Proof of Semicontinuity via Sublevel and Superlevel Sets

Throughout, $B_{d_A}(x,\delta)$ denotes the open ball in $(A,d_A)$, so that $B_{d_A}(x,\delta)=\{y\in A: d(x,y)<\delta\}$ because $d_A$ agrees with $d$ on $A$. We use the order arithmetic of Elementary Order Arithmetic in an Ordered Field and the fact that the order $\le$ of $\mathbb{R}$ is a total order, hence reflexive, antisymmetric and comparing any two elements. We write $s-t$ for $s+(-t)$.

Step 0: complements. Let $y\in A$ and $c\in\mathbb{R}$. The conditions $u(y)<c$ and $c\le u(y)$ cannot both hold, since together with antisymmetry they would give $u(y)=c$, contradicting $u(y)\ne c$. If $u(y)<c$ fails, then either $u(y)\le c$ fails, in which case comparability gives $c\le u(y)$, or $u(y)\le c$ holds and $u(y)=c$, in which case $c\le u(y)$ by reflexivity. Hence exactly one of the two conditions holds, and

the second identity following in the same way.

Claim 1, necessity. Suppose $u$ is upper semicontinuous on $A$, and let $c\in\mathbb{R}$ and $x\in A_{u<c}$. From $u(x)<c$, claim 1 of Elementary Order Arithmetic in an Ordered Field, applied by adding $-u(x)$, gives $0<c-u(x)$. Applying the defining condition of upper semicontinuity at $x$ with $\varepsilon=c-u(x)$ produces $\delta$ with $0<\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies

Thus $B_{d_A}(x,\delta)\subseteq A_{u<c}$. As $x$ was arbitrary, $A_{u<c}$ is open in $(A,d_A)$.

Claim 1, sufficiency. Suppose $A_{u<c}$ is open in $(A,d_A)$ for every $c\in\mathbb{R}$. Let $x\in A$ and let $\varepsilon$ satisfy $0<\varepsilon$. Put $c=u(x)+\varepsilon$; adding $u(x)$ to $0<\varepsilon$ gives $u(x)<c$ by claim 1 of Elementary Order Arithmetic in an Ordered Field, so $x\in A_{u<c}$. By openness there is $\delta$ with $0<\delta$ and $B_{d_A}(x,\delta)\subseteq A_{u<c}$, that is, every $y\in A$ with $d(x,y)<\delta$ satisfies $u(y)<u(x)+\varepsilon$. Hence $u$ is upper semicontinuous at $x$ relative to $A$, and, $x$ being arbitrary, on $A$.

Claim 2, necessity. Suppose $u$ is lower semicontinuous on $A$, and let $c\in\mathbb{R}$ and $x\in A_{c<u}$. From $c<u(x)$, adding $-c$ gives $0<u(x)-c$. Applying the defining condition of lower semicontinuity at $x$ with $\varepsilon=u(x)-c$ produces $\delta$ with $0<\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies

Thus $B_{d_A}(x,\delta)\subseteq A_{c<u}$, so $A_{c<u}$ is open in $(A,d_A)$.

Claim 2, sufficiency. Let $x\in A$ and $0<\varepsilon$. Put $c=u(x)-\varepsilon$; adding $u(x)$ to $-\varepsilon<0$, which holds by claim 4 of Elementary Order Arithmetic in an Ordered Field, gives $c<u(x)$, so $x\in A_{c<u}$. By openness there is $\delta$ with $0<\delta$ and $B_{d_A}(x,\delta)\subseteq A_{c<u}$, that is, every $y\in A$ with $d(x,y)<\delta$ satisfies $u(x)-\varepsilon<u(y)$. Hence $u$ is lower semicontinuous at $x$ relative to $A$, and on $A$.

Claim 3. Suppose $u$ is upper semicontinuous on $A$ and let $c\in\mathbb{R}$. By claim 1 the set $A_{u<c}$ is open in $(A,d_A)$, hence belongs to the topology of $A$, and by Step 0 the set $A_{c\le u}$ is its complement relative to $A$. By the definition of a closed subset, $A_{c\le u}$ is closed in $A$. The lower semicontinuous case is identical, using claim 2 and the second identity of Step 0.