MHB Why Is the Function Bounded in the Extreme Value Theorem Proof?

[Amer]

I was reading Wiki, I met a problem in understanding the the proof of boundedness theorem exactly when they said

"Because [a,b] is bounded, the Bolzano–Weierstrass theorem implies that there exists a convergent subsequence"

but Bolzano theorem state that if the sequence is bounded, which is not necessary in our case.
What I miss here

And in the alternative proof they said
"The set {y ∈ R : y = f(x) for some x ∈ [a,b]} is a bounded set."
f is continuous at [a,b] but how should it be bounded it is clear but how to prove that ?

Thanks

 

[Fantini]

Re: Extreme vlaue theore Proof

but Bolzano theorem state that if the sequence is bounded, which is not necessary in our case.

Of course it is! If the sequence is defined in $[a,b]$, this means that for all $n \in \mathbb{N}$ we have $x_n \in [a,b]$, which in turn means that $a \leq x_n \leq b$.

As for the other, it is using the fact that if a function $f: X \to \mathbb{R}$ is continuous, then if $X$ is compact you have that $f(X)$ is compact. This of course means that $f(X) = \{ y \in \mathbb{R} : y = f(x) \text{ for some }x \in [a,b] \}$ is closed and bounded.