What is the significance of compactness?

[pob1212]

Hi

I'm trying to determine the significance of a compact subset of a metric space in relation to calculus in general.

I know the definition: A subset \small K of a metric space \small X is said to be compact if every open cover of \small K contains a finite subcover.
But what is the importance of every open cover containing a finite subcover? For example suppose I have a subset \small E of a metrics space \small X, which is open relative to \small X. I assign to each p \in E a positive number r_{p} such that the conditions d(p,q) < r_{p}, q \in X imply q \in E. This is an open cover of \small E, which does not have a finite subcover which contains \small E, correct? If this is true, then I'm thinking this must be why no open set in \small X can be compact, because we can find an open cover with no finite subcover. (the one above)
Additionally, can't we obtain an open cover of this same \small E with a finite subcover if we simply pick neighborhoods of a finite set of points in \small E with radius large enough that \small E is contained in the union of these neighborhoods? Hence the importance of every, as noted before
If this is all true, then my point is I don't see the 'big picture' importance of these observations as they relate to the development of calculus? What are the implications?

Furthermore, I know things such as every sequence in a compact metric space \small X has a subsequence which converges to a point in \small X, and the theorems about all Cauchy sequences converging if in a compact metric space.

I'm struggling to tie things together...

As always, thanks
pob

 

[algebrat]

Your definition of compact is rather abstract, and in fact, it is the definition of compact for general topologcial spaces, not just in the topci of metric spaces you are studying here.

In metric spaces, this defnition, if i recall, is equivalent to closed and bounded.

In earlier developments of math, it was found that closed and bounded sets had some desirable properties, for instance continuous functions acting on these preserved closed and boundedness.

When more general topologies were studied later, they realized the notion of closed and bounded was not generalizing desirably, so someone invented the more abstract definition you mention, where all the amazing (practical) theorems they had for closed and bounded sets, generalized to this more abstract notion of compact.

Maybe this was all sitting in the back of my head somewhere from my schooling, but looking up compact on wikipedia helped me a bit. Just sayin', check it out, you can confimr some of this, and find a few more details.

Another aspect of your question, is your discomfort with the details of the abstract definition. You might play around with the intervals (0,1), [0,∞) and [0,1], and think about why the definition works for these, to see why you want to consider an *arbitrary* open cover, and pick a fintite subcover.

 

[HallsofIvy]

I've said this before but it was a while ago: "compact" is the "next best thing to finite"!

That is because the condition that "every open cover contains a finite subcover" allows us to reduce to finiteness and so have most, if not all, of the properties of finite sets: Every finite set is closed and compact sets are closed; ever finite set is bounded (in metric spaces where we can define "bounded") and compact sets in metric spaces are bounded; every finite set in a linearly ordered space has both maximum and minimum values and so does a compact set, etc.

 

[mathwonk]

I agree with Halls, compact is a generalization of finiteness, the easiest case to deal with in any problem. the importance of finiteness is the theorems that come from it, like all continuous functions have a maximum, which is true for finite sets of values. all sequences have convergent subsequences, which also holds for finite sets. Halls already said it more clearly, just agreeing. another nice property, any continuous bijection of compact hausdorff spaces is a homeomorphism, i.e. an isomorphism. in fact it seems to me that any continuous injection of a compact n manifold to itself should be a homeomorphism, another property reminiscent of finite sets.

 

[pob1212]

I've said this before but it was a while ago: "compact" is the "next best thing to finite"!

That is because the condition that "every open cover contains a finite subcover" allows us to reduce to finiteness and so have most, if not all, of the properties of finite sets: Every finite set is closed and compact sets are closed; ever finite set is bounded (in metric spaces where we can define "bounded") and compact sets in metric spaces are bounded; every finite set in a linearly ordered space has both maximum and minimum values and so does a compact set, etc.

So what if we take the extended real line (which is compact) as the set X. And define the function f(x) = x, for all x in X. Then is the max of f ∞? If so does this not work for the real line? And is f bounded by -∞ and ∞?

In metric spaces, this defnition, if i recall, is equivalent to closed and bounded.

This is only true of R^{k}, not metric spaces in general. (The Heine-Borel Theorem)

Another aspect of your question, is your discomfort with the details of the abstract definition. You might play around with the intervals (0,1), [0,∞) and [0,1], and think about why the definition works for these, to see why you want to consider an *arbitrary* open cover, and pick a fintite subcover.

I may be misunderstanding you but (0,1) is not compact in \small R since the union of open sets V_{n} := {(1/n, 1): n≥2}, which is an open cover of (0,1), contains no finite subcover.

 

[dylanbyte]

So what if we take the extended real line (which is compact) as the set X. And define the function f(x) = x, for all x in X. Then is the max of f ∞? If so does this not work for the real line? And is f bounded by -∞ and ∞?

The extended real line is homeomorphic (isomorphic) to the unit interval [0,1] so there are no surprises there, the maximum is infinity and it achieves this maximum.

This does not work for the real line, because plus and minus infinity do not exist there.
It is equivalent to asking about the open unit interval (0,1) with the identity function, 1 does not exist in the interval, and it is not compact.

f is not really bounded by infinity, as this is not a point in the space, if we consider the real line to be a subspace of the extended reals then we can say that it is bounded, but then consider that the metric on the extended reals is not an extension of the real metric (what is an open ball around plus infinity?).