# Field Theory/The real numbers

Proposition (supremum commutes with continuous monotone function):

Let ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ be a continuous and monotonely increasing function, and let ${\displaystyle S\subseteq \mathbb {R} }$ be a set. Then

if ${\displaystyle S}$ is bounded from above, ${\displaystyle f(\sup S)=\sup f(S)}$ and if ${\displaystyle S}$ is bounded from below, ${\displaystyle f(\inf S)=\inf f(S)}$.

If ${\displaystyle f}$ instead is decreasing, then

if ${\displaystyle S}$ is bounded from above, ${\displaystyle f(\sup S)=\inf f(S)}$ and if ${\displaystyle S}$ is bounded from below, ${\displaystyle f(\inf S)=\sup f(S)}$

Proof: We first prove that if ${\displaystyle f}$ is increasing, then ${\displaystyle f(\sup S)=\sup f(S)}$ and ${\displaystyle f(\inf S)=\inf f(S)}$. Indeed, suppose that ${\displaystyle b=\sup S}$ and ${\displaystyle a=\inf S}$. By definition of supremum and infimum, for each ${\displaystyle \delta >0}$ the sets ${\displaystyle B_{\delta }(b)\cap S}$ and ${\displaystyle B_{\delta }(a)\cap S}$ contain some points. Hence, so do the sets ${\displaystyle f(B_{\delta }(b)\cap S)}$ and ${\displaystyle f(B_{\delta }(a)\cap S)}$. By continuity of ${\displaystyle f}$, whenever ${\displaystyle \epsilon >0}$ is arbitrary and ${\displaystyle \delta >0}$ is sufficiently small, ${\displaystyle f(B_{\delta }(a)\cap S)\subseteq B_{\epsilon }(f(a))}$ and ${\displaystyle f(B_{\delta }(b)\cap S)\subseteq B_{\epsilon }(f(b))}$. Since ${\displaystyle f(B_{\delta }(a)\cap S),f(B_{\delta }(b)\cap S)\subseteq f(S)}$, we obtain ${\displaystyle \sup f(S)\geq f(b)}$ and ${\displaystyle \inf f(S)\leq f(a)}$. On the other hand, for ${\displaystyle s\in S}$ we have ${\displaystyle f(a)\leq f(s)\leq f(b)}$ by monotonicity, so that ${\displaystyle \sup f(S)\leq f(b)}$ and ${\displaystyle \inf f(S)\geq f(a)}$.

If ${\displaystyle f}$ is decreasing instead, then ${\displaystyle g(x):=-f(x)}$ is increasing, so that ${\displaystyle -f(\sup S)=g(\sup S)=\sup g(S)=\sup -f(S)=-\inf f(S)}$. Similarly ${\displaystyle f(\inf S)=\sup f(S)}$. ${\displaystyle \Box }$