Let $f\ [0,1]\longrightarrow [0,1]$ be increasing function. let: $$E=\{x\in [0,1] \mid f(x)\geq x \} $$
Show that $E$ has a supermum $b$ and that $f(b)= b$.
we have $x\leq 1$ since $f$ is inceasing $f(x)\leq f(1)$ and $x\leq f(x)$
i don't know how to answer that question any help would be apppreciated
First note that $0\in E$, since $f(0)\geq0$, so $E$ is not empty. Moreover, since $E\subset[0,1]$, it is bounded. Thus, $b:=\sup E$ exists. Then we find a sequence $(x_n)\subset E$ with $x_n\to b$ and $x_n\leq b$. Since $f$ is increasing, $f(b)\geq f(x_n)\geq x_n\to b$, so $f(b)\geq b$. If $b=1$, then $f(1)=1$, since $f(1)\leq 1$. Otherwise ($b<1$) we have $f(x)<x$ for each $b<x<1$, and again since $f$ is increasing, we have for these $x$ that $f(b)\leq f(x)<x$. Letting $x\to b$ finally yields $f(b)=b$.