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f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,

f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS

3) Will show that : a) $f(\xi)\geq\eta $ ,b) $ f(\xi)\leq\eta $ and hence : $f(\xi)=\eta$

3a) By the definition of supremum, for every positive ε, there is a number x' of S with

$\xi-\epsilon\leq x'\leq\xi$ .

For this x', f(x')<η.

Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta $ in part 3b and $f(\xi)\leq\eta$ in part 3a