A metric space/topology question

[Phillips101]

The question I'm doing is as follows:

(a) Show that every compact subset of a Hausdorff space is closed. I've done this.

(b) F is a compact metric space. F is a closed subset of X, and p is any point of X. Show there is a point q in F such that d(p,q)=infimum(d(p,q')) such that q' is in F.

Suppose that for every x and y in X there is a point m such that d(x,m)=0.5*d(x,y) and d(y,m)=0.5*d(x,y). Show that X is connected.

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I don't really know how to do (b). This is my attempt at the first part of (b):

Take a sequence qn in F such that d(p,qn) decreases as n increases. Then by closedness of F, the limit of qn, say q.hat, is in F. This q.hat has the property that d(p,q.hat)=inf d(p,q') as required.

But, in the above solution, I didn't use the fact that X is compact? And I can't do the second part of (b) at all.

Thanks for any help!

 

[fourier jr]

i don't quite understand the first part as it's written there but i think i can see that it would be useful to know that in a compact metric space any sequence has a convergent subsequence. for the second part the geometrical interpretation is that there is always a point halfway between x & y, so if X is *not* connected there are disjoint open sets (etc) & get a contradiction.

 

[some_dude]

Because what you claim is not true of a closed set, the limit need not exist. However it is true for a compact set.

(Try to use a little Tex, it would make your question a lot more readable and more likely to be answered).

 

[Phillips101]

Thanks for the quick reply, and sorry, you're absolutely right the first part of (b) doesn't make sense as there's a typo. I'll make it clearer:

(b)X is a compact metric space. F is a closed subset of X, and p is any point of X. Show there is a point q in F such that d(p,q)=infimum(d(p,q')), where the infimum runs over all points q' in F.

Yeah, as you say, I thought that fact would be useful. But I don't really see any holes in my reasoning as it is *shrug*

And for the second part:
Suppose X is disconnected. Pick a connected component. This is then open and closed, as its complement is a union of open sets. Call this connected component F. Then, for every x in X\F, we have d(x,p)>0 for all p in F. So, there exists an x' and a p(x') such that this distance is minimal - ie d(x',p(x')) is the 'closest these two sets get'. And by (a), we see that p(x') is indeed in F, so d(x',p(x')) is strictly greater than zero.

I think that isn't sufficient, but I can't formalise it any more.

 

[Phillips101]

To some_dude, thanks for that clarification. And I haven't had the time to learn any Latex or Tex, but it is high on my to-do list. Thanks for the help

 

[Phillips101]

I've formalised everything and I'm very happy with the answers now - I appreciate all the help