# How can I use compactness?

## Homework Statement

Let K be a compact subset of a metric space M, and let {U_α}_α∈I be an open cover of
K. Show that there is a positive real number δ with the property that for every x ∈ K there is
some α ∈ A with B_δ (x) ⊆ U_α.

n/a

## The Attempt at a Solution

Since U_α are open, every point of them is a an interior point. For any positive δ, B_δ (x) $\subset$ U_α, for some α?

I'm lost here, anything I think seems trivial...

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Dick
Homework Helper
Define f(x) for x in K to be the diameter of the largest ball around x that is contained in one of the U's. Can you tell me why f(x) is positive? Now define the set A_n to be the set of all x such that f(x)>1/n. Do the A_n cover K?

Define f(x) for x in K to be the diameter of the largest ball around x that is contained in one of the U's. Can you tell me why f(x) is positive? Now define the set A_n to be the set of all x such that f(x)>1/n. Do the A_n cover K?
f(x) is positive because it's a distance?

I can't decide if A_n cover K because K may be the set of all integers and n may be too large to cover the gaps between the integers.

Actually I'm still not sure what's wrong with my proof

Oh I was totally wrong, but I still don't understand your argument

Bacle2
I think Dick was going for finding a finite subcover of the cover of all such balls,all

of which have positive diameter. Re A_n covering: is the diameter of any given U_n

larger than 1/n for given n? (to be very formal, this is the Archimedean Property.)

I've started to learn these kind of things a month ago but I still don't get the Archimedean property and can't use it in proofs. Maybe mathematics is unfortunately not my thing afterall...

Bacle2
I've started to learn these kind of things a month ago but I still don't get the Archimedean property and can't use it in proofs. Maybe mathematics is unfortunately not my thing afterall...
Don't take it so hard, we all have problems at some point. Besides, I think it was overkill

on my part, sorry. Try to see why the diameter is larger than 1/n for some n.

HallsofIvy
Homework Helper
The infinite sequence 1/n, for n a positive integer, goes to 0 so it is NOT true that there is a positive lower bound. However, since the set is compact, there exist a finite subcover, giving a finite subset of {1/n} and every finite set of numbers has a smallest number.

Dick
Homework Helper
For a given x, f(x)>0 because x is contained in SOME U_a. So there is a ball of some diameter d>0 centered on x and contained in U_a (definition of an open set). So f(x) must be greater than or equal to d>0. Now think about how the sets A_n are related to each other. Which is larger A_2, or A_3?

Would not it be easier to just say that since f(x) is never zero on the compact set, it reaches a positive minimum there?

Dick
Homework Helper
Would not it be easier to just say that since f(x) is never zero on the compact set, it reaches a positive minimum there?
If you have that theorem, that would be a nice way to prove it.

Define f(x) for x in K to be the diameter of the largest ball around x that is contained in one of the U's. Can you tell me why f(x) is positive? Now define the set A_n to be the set of all x such that f(x)>1/n. Do the A_n cover K?
How do we know that such a maximum diameter exists? If K is a compact subset of the real line, and one of the U's is the entire space, then there is no such maximum, right? (maybe not, I don't know, but clarification would be helpful for me).

Can we just exclude this case (generalized to metric spaces), because the theorem is clear for the case when {Ua} contains such a set in which any ball around x is contained (I think, because any real-valued delta will do the job)?

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Can we just exclude this case (generalized to metric spaces), because the theorem is clear for the case when {Ua} contains such a set in which any ball around x is contained (I think, because any real-valued delta will do the job)?
Exactly.