Proof of The Maximum of Two Upper Semicontinuous Functions

Conventions. Elementary order facts are those of Elementary Order Arithmetic in an Ordered Field, whose claim 1 is the compatibility of the strict order with addition; the non-strict law is an axiom of the ordered field $\mathbb{R}$, whose order $\le$ is a total order, so that its reflexivity, antisymmetry, transitivity and totality are axioms of that definition.

Proof of claim 1. Let $a,b\in\mathbb{R}$; recall that $-(-a)=a$ by the additive inverse axioms of the ordered field $\mathbb{R}$, and that by claim 4 of Elementary Order Arithmetic in an Ordered Field the inequality $a\le b$ holds if and only if $-b\le-a$.

Suppose first that $a\le b$, so that $\min\{a,b\}=a$ by the definition of the minimum. If $-a\le-b$, then $b\le a$ by claim 4, hence $a=b$ by antisymmetry, and $\max\{-a,-b\}=-b=-a$ by the definition of the maximum, so $-\max\{-a,-b\}=a$. If instead $-a\le-b$ fails, then $\max\{-a,-b\}=-a$, so again $-\max\{-a,-b\}=a$. In both cases $\min\{a,b\}=-\max\{-a,-b\}$.

Suppose now that $a\le b$ fails, so that $\min\{a,b\}=b$. By totality $b\le a$, hence $-a\le-b$ by claim 4, so $\max\{-a,-b\}=-b$ and $-\max\{-a,-b\}=b=\min\{a,b\}$. This proves the identity, and applying it at each $y\in A$ to the pair $u(y),v(y)$ gives $u\wedge v=-\bigl((-u)\vee(-v)\bigr)$.

Proof of claim 2. Let $\varepsilon\in\mathbb{R}$ be positive. By upper semicontinuity of $u$ at $x$ relative to $A$ there is a positive $\delta_{u}$ such that every $y\in A$ with $d(x,y)<\delta_{u}$ satisfies $u(y)<u(x)+\varepsilon$, and likewise a positive $\delta_{v}$ for $v$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is $\delta\in\mathbb{R}$ with $\delta\le\delta_{u}$, $\delta\le\delta_{v}$ and $\delta=\delta_{u}$ or $\delta=\delta_{v}$; in either case $\delta$ is positive.

Let $y\in A$ satisfy $d(x,y)<\delta$. By the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field we get $d(x,y)<\delta_{u}$ and $d(x,y)<\delta_{v}$, hence $u(y)<u(x)+\varepsilon$ and $v(y)<v(x)+\varepsilon$. By claim 1 of Elementary Properties of the Maximum of Two Elements we have $u(x)\le(u\vee v)(x)$ and $v(x)\le(u\vee v)(x)$; adding $\varepsilon$ to both sides, which preserves $\le$ by the compatibility axiom of the ordered field $\mathbb{R}$, gives

By the mixed transitivity of claim 2 we conclude $u(y)<(u\vee v)(x)+\varepsilon$ and $v(y)<(u\vee v)(x)+\varepsilon$. By claim 2 of Elementary Properties of the Maximum of Two Elements the value $(u\vee v)(y)$ equals $u(y)$ or $v(y)$, so in either case

As $\varepsilon$ was an arbitrary positive real, $u\vee v$ is upper semicontinuous at $x$ relative to $A$. If $u$ and $v$ are upper semicontinuous on $A$, applying this at every point of $A$ shows that $u\vee v$ is upper semicontinuous on $A$.

Proof of claim 3. Assume $u$ and $v$ are lower semicontinuous at $x$ relative to $A$. By claim 1 of Negation, Restriction, and Separated Differences of Semicontinuous Functions the functions $-u$ and $-v$ are upper semicontinuous at $x$ relative to $A$, so $(-u)\vee(-v)$ is upper semicontinuous at $x$ relative to $A$ by claim 2 above, and therefore $-\bigl((-u)\vee(-v)\bigr)$ is lower semicontinuous at $x$ relative to $A$, again by claim 1 of Negation, Restriction, and Separated Differences of Semicontinuous Functions. By claim 1 this function is $u\wedge v$. The statement on $A$ follows by applying this at every point of $A$. $\blacksquare$