
#1
Apr312, 02:16 PM

P: 240

I'm seeking a neater proof for the following:
Let E [itex]\subseteq[/itex] R. Let every continuous realvalued function on E be bounded. Show that E is compact. I tried to argue based on HeineBorel theorem as follows: E cannot be unbounded because if it is the case, define f(x)=x on E and f(x) is continuous but unbounded. E cannot be not closed, because E=(0,1] is not closed and we can define f(x)=1/x which is continuous but still unbounded. Thus, E has to be both bounded and closed and hence compact. I think this argument is not very solid and I'd appreciate any hints. 



#2
Apr312, 02:38 PM

PF Gold
P: 274

Here's a hint: if a set is closed, then there is a limit point c not in the set. What can you say about 1/(xc)? 



#3
Apr312, 02:51 PM

P: 240

I see your point. So it would be better to say the following:
Let E be not closed by missing the limit point c. Then f(x)=1/(xc) is continuous on E but clearly undbounded. I realize now that the argument has improved a lot, but is there a neater way to avoid the the dichotomy I made. 



#4
Apr312, 03:09 PM

Sci Advisor
P: 1,168

Continuity and Compactness
What dichotomy are you referring to? Do you mean treating the cases of E being
closed and E being bounded? If so, only way of avoiding it I can think of is using covers and subcovers, but I don't see how that would work. The arguments, yours and Uzuki's, are perfectly fine IMHO, tho. 



#5
Apr312, 03:16 PM

P: 240

By dichotomy I refer to examining each case of E (according to HeineBorel theorem) and giving a counter example for each.




#6
Apr312, 03:30 PM

Sci Advisor
P: 1,168

Didn't you just do that?
Using the arguments you and Uzuki gave: i)Assume E not bounded. Then f(x)=x is unbounded ii)Assume E not closed , then there is a limit point c of E not contained in E. Then f(x)=1/(xc) is unbounded in E. 



#7
Apr312, 03:38 PM

P: 240

May be classifying this proof as proof by dichotomy was not the best way to describe it. 



#8
Apr312, 05:17 PM

PF Gold
P: 274

As it stands now your proof is fine. In fact, this is the proof that I would use . The fact that you have to divide into two cases is built right into the statement of the HeineBorel theorem (compactness is equivalent to closed and bounded). So the only way to avoid it would be to avoid using the HeineBorel theorem altogether. But the result you're trying to prove (that pseudocompactness implies compactness) is not true for general topological spaces, so
Edit: Actually, I'm wrong about that. You do have to use the subset of R condition, but you don't have to go through HeineBorel to do it. You could use the equivalence (in metric spaces) of compactness and limit point compactness, combined with the Tietze extension theorem. This has the nice advantage of generalizing quickly to all metric spaces. Of course if you only want to prove it for subsets of R, invoking the Teitze extension theorem is like using a sledgehammer to kill a mosquito, but sometimes that can be immensely satisfying. 



#9
Apr412, 01:06 AM

P: 240

Thank you for your comment. I'm not familiar yet with Teitze extension theorem, but hopefully in the near future I will be. 



#10
Apr412, 02:28 AM

P: 799

I happen to remember something obscure about this subject. If E is a topological space with the property that every continuous function from E to the reals is bounded, is E compact?
The answer is no. A space with the above property is pseudocompact. http://en.wikipedia.org/wiki/Pseudocompact_space An example of a psuedocompact space that's not compact is any infinite set with the particular point topology. http://en.wikipedia.org/wiki/Particular_point_topology (edit) I see that Citan Uzuki mentioned psuedocompactness earlier. Good knowledge! 


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