Proof of Elementary Properties of Sequentially Strict Extrema

Claim 1. Suppose $u$ attains a sequentially strict maximum on $A$ at $\bar{x}$. For $x\in A$ we have $u(x)\le u(\bar{x})$, hence $-u(\bar{x})\le-u(x)$ by claim 3 of Elementary Arithmetic in an Ordered Field applied twice, so $(-u)(\bar{x})\le(-u)(x)$. If $(x_{m})_{m\in\mathbb{N}}$ is a sequence in $A$ such that $((-u)(x_{m}))$ converges to $(-u)(\bar{x})$, then multiplying by $-1$ and using claim 3 of Arithmetic of Limits of Real Sequences shows that $(u(x_{m}))$ converges to $u(\bar{x})$, so $(x_{m})$ converges to $\bar{x}$. Hence $-u$ attains a sequentially strict minimum on $A$ at $\bar{x}$. The converse follows by the same argument applied to $-u$ in place of $u$, since $-(-u)=u$.

Claim 2. Let $x\in A$ satisfy $u(x)=u(\bar{x})$, and let $(x_{m})_{m\in\mathbb{N}}$ be the sequence in $A$ with $x_{m}=x$ for every $m$. Then $(u(x_{m}))$ is the constant sequence with value $u(\bar{x})$, which converges to $u(\bar{x})$; hence $(x_{m})$ converges to $\bar{x}$. But $(x_{m})$ is constant with value $x$ and therefore converges to $x$, so $x=\bar{x}$ by Uniqueness of Limits in a Metric Space.

Now let $x\in A$ with $x\ne\bar{x}$. Then $u(x)\le u(\bar{x})$ and, by what we have just shown, $u(x)\ne u(\bar{x})$; hence $u(x)<u(\bar{x})$. If $u$ attains its maximum value at some $x\in A$, then $u(\bar{x})\le u(x)\le u(\bar{x})$, so $u(x)=u(\bar{x})$ and therefore $x=\bar{x}$. Since a point at which $u$ attains a sequentially strict maximum is in particular a point at which $u$ attains its maximum value, $\bar{x}$ is also the only such point.

Claim 3. Consider first $u+c$. For $x\in A$, claim 3 of Elementary Arithmetic in an Ordered Field gives $u(x)+c\le u(\bar{x})+c$. Let $(x_{m})$ be a sequence in $A$ such that $((u+c)(x_{m}))$ converges to $(u+c)(\bar{x})$. The constant sequence with value $-c$ converges to $-c$, so by claim 1 of Arithmetic of Limits of Real Sequences the sequence $(u(x_{m}))=((u+c)(x_{m})+(-c))$ converges to $u(\bar{x})+c+(-c)=u(\bar{x})$; hence $(x_{m})$ converges to $\bar{x}$.

Now consider $\lambda u$ with $0<\lambda$. For $x\in A$, claim 5 of Elementary Arithmetic in an Ordered Field gives $\lambda u(x)\le\lambda u(\bar{x})$. If $(\lambda u(x_{m}))$ converges to $\lambda u(\bar{x})$, then, $\lambda$ being nonzero, claim 3 of Arithmetic of Limits of Real Sequences applied with the factor $\lambda^{-1}$ shows that $(u(x_{m}))$ converges to $u(\bar{x})$, so $(x_{m})$ converges to $\bar{x}$.

Claim 4. For $x\in A$ we have $u(x)\le u(\bar{x})$ and $v(x)\le v(\bar{x})$, hence $(u+v)(x)\le(u+v)(\bar{x})$ by two applications of claim 3 of Elementary Arithmetic in an Ordered Field. Let $(x_{m})$ be a sequence in $A$ such that $((u+v)(x_{m}))$ converges to $(u+v)(\bar{x})$, and put

Then $\alpha_{m}\le0$ and $\beta_{m}\le0$ for every $m$, and $(\alpha_{m}+\beta_{m})$ converges to $0$ by claims 1 and 3 of Arithmetic of Limits of Real Sequences. From $\beta_{m}\le0$ and claim 3 of Elementary Arithmetic in an Ordered Field we get $\alpha_{m}+\beta_{m}\le\alpha_{m}\le0$, so claim 2 of Order Properties of Limits of Real Sequences, applied with the convergent sequences $(\alpha_{m}+\beta_{m})$ and the constant sequence $0$, shows that $(\alpha_{m})$ converges to $0$. By claims 1 and 3 of Arithmetic of Limits of Real Sequences again, $(u(x_{m}))$ converges to $u(\bar{x})$, hence $(x_{m})$ converges to $\bar{x}$.

Claim 5. For $x\in B$ we have $x\in A$, so $u|_{B}(x)=u(x)\le u(\bar{x})=u|_{B}(\bar{x})$. If $(x_{m})$ is a sequence in $B$ such that $(u|_{B}(x_{m}))$ converges to $u|_{B}(\bar{x})$, then $(x_{m})$ is a sequence in $A$ and $(u(x_{m}))$ converges to $u(\bar{x})$, so $(x_{m})$ converges to $\bar{x}$.