Proof about compact metric spaces.

[Demon117]

1. Let M be a compact metric space. If r>0, show that there is a finite set S in M such that every point of M is within r units of some point in S.



2. Relevant theorems & Definitions:

-Every compact set is closed and bounded.

-A subset S of a metric space M is sequentially compact if every sequence in S has a subsequence that converges to a limit in S.

-If every open covering of S reduces to a finite subcovering then we say that S is covering compact.



3. Attempt at the solution

To begin with I don't completely understand coverings. So I am unsure whether I should use the sequential definition or the covering definition. Could someone please explain coverings?

I also wondered if S is a subset of a compact metric space, does it follow that S is compact?

INFORMAL PROOF: Since M is compact, it follows that every sequence in M has a convergent subsequence in M. Need to show that for any point in M is within some distance from a point in S. Let $$m_{k_{j}}$$ be such a subsequence. So for $$\epsilon>0$$ there exists N such that $$m_{k_{j}}$$ is within $$\epsilon$$ units of some point, call it m.

Not sure where to go from here.

 

[radou]

What's the definition of compactness? Consider the collection {B(x, r) : x is in M and r > 0}.

 

[radou]

Regarding your second question, this is not true. But it is true that a closed subspace of a compact space is itself compact.

Edit: a simple counterexample is the closed unit interval [0, 1] which is compact, but <0, 1> is a subset of [0, 1] which is not compact.

 

[Demon117]

What's the definition of compactness? Consider the collection {B(x, r) : x is in M and r > 0}.

The definition for sequential compactness is that for any sequence in M there is a subsequence which converges to some point in M.

If the collection {B(x,r)|x in M and r>0} is contained in M, could we construct some sequence which converges to the collection? I am not sure what you mean by this collection entirely.