Proof of Coordinatewise Criterion for Continuity of a Map Between Euclidean Spaces

All sums below are finite sums in the field of real numbers, and $|\,\cdot\,|$ is the absolute value. Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm. Unwinding Continuity at a Point for Maps Between Euclidean Spaces for a single coordinate function, $f_j$ is continuous at $a$ precisely when for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $x\in E$ with $\sum_{i=1}^{n}(x_i-a_i)^2<\delta^2$ satisfies $(f_j(x)-f_j(a))^2<\varepsilon^2$.

Necessity. Suppose $f$ is continuous at $a$ and fix $j$. Let $\varepsilon>0$ be real and let $\delta>0$ be as in Continuity at a Point for Maps Between Euclidean Spaces for $f$ and $\varepsilon$. Let $x\in E$ satisfy $\sum_{i=1}^{n}(x_i-a_i)^2<\delta^2$ and put $z=f(x)-f(a)\in\mathbb{R}^m$, so that $z_l=f_l(x)-f_l(a)$ for every $l$. Then $\sum_{l=1}^{m}z_l^2<\varepsilon^2$, and by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square this sum is $\lVert z\rVert^2$, with $0\le\lVert z\rVert$. Since $0<\varepsilon$, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\lVert z\rVert<\varepsilon$, and claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square gives $|z_j|\le\lVert z\rVert$, so $|z_j|<\varepsilon$ and, by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field again, $|z_j|\cdot|z_j|<\varepsilon^2$. Finally $z_j^2\le|z_j^2|=|z_j|\cdot|z_j|$ by claims 3 and 4 of Properties of the Absolute Value in an Ordered Field, so $(f_j(x)-f_j(a))^2=z_j^2<\varepsilon^2$. Hence $f_j$ is continuous at $a$.

Sufficiency. Suppose every $f_j$ is continuous at $a$. Put $S=\sum_{l=1}^{m}1$. Every summand equals $1$, which is nonnegative, so claim 6 of Properties of Finite Sums gives $1\le S$; in particular $0<S$, and $0<S\cdot S$ by claim 5 of Elementary Order Arithmetic in an Ordered Field.

Let $\varepsilon>0$ be real and put $\varepsilon_0=\varepsilon/(1+S)$, a positive real number. For each $l\in\{1,\dots,m\}$ choose, by the continuity of $f_l$ at $a$, a real $\delta_l>0$ such that every $x\in E$ with $\sum_{i=1}^{n}(x_i-a_i)^2<\delta_l^2$ satisfies $(f_l(x)-f_l(a))^2<\varepsilon_0^2$. Let $\delta$ be the least of $\delta_1,\dots,\delta_m$, a positive real number; since $0<\delta\le\delta_l$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\delta^2\le\delta_l^2$ for every $l$.

Let $x\in E$ satisfy $\sum_{i=1}^{n}(x_i-a_i)^2<\delta^2$. Then that sum is smaller than $\delta_l^2$ for every $l$, so $(f_l(x)-f_l(a))^2<\varepsilon_0^2$ for every $l$. By claim 1 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, applied with the summands $(f_l(x)-f_l(a))^2$ and the constant summands $\varepsilon_0^2$, and by the homogeneity of finite sums (claim 3 of Properties of Finite Sums),

$$\sum_{l=1}^{m}\bigl(f_l(x)-f_l(a)\bigr)^2\ \le\ \sum_{l=1}^{m}\varepsilon_0^2\ =\ \varepsilon_0^2\,S .$$

Now $(1+S)\cdot(1+S)=1+S+S+S\cdot S$, and $1+S+S\cdot S$ is positive, so $S<(1+S)\cdot(1+S)$; multiplying by the positive number $\varepsilon^2/\bigl((1+S)(1+S)\bigr)$ and using $\varepsilon_0^2=\varepsilon^2/\bigl((1+S)(1+S)\bigr)$ gives $\varepsilon_0^2\,S<\varepsilon^2$, the multiplication of a strict inequality by a positive number being claim 10 of Elementary Order Arithmetic in an Ordered Field. Hence

$$\sum_{l=1}^{m}\bigl(f_l(x)-f_l(a)\bigr)^2<\varepsilon^2,$$

which is the condition of Continuity at a Point for Maps Between Euclidean Spaces for $f$ at $a$ with the tolerance $\varepsilon$ and the threshold $\delta$. Therefore $f$ is continuous at $a$. $\blacksquare$