Proof of Semicontinuity Under Negation and Characterization of Continuity

By The Absolute Value Metric on the Real Line the metric $d_{\mathbb{R}}$ is given by $d_{\mathbb{R}}(s,t)=|s-t|$, with the absolute value $|\cdot|$ on $\mathbb{R}$. We use the elementary order arithmetic of the ordered field $\mathbb{R}$ and the properties of the absolute value; each claim number below is attributed to one of these two lemmas explicitly.

Step 0 (a strict two-sided bound). For $a,c\in\mathbb{R}$, the inequality $|a|<c$ holds if and only if both $-c<a$ and $a<c$ hold.

Suppose $|a|<c$. Claim 3 of Properties of the Absolute Value in an Ordered Field gives $a\le|a|$, so claim 2 of Elementary Order Arithmetic in an Ordered Field gives $a<c$. The same claim 3 gives $-|a|\le a$, while claim 4 of Elementary Order Arithmetic in an Ordered Field applied to $|a|<c$ gives $-c<-|a|$; claim 2 of Elementary Order Arithmetic in an Ordered Field then gives $-c<a$.

Conversely suppose $-c<a$ and $a<c$. Claim 1 of Properties of the Absolute Value in an Ordered Field gives that $|a|$ equals $a$ or $-a$. If $|a|=a$, then $|a|=a<c$. If $|a|=-a$, then claim 4 of Elementary Order Arithmetic in an Ordered Field applied to $-c<a$ gives $-a<c$, so $|a|=-a<c$.

Step 1 (claim 1). Fix $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, fix $\delta\in\mathbb{R}$ with $0<\delta$, and let $y\in A$. Adding the element $u(x)+u(y)-\varepsilon$ to both sides of an inequality is an equivalence by claim 1 of Elementary Order Arithmetic in an Ordered Field. Adding it to the left-hand side of

gives $-u(y)+u(x)+u(y)-\varepsilon=u(x)-\varepsilon$, and adding it to the right-hand side gives $-u(x)+\varepsilon+u(x)+u(y)-\varepsilon=u(y)$. Hence that inequality holds if and only if

holds. Consequently, for this $\varepsilon$ and this $\delta$, the condition that every $y\in A$ with $d(x,y)<\delta$ satisfies $(-u)(y)<(-u)(x)+\varepsilon$ is the very same condition as the condition that every such $y$ satisfies $u(x)-\varepsilon<u(y)$. Quantifying over all $\varepsilon$ with $0<\varepsilon$ and all $\delta$ with $0<\delta$, the assertion that $-u$ is upper semicontinuous at $x$ relative to $A$ is the assertion that $u$ is lower semicontinuous at $x$ relative to $A$.

Step 2 (claim 2). By the definition of a continuous map applied to $u$ as a map from $A$ into $(\mathbb{R},d_{\mathbb{R}})$, continuity of $u$ at $x$ relative to $A$ says that for every $\varepsilon$ with $0<\varepsilon$ there is $\delta$ with $0<\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $|u(y)-u(x)|<\varepsilon$. By Step 0 with $a=u(y)-u(x)$ and $c=\varepsilon$, this last inequality holds if and only if both $-\varepsilon<u(y)-u(x)$ and $u(y)-u(x)<\varepsilon$ hold; adding $u(x)$ to both sides of each and using claim 1 of Elementary Order Arithmetic in an Ordered Field, these two are equivalent to $u(x)-\varepsilon<u(y)$ and $u(y)<u(x)+\varepsilon$ respectively.

Suppose $u$ is continuous at $x$ relative to $A$, and let $\varepsilon$ with $0<\varepsilon$ be given. The $\delta$ furnished by continuity for this $\varepsilon$ witnesses simultaneously the defining condition of upper semicontinuity and that of lower semicontinuity at $x$ for this $\varepsilon$. Hence $u$ is both upper and lower semicontinuous at $x$ relative to $A$.

Conversely suppose $u$ is both upper and lower semicontinuous at $x$ relative to $A$, and let $\varepsilon$ with $0<\varepsilon$ be given. Upper semicontinuity provides $\delta_1$ with $0<\delta_1$, and lower semicontinuity provides $\delta_2$ with $0<\delta_2$, for this $\varepsilon$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\delta\in\mathbb{R}$ with $\delta\le\delta_1$, $\delta\le\delta_2$, and $\delta$ equal to $\delta_1$ or to $\delta_2$; in either case $0<\delta$. Let $y\in A$ satisfy $d(x,y)<\delta$. Claim 2 of Elementary Order Arithmetic in an Ordered Field gives $d(x,y)<\delta_1$ and $d(x,y)<\delta_2$, so $u(y)<u(x)+\varepsilon$ and $u(x)-\varepsilon<u(y)$, and therefore $|u(y)-u(x)|<\varepsilon$ by the equivalences of the previous paragraph. Hence $u$ is continuous at $x$ relative to $A$.

Step 3 (the final assertion). Continuity on $A$, upper semicontinuity on $A$, and lower semicontinuity on $A$ are each defined as the corresponding property at every point of $A$ relative to $A$. So applying claim 2 at each $x\in A$ gives that $u$ is continuous on $A$ if and only if $u$ is both upper semicontinuous on $A$ and lower semicontinuous on $A$.