Is Every Continuous Real-Valued Function on a Subset of R Bounded?

[Useful nucleus]

I'm seeking a neater proof for the following:
Let E \subseteq R. Let every continuous real-valued function on E be bounded. Show that E is compact.
I tried to argue based on Heine-Borel 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.

 

[Citan Uzuki]

E cannot be unbounded because if it is the case, define f(x)=x on E and f(x) is continuous but unbounded.

This part is okay.

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.

This part is not. You've shown that the specific non-closed set (0, 1] admits an unbounded continuous real-valued function, but you have said nothing about the other \beth_2 non-closed subsets of R.

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/(x-c)?

 

[Useful nucleus]

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/(x-c) 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.

 

[Bacle2]

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.

 

[Useful nucleus]

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

 

[Bacle2]

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/(x-c) is unbounded in E.

 

[Useful nucleus]

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/(x-c) is unbounded in E.

Exactly! What I'm looking for is a more elegant way to prove the statement (if any) and to see whether there is a flaw in current proof we have in hand.

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

 

[Citan Uzuki]

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 Heine-Borel theorem (compactness is equivalent to closed and bounded). So the only way to avoid it would be to avoid using the Heine-Borel theorem altogether. But the result you're trying to prove (that pseudocompactness implies compactness) is not true for general topological spaces, so [strike]you can't avoid invoking Heine-Borel[/strike]. As such, I think that this is actually the most elegant proof possible.

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 Heine-Borel 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.

 

[Useful nucleus]

but sometimes that can be immensely satisfying.

Completely agree!

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

 

[SteveL27]

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!