The Maximum of Two Upper Semicontinuous Functions

Let $(X,d)$ be a metric space, let $A\subseteq X$, and let $\mathbb{R}$ be the set of real numbers with the addition and the order $\le$ of its ordered field structure, which is a total order by that definition; for $a,b\in\mathbb{R}$ we write $a<b$ to mean that $a\le b$ and $a\ne b$, and $a-b$ for $a+(-b)$. For $a,b\in\mathbb{R}$ let $\max\{a,b\}$ be their maximum and $\min\{a,b\}$ their minimum.

Let $u,v:A\to\mathbb{R}$, and let $u\vee v:A\to\mathbb{R}$ and $u\wedge v:A\to\mathbb{R}$ be the functions whose values at $y\in A$ are

Then the following hold.

1. (Minimum through the maximum)¶ $\min\{a,b\}=-\max\{-a,-b\}$ for all $a,b\in\mathbb{R}$; consequently $u\wedge v=-\bigl((-u)\vee(-v)\bigr)$, where $-u$ is the function whose value at $y$ is $-u(y)$.

2. (Maximum of two upper semicontinuous functions)¶ Let $x\in A$. If $u$ and $v$ are upper semicontinuous at $x$ relative to $A$, then so is $u\vee v$. In particular, if $u$ and $v$ are upper semicontinuous on $A$, then so is $u\vee v$.

3. (Minimum of two lower semicontinuous functions)¶ Let $x\in A$. If $u$ and $v$ are lower semicontinuous at $x$ relative to $A$, then so is $u\wedge v$. In particular, if $u$ and $v$ are lower semicontinuous on $A$, then so is $u\wedge v$.