Proof question - extremes of continous function on compact

1. Jan 20, 2006

twoflower

Hi all,

you all sure know this theorem. We've it as follows:

Let

$$(P,\rho)$$

be metric space, let

$$K \subset P$$

be compact and let

$$f:\ K \rightarrow \mathbb{R}$$

is continuous with respect to

$$K$$.

Then

$$f$$

has its maximum and minimum on

$$K$$.

Proof:

$$f(K)$$

is compact (we know from previous theorem) in

$$\mathbb{R} \Rightarrow f(K)$$

is closed and bounded in

$$\mathbb{R}$$.

Lets put

$$s := \inf f(K)$$.

So there exists sequence

$$\left\{y_n\right\} \subset K$$

such that

$$y_n \rightarrow s$$
.

$$f(K)$$

is closed

$$\Rightarrow s \in f(K)$$.

Thus

$$\exists\ x:\ f(x) = s$$

and

$$f$$

has its minimum in $s$.

I don't understand why the bold part of the proof is necessary there..What I actually want is that $s \in f(K)$. But I think that it's assured by the fact that $f(K)$ is closed. Am I wrong?

Could someone explain this to me?

Thank you!

Last edited: Jan 20, 2006
2. Jan 20, 2006

matt grime

fix your latex. separate lines for each entry seems to work best.

it doesn't seem necessary to pick a sequence: any compact set in R has a max, M and necessarily there is an x such that f(x)=M, however this seems to be simply what you're proving, quite possibly unnecessarily. Have you actually proved that a closed bounded subset of Rhas a maximum (or minimum)?

But then I'm supposing that this is the proof given in lectures. It's nice that students think lecturers are infallible, but it is certainly not true.

Last edited: Jan 20, 2006
3. Jan 20, 2006

StatusX

The idea is that the infimum of the set has points in the set arbitrarily close to it, ie, it is a limit point of the set, and since the set is closed, the infimum must therefore belong to it.

4. Jan 20, 2006

twoflower

I think the proof is ok even with the note about the sequence, anyway, I found it kind of superfluous here and rather confusing. So I just wanted to ask whether it will be ok to exclude it while not breaking the validity at the same time.

Yes, that's it. Since the set is closed, it must belong to it. If it didn't, it would be contradiction with the closedness property. That's why I found it not necessary.

Ok, maybe I'm just making too much science of it (sorry for terrible expression, I just don't know how to translate this phrase into English).

5. Jan 21, 2006

matt grime

How can you find a proof ok yet think it contains an unecessary, superfluous and confusing segment? I was pointing out that it is only unnecessary if you cite the result

"compact means closed and bounded, hence it contains its sup and inf'

in the process of his proof he proved that it contains the sup and inf directly, ie he didn't leave any gaps in his argument.