# Field Theory/The real numbers

Proposition (supremum commutes with continuous monotone function):

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

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

If $f$ instead is decreasing, then

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

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