# Are the real numbers compact?

1. Oct 24, 2005

### *melinda*

This is something that I think I should already know, but I am confused.

It really seems to me that the set of all real numbers, $\Re$ should be compact.

However, this would require that $\Re$ be closed and bounded, or equivalently,

that every sequence of points in $\Re$ have a limit point in $\Re$.

But I don't see how $\Re$ can be closed and bounded, unless we somehow decide that infinity is a real number.

So, are the real numbers compact?

thanks

2. Oct 24, 2005

### shmoe

Just think about the sequential version for the moment, can you find a sequence in R that has no limit point in R?

3. Oct 24, 2005

### fourier jr

.... or find an open cover that has no finite subcover. (i can think of two)

4. Oct 24, 2005

### *melinda*

That's just the problem...

I cannot think of a sequence in $\Re$ that does not have a limit point in $\Re$.

Most of my confusion is that the two statements are equivalent, but seem to imply two different things about the real numbers.

I'm assuming at this point that the real numbers are in fact compact.

But if the real numbers are compact, that means they are closed and bounded.

In every math class prior to this the real numbers were always given by:

$$\Re = (-\infty , \infty)$$.

Because of this, I keep thinking that $\Re$ is open.

What's my problem?

5. Oct 24, 2005

### shmoe

Think back to your calculus classes. Can you give a few examples of divergent sequences? Do they have limit points? (note that the idea of a "limit point" you've recently met is more general than the idea of the limit of a convergent sequence from your calc class)

They are open. Next question: are they closed?

6. Oct 25, 2005

### HallsofIvy

Staff Emeritus
Why does it seem to you that the set of all real numbers should be compact? The basic concept of "compact" is that a compact set is "the next best thing to finite". Compact sets, precisely because "every open cover has a finite subcover", have many of the properties of finite sets. In particular, every compact set of real numbers contains a largest and a smallest number.

(The set of all real numbers is both closed and open.)

7. Oct 25, 2005

### *melinda*

I think my confusion was with a theorem which states:

A continuous function on a compact set is uniformly continuous.

It led me to think that the real numbers (where continuous functions in previous courses have been defined) should be compact.

For some reason I was thinking I needed uniform continuity for real valued functions, which obviously is not the case.

Thanks for all the feedback.

I think I'm ok... for the moment

8. Oct 25, 2005

### HallsofIvy

Staff Emeritus
If it were the case that every continuous function on the real numbers were uniformly continuous, the concept of "uniform continuity" would never have been distinguished from "continuity"!

9. Oct 25, 2005

### Oxymoron

I thought that you cannot say whether or not the real line, $\mathbb{R}$, or any space for that matter, is compact or not without stating the topology.

For example, (I think Fourier Jr was hinting toward this), the real line is compact with respect to the finite complement topology. But under the usual topology, $\mathbb{R}$ is not compact because there does not exist an open cover with a finite subcover.

In fact, you can generalize this into $\mathbb{R}^n$ (with the usual topology) and once again there is no open cover which has a finite subcover.

No, not quite. A subspace of $\mathbb{R}$ is compact if and only if that subspace is closed and bounded - not real line itself. This comes from the fact that if you have some compact topological space (ie. a space which is compact and you have a good idea what the open sets in it are...) then any closed subset $A$ of that space is compact. Note: I think it is also a corollary for the Heine-Borel theorem - but I cant be sure, maybe someone can check)

Last edited: Oct 25, 2005
10. Oct 25, 2005

### Oxymoron

No, the real numbers are not compact. And you cannot say that $\mathbb{R}$ is compact if it is closed and bounded - only a subset of $\mathbb{R}$ is compact if it is closed and bounded.

Think carefully to what Halls said about compact sets being the next best thing to being finite (I found that statement particularly enlightening). Because determining whether a set is compact requires you to find a finite subcover given an open cover. If your set is infinite, it is going to be hard to find a finite subcover (unless of course your topology takes care of the open cover - just as the finite complement topology does).

So, with respect to the usual topology, the real line is definately not compact because it is far from being finite (this is the short answer!)

11. Oct 26, 2005

### fourier jr

since melinda is (evidently) in a basic course on topology or analysis i guessed that she hasn't met many other topologies & meant R with the usual topology. the two coverings i thought of were
a) {$$(n, n+2) | n\in \mathBB{Z}$$}
b) {$$(-n,n) | n\in \mathbb{N}$$}
neither of which have finite subcovers. that's using the definition of compactness, not sequential compactness, which needs metrizability for the two to be equivalent.

Last edited: Oct 26, 2005
12. Oct 26, 2005

### fourier jr

13. Oct 26, 2005

### HallsofIvy

Staff Emeritus
But $\mathbb{R}$ is a subspace of itself! It is, of course, trivially true that "The set of all real numbers (with the usual topology) is compact if and only if it is both closed and bounded" since neither part of that is true! In any metric topology it is true that a compact set is bounded. In any Hausdorf topology it is true that a compact set is closed. The converse, that any closed and bounded set is compact (the Heine-Borel theorem) requires (and is equivalent to) the Least Upper Bound property- which is itself equivalent to Monotone Convergence, Cauchy Criterion, Bolzano-Weierstrass, etc.

By the way, *melinda*, an example of a "sequence which does not have a limit point" is 1, 2, 3, 4, ..., n,... !

Yes, it is true that if you "somehow decide that infinity is a real number" you can make $\mathbb{R}$ compact.
The "Stone-Cech compactification" adds two points (infinity and -infinity) together with open sets containing them so that $\mathbb{R}$ is topologically equivalent to a closed, bounded interval.
The "one point compactification" adds a single point (infinity) together with open sets containing it so that $\mathbb{R}$ is topologically equivalent to a circle.

Last edited: Oct 26, 2005
14. Oct 26, 2005

### HallsofIvy

Staff Emeritus
But $\mathbb{R}$ is a subspace of itself! It is, of course, trivially true that "The set of all real numbers (with the usual topology) is compact if and only if it is both closed and bounded" since neither part of that is true! In any metric topology it is true that a compact set is bounded. In any Hausdorf topology it is true that a compact set is closed. The converse, that any closed and bounded set is compact (the Heine-Borel theorem) requires (and is equivalent to) the Least Upper Bound property- which is itself equivalent to Monotone Convergence, Cauchy Criterion, Bolzano-Weierstrass, etc.

By the way *melinda*, an example of a "sequence which does not have a limit point" is 1, 2, 3, 4, ..., n,... ! Yes, it is true that if you "somehow decide that infinity is a real number."

The "Stone-Cech compactification" adds two points (infinity and -infinity) together with open sets containing them so that $\mathbb{R}$ is topologically equivalent to a closed, bounded interval.
The "one point compactification" adds a single point (infinity) together with open sets containing it so that $\mathbb{R}$ is topologically equivalent to a circe.