Proof of Sequential Form of the Continuity of a Second-Order Equation Operator

Conventions. From the setting we use the real numbers with their order, Euclidean space with its norm and distance, and $\mathcal{S}(n)$ with its distance. Recall $d_{E}(z,w)=\lVert z-w\rVert$ and, by The Absolute Value Metric on the Real Numbers, $d_{\mathbb{R}}(s,t)=|s-t|$; the symmetry of a metric is one of the metric axioms. The order $\le$ on $\mathbb{N}$ is a total order by claims 1, 2 and 3 of Properties of the Order on the Natural Numbers, so the maximum of two natural numbers is defined and Elementary Properties of the Maximum of Two Elements applies to it.

Proof. Let $\varepsilon\in\mathbb{R}$ be positive. By clause Continuity of a Second-Order Equation Operator (i) there is a positive $\delta\in\mathbb{R}$ such that all $y\in U$, $s\in\mathbb{R}$, $q\in\mathbb{R}^{n}$ and $Y\in\mathcal{S}(n)$ satisfying

$$d_{E}(y,x_{0})<\delta,\qquad|s-r_{0}|<\delta,\qquad\lVert q-p_{0}\rVert<\delta,\qquad d_{\mathcal{S}(n)}(Y,X_{0})<\delta$$

also satisfy $\bigl|F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon$.

By convergence of the four given sequences there are $N_{1},N_{2},N_{3},N_{4}\in\mathbb{N}$ such that: $d_{E}(y_{k},x_{0})<\delta$ for every $k\ge N_{1}$; $d_{\mathbb{R}}(s_{k},r_{0})=|s_{k}-r_{0}|<\delta$ for every $k\ge N_{2}$; $d_{E}(q_{k},p_{0})=\lVert q_{k}-p_{0}\rVert<\delta$ for every $k\ge N_{3}$; and $d_{\mathcal{S}(n)}(Y_{k},X_{0})<\delta$ for every $k\ge N_{4}$.

Put $N=\max\bigl\{\max\{N_{1},N_{2}\},\max\{N_{3},N_{4}\}\bigr\}$. By claim 1 of Elementary Properties of the Maximum of Two Elements, $N_{1}\le\max\{N_{1},N_{2}\}\le N$ and likewise for $N_{2}$, $N_{3}$ and $N_{4}$, so $N_{i}\le N$ for $i=1,2,3,4$ by the transitivity of claim 1 of Properties of the Order on the Natural Numbers. Let $k\in\mathbb{N}$ satisfy $k\ge N$. Then $k\ge N_{i}$ for each $i$, again by transitivity, so all four displayed bounds hold with $y=y_{k}$, $s=s_{k}$, $q=q_{k}$ and $Y=Y_{k}$, and therefore

$$d_{\mathbb{R}}\bigl(F(y_{k},s_{k},q_{k},Y_{k}),F(x_{0},r_{0},p_{0},X_{0})\bigr)=\bigl|F(y_{k},s_{k},q_{k},Y_{k})-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon .$$

Since $\varepsilon$ was an arbitrary positive real, $\bigl(F(y_{k},s_{k},q_{k},Y_{k})\bigr)_{k\in\mathbb{N}}$ converges to $F(x_{0},r_{0},p_{0},X_{0})$ in $(\mathbb{R},d_{\mathbb{R}})$. $\blacksquare$