Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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

Let

[tex](P,\rho)[/tex]

be metric space, let

[tex]K \subset P[/tex]

be compact and let

[tex]f:\ K \rightarrow \mathbb{R}[/tex]

is continuous with respect to

[tex]K[/tex].

Then

[tex]f[/tex]

has its maximum and minimum on

[tex]K[/tex].

Proof:

[tex]f(K)[/tex]

is compact (we know from previous theorem) in

[tex]\mathbb{R} \Rightarrow f(K)[/tex]

is closed and bounded in

[tex]\mathbb{R}[/tex].

Lets put

[tex]s := \inf f(K)[/tex].

So there exists sequence.

[tex]\left\{y_n\right\} \subset K[/tex]

such that

[tex]y_n \rightarrow s[/tex]

[tex]f(K)[/tex]

is closed

[tex]\Rightarrow s \in f(K)[/tex].

Thus

[tex]\exists\ x:\ f(x) = s[/tex]

and

[tex]f[/tex]

has its minimum in [itex]s[/itex].

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

Could someone explain this to me?

Thank you!

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# Homework Help: Proof question - extremes of continous function on compact

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