Prove that $f(s) = \sup\{x \in [a,b] : f(x) > x\}$, where $f$ is increasing

Question

Given $f: [a,b] \rightarrow [a,b]$ increasing, where $f(a) > a$ and $A = \{x \in [a,b] : f(x) > x\}. \ $ If $s = \sup(A), $ prove that $f(s) = s$.

In order to show that $f(s) = s$, I was trying to prove that $f(s) \leq s$ and $f(s) \geq s$.

Can I simply state that since $f$ is an increasing function then $f(s) \geq s$? How can I prove that $f(s) \leq s$?

Answer 1

You've shown $f(s)\geq s.$

To summarize, since it is in comments only:

For $x\in A,$ $s\geq x$ and thus $f(s)\geq f(x)>x,$ so $f(s)\geq \sup A=s.$

If $f(s)>s,$ then in particular, $s<b,$ since $f(b)\leq b.$ Let $\delta=\frac{f(s)-s}2>0.$ Then, since $f(s)\leq b,$ we have $$s+\delta\leq s+\frac{b-s}2=\frac{s+b}2<b.$$ So $$f(s+\delta)\geq f(s)=s+(f(s)-s)> s+\delta,$$ which means $s+\delta\in A,$ contradicting $s=\sup A.$

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Essentially, we are proving that for each $x\in A,$ there is an $x'>x$ with $x'\in A,$ so the supremum cannot be in $A.$