Author |
Topic: Compact??? (Read 781 times) |
|
desi
Newbie
Gender: 
Posts: 5
|
Easy one.
Prove/Disprove
If X is a metric space such that every continuous function f:X->R is bounded then X is compact.
|
|
 IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
 |
Re: Compact???
« Reply #1 on: Mar 19th, 2006, 3:19pm » |
 Quote  Modify
|
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?
|
| « Last Edit: Mar 19th, 2006, 3:27pm by Obob » |
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Compact???
« Reply #2 on: Mar 20th, 2006, 3:23pm » |
Quote Modify
|
on Mar 19th, 2006, 3:19pm, 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.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: Compact???
« Reply #3 on: Mar 20th, 2006, 5:38pm » |
Quote Modify
|
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: Urysohn's Lemma.
|
| « Last Edit: Mar 20th, 2006, 6:11pm by Obob » |
IP Logged |
|
|
|
desi
Newbie
Gender:
Posts: 5
|
|
Re: Compact???
« Reply #4 on: Mar 20th, 2006, 8:53pm » |
Quote Modify
|
A hint : | hidden: | | use tietze's extension theorem |
|
|
IP Logged |
|
|
|
Obob
Senior Riddler
Gender:
Posts: 489
|
|
Re: Compact???
« Reply #5 on: Mar 20th, 2006, 9:11pm » |
Quote Modify
|
Eh, is an awfully strong result to solve such a simple problem. It does make this totally trivial, though.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print