wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> Compact???
        
(Message started by: desi on Mar 19^th, 2006, 2:41pm)

Title: Compact???
Post by desi on Mar 19^th, 2006, 2:41pm

Easy one.
Prove/Disprove
If X is a metric space such that every continuous function f:X->R is bounded then X is compact.

Title: Re: Compact???
Post by Obob on Mar 19^th, 2006, 3:19pm

Consider the metric space of rational numbers between 0 and 1. Any continuous function from this space to R has a unique extension to a continuous function [0,1] -> R, and is therefore bounded. However, this space is not compact since a metric space is compact iff it is complete and totally bounded, and it is obviously not complete.

Follow up, also easy question: If every continuous real-valued function on X is bounded, must X be bounded?

Maybe slightly harder: must X be totally bounded?

Title: Re: Compact???
Post by Icarus on Mar 20^th, 2006, 3:23pm

on 03/19/06 at 15:19:37, Obob wrote:

Consider the metric space of rational numbers between 0 and 1. Any continuous function from this space to R has a unique extension to a continuous function [0,1] -> R, and is therefore bounded.

f(x) = 1/(x-r), where r is an irrational number between 0 and 1, is continuous on Q, but unbounded.

Title: Re: Compact???
Post by Obob on Mar 20^th, 2006, 5:38pm

Woops, you are of course correct Icarus. We can only extend every uniformly continuous map, and it is already obvious that the image of a bounded metric space under a uniformly continuous map is bounded.

We can actually turn Icarus' idea into a proof that X must be complete. For suppose that X is not complete, and that Y is the completion of X. Pick y in Y \ X, and define f(x)=1/d(x,y). This can be viewed as the composition Y \ {y}->R->R of the maps x->d(x,y) and a->1/a, so it is continuous on Y \ {y}. In particular it is continuous on the dense subset X of Y, and it is not bounded since there is a sequence in X converging to y.

Therefore we are reduced to the earlier question I posed, about whether or not X must be totally bounded. Any takers? Hint: [hide]Urysohn's Lemma.[/hide]

Title: Re: Compact???
Post by desi on Mar 20^th, 2006, 8:53pm

A hint : [hideb] use tietze's extension theorem [/hideb]

Title: Re: Compact???
Post by Obob on Mar 20^th, 2006, 9:11pm

Eh, [hideb]Tietze[/hideb] is an awfully strong result to solve such a simple problem. It does make this totally trivial, though.