Is there a compact subspace of real numbers?

In summary, the conversation discusses the concept of compactness in relation to the set of real numbers. It is stated that in order for the real numbers to be compact, they must be both closed and bounded, and every sequence of points in the set must have a limit point within the set. However, it is mentioned that the real numbers cannot be both closed and bounded unless infinity is considered a real number. The conversation also touches on the concept of uniform continuity and its relationship to compactness. Ultimately, it is concluded that the real numbers are not compact in the usual topology, but can be compact in other topologies.
  • #1
*melinda*
86
0
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, [itex]\Re[/itex] should be compact.

However, this would require that [itex]\Re[/itex] be closed and bounded, or equivalently,

that every sequence of points in [itex]\Re[/itex] have a limit point in [itex]\Re[/itex].

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

So, are the real numbers compact?

thanks :blushing:
 
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  • #2
Just think about the sequential version for the moment, can you find a sequence in R that has no limit point in R?
 
  • #3
... or find an open cover that has no finite subcover. (i can think of two)
 
  • #4
That's just the problem...

I cannot think of a sequence in [itex]\Re[/itex] that does not have a limit point in [itex]\Re[/itex].

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:

[tex]\Re = (-\infty , \infty)[/tex].

Because of this, I keep thinking that [itex]\Re[/itex] is open.

What's my problem?
 
  • #5
*melinda* said:
I cannot think of a sequence in [itex]\Re[/itex] that does not have a limit point in [itex]\Re[/itex].

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)

*melinda* said:
Because of this, I keep thinking that [itex]\Re[/itex] is open.

They are open. Next question: are they closed?
 
  • #6
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
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 :rolleyes:
 
  • #8
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
I thought that you cannot say whether or not the real line, [itex]\mathbb{R}[/itex], 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, [itex]\mathbb{R}[/itex] is not compact because there does not exist an open cover with a finite subcover.

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

However, this would require that [itex]\mathbb{R}[/itex] be closed and bounded, or equivalently

No, not quite. A subspace of [itex]\mathbb{R}[/itex] 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 [itex]A[/itex] of that space is compact. Note: I think it is also a corollary for the Heine-Borel theorem - but I can't be sure, maybe someone can check)
 
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  • #10
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.

No, the real numbers are not compact. And you cannot say that [itex]\mathbb{R}[/itex] is compact if it is closed and bounded - only a subset of [itex]\mathbb{R}[/itex] 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 definitely not compact because it is far from being finite (this is the short answer!)
 
  • #11
Oxymoron said:
I thought that you cannot say whether or not the real line, [itex]\mathbb{R}[/itex], or any space for that matter, is compact or not without stating the topology.
For example, (I think Fourier Jr was hinting toward this),
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) {[tex] (n, n+2) | n\in \mathBB{Z}[/tex]}
b) {[tex] (-n,n) | n\in \mathbb{N}[/tex]}
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.
 
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  • #12
*melinda* said:
But I don't see how [itex]\Re[/itex] can be closed and bounded, unless we somehow decide that infinity is a real number.
there you go... answered your own question
 
  • #13
Oxymoron said:
No, not quite. A subspace of 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 of that space is compact. Note: I think it is also a corollary for the Heine-Borel theorem - but I can't be sure, maybe someone can check)
But [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] compact.
The "Stone-Cech compactification" adds two points (infinity and -infinity) together with open sets containing them so that [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] is topologically equivalent to a circle.
 
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  • #14
Oxymoron said:
No, not quite. A subspace of 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 of that space is compact. Note: I think it is also a corollary for the Heine-Borel theorem - but I can't be sure, maybe someone can check)
But [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] 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 [itex]\mathbb{R}[/itex] is topologically equivalent to a circe.
 

1. What does it mean for the real numbers to be compact?

The concept of compactness refers to the property of a set that allows it to be covered by a finite number of subsets. In the context of real numbers, this means that any infinite set of real numbers can be contained within a finite range, or interval.

2. Are the real numbers compact?

No, the real numbers are not compact. This is because they are an infinite set and cannot be contained within a finite range or interval. In other words, the real numbers have no upper or lower bound, making them non-compact.

3. How does compactness relate to continuity?

Compactness and continuity are closely related concepts in mathematics. For a function to be continuous, it must preserve the compactness of sets. This means that if a set of real numbers is compact, the image of that set under a continuous function will also be compact.

4. Can subsets of the real numbers be compact?

Yes, subsets of the real numbers can be compact. In fact, many commonly used subsets such as closed intervals, closed and bounded sets, and finite sets are all examples of compact subsets of the real numbers.

5. Why is compactness an important concept in mathematics?

Compactness is a fundamental concept in mathematics with applications in many areas such as analysis, topology, and geometry. It allows us to study infinite sets in a more manageable way and has many useful properties that help us prove theorems and solve problems.

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