域论/实数

命题（上确界与连续单调函数交换）:

设 $f:\mathbb {R} \to \mathbb {R}$ 为连续且单调递增函数，设 $S\subseteq \mathbb {R}$ 为一个集合。则

如果 $S$ 有上界，$f(\sup S)=\sup f(S)$；如果 $S$ 有下界，$f(\inf S)=\inf f(S)$.

如果 $f$ 是递减函数，则

如果 $S$ 有上界，$f(\sup S)=\inf f(S)$；如果 $S$ 有下界，$f(\inf S)=\sup f(S)$

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 $\inf f(S)\geq f(a)$.

如果 $f$ 是递减函数，则 $g(x):=-f(x)$ 是递增函数，因此 $-f(\sup S)=g(\sup S)=\sup g(S)=\sup -f(S)=-\inf f(S)$。类似地 $f(\inf S)=\sup f(S)$。 $\Box$