# Extreme value theorem Proof

#### Amer

##### Active member
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 {yR : 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

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#### Fantini

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.