I want proof of this theorem:
Let $A$ and $B$ non empty subsets of $\mathbb{R}$ with $A\subseteq B$; suppose that $\sup(A)$ and $\sup f(A)$ exist and $\sup A\in B$, let $f\colon B\longrightarrow\mathbb{R}$ a function continuous in $\sup(A)$. Proof that $f(\sup(A))\leq\sup (f(A))$.
Here is my attempt:
There are two cases:
- Let $\sup(A)\in A$; then $f(\sup(A))\in f(A)$, so $f(\sup(A))\leq\sup (f(A))$.
- Let $\sup(A)\not\in A$. By hypothesis $f$ is continuous in $\sup A$. I call $c=\sup A$. By continuity $\forall\epsilon>0 \exists\delta>0 : \forall x\in B : |x-c|<\delta\Rightarrow |f(c)-f(x)|<\epsilon$.
By $\sup$ properties i know that $\forall\delta>0\exists a\in A : c-\delta<a\leq c$; so obviously $\forall\delta>0\exists a\in A : a\in (c-\delta,c+\delta)$. Using continuity we have:
$\forall\epsilon>0 \exists\delta>0 : \forall a\in A : |x-c|<\delta\Rightarrow |f(c)-f(a)|<\epsilon$
So: $f(c)-f(a)<\epsilon$, and $f(c)<f(a)+\epsilon$
remember that $c=\sup A$, we have:
$f(\sup A)<f(a)+\epsilon$; this inequality is true $\forall a\in A\cap (c-\delta,c+\delta)$; so:
$f(\sup A)<\sup(f(A))+\epsilon$
But $\epsilon>0$ is chosen arbitrarily, so finally:
$f(\sup A)<\sup(f(A))$
Thanks for your replies