weierstrass theorem

[woolley]

Prove or counter the following statement:

If f is a continuous real-valued function on a bounded (but not necessarily closed) set A, then sup f(A) is finite. (nb. sup f(A) = {y element of R : y = f(x) for some x element of A}).

I don't know where to begin. I don't even know if I should prove or counter...I suspect that if it is closed then should prove, since weierstrass would apply. But if it is not closed, then weierstrass does not apply...so counter, right?

Here is what I know:
Existence of supremum does not guarantee existence of maximum.
If closed, bounded, and continuous, then maximum guaranteed.

That's all I got, please help!

 

[pka]

Think about this.
\(\displaystyle \L f(x) = \frac{1}{x}\;\mbox{on}\;A = (0,1]\)

 

[woolley]

supremum = 1 is finite, correct?

Is this a couterexample since it is not closed?

 

[pka]

Prove or counter the following statement:
If f is a continuous real-valued function on a bounded (but not necessarily closed) set A, then sup f(A) is finite. (nb. sup f(A) = {y element of R : y = f(x) for some x element of A}).

Look at what you posted.
A is a bounded (not necessarily closed) set. Well that is true of \(\displaystyle A = (0,1]\).
You wrote that f is a continuous function on A. Well that is true of \(\displaystyle f(x) = \frac{1}{x}\).

Now \(\displaystyle f(A) = [1,\infty )\). But what about \(\displaystyle \sup \left\{ {f(A)} \right\} = ?\).

 

[woolley]

OK, I see

sup f(A) = infinity...which is not finite.

So, this would be a sufficient counterexample?

Would it make any difference if A was closed, say [0,1]?

 

[pka]

Re: OK, I see

Would it make any difference if A was closed, say [0,1]?

Yes certainly, f could not be continuous.
The continuous image of a compact set is compact.