution
En route to this problem I've shown that the image of a compact set by a continuous function is compact and that the image of a connected function by a continuous function is connected. Because an [a,b] is connected its image by f has to be an interval as well, by the second definition above. This image is compact also since [a,b] is compact. And we know that intervals are bounded.
All in all, I know that the image of [a,b] by f is a compact, bounded interval. I'm trying to use the definition of compact to show that the image has to also be closed, since my feeling is that will help me prove the EVT.
Right now I'm trying contradiction, supposing the image is compact but has a limit point outside of it. I'm not sure where to go from there in terms of arriving at a contraction.
Does it seem like I am on at all on the right track?

2. Oct 17, 2012

### jbunniii

It will suffice to prove that any compact subset of $\mathbb{R}$ is closed and bounded. (The converse is also true: any closed, bounded subset of $\mathbb{R}$ is compact. You have presumably already used this fact to conclude that $[a,b]$ is compact.)

By the way, it is not true that all intervals are bounded. $\mathbb{R} = (-\infty,\infty)$ is an unbounded interval, and so are $(a,\infty)$ and $(-\infty, a)$ for any real $a$.

Fortunately, proving that compact => bounded is easy. If a set $A$ is not bounded, then it's easy to find a sequence that diverges to either $+\infty$ or $-\infty$, with no convergent subsequence.

So that leaves compact => closed. Suppose the set $A$ has a limit point $a$ that is not contained in $A$. Can you construct a sequence of points in $A$ which converges to $a$? If so, can that sequence have a subsequence that converges to a point in $A$?

3. Oct 17, 2012

### Yagoda

Yes, you're right that not all intervals are bounded. I guess I got carried away since we are mapping a closed bounded interval by a continuous function so the image being bounded seemed intuitive.

Is there a strategy for proving that compactness implies that a set is closed and bounded without using sequences? (We haven't covered that in our class).
Using definitions involving balls/neighborhoods worked well to prove the converse. But even after consulting with a classmate I couldn't get further than the fact that since A is compact, each of its infinite subsets has a limit point in A. And these limit points are also limit points of A since they are limit points of subsets of A. This tells us that A contains lots of its limit points, but I'm not seeing how that implies that it contains all of them or that it would be bounded.

4. Oct 17, 2012

### jbunniii

Well, it's possible to map an OPEN bounded interval to all of $\mathbb{R}$; for example, $\tan$ is a continuous mapping from $(-\pi/2, \pi/2)$ onto all of $\mathbb{R}$. So closed bounded intervals are rather special in this regard.
OK. There are actually three types of compactness, all of which turn out to be equivalent in a metric space (of which $\mathbb{R}$ is an example).

My proof sketches were using what is called sequential compactness. A set X is sequentially compact if every sequence of elements of X has a subsequence that converges to a limit in X.

It appears that you are using limit point compactness. A set X is limit-point compact if every infinite subset of X has a limit point contained in X. I incorrectly assumed that you already knew that limit point compactness and sequential compactness are equivalent in $\mathbb{R}$.

It's not too hard to reformulate what I wrote earlier in terms of limit point compactness.

limit point compact => bounded: If X is unbounded, you can construct an infinite subset of X with the property that no two points in the subset are within distance 1 of each other. This subset will have no limit points. I'll let you fill in the details.

limit point compact => closed: If X is not closed, then X has a limit point x that is not contained in X, then you can construct a sequence of DISTINCT elements of X which converges to x. Now forget about the ordering of the sequence and just consider the subset Y containing the elements of the sequence (i.e., Y is the image of the sequence). Since the elements of the sequence are distinct, Y is an infinite subset. I claim that Y cannot have any limit point except x, and since x is not in X, we have shown that X is not limit point compact. Once again, I'll let you fill in the details.

Last edited: Oct 17, 2012