Proof of Coordinatewise Characterization of Continuity for Euclidean Maps

We first prove $(1)\Rightarrow(2)$. Assume that $f$ is continuous at $a$. Fix an index $j\in\{1,\dots,m\}$ and let $\varepsilon>0$. By continuity of $f$ at $a$, there exists $\delta>0$ such that whenever $x\in E$ and

$$\sum_{i=1}^n (x_i-a_i)^2<\delta^2,$$

one has

$$\sum_{r=1}^m \bigl(f_r(x)-f_r(a)\bigr)^2<\varepsilon^2.$$

Since the $j$th summand is nonnegative, it follows that

$$\bigl(f_j(x)-f_j(a)\bigr)^2<\varepsilon^2.$$

Hence $|f_j(x)-f_j(a)|<\varepsilon$. This is exactly continuity of $f_j$ at $a$ in the sense of Continuity at a Point. Since $j$ was arbitrary, every coordinate function is continuous at $a$.

Now prove $(2)\Rightarrow(1)$. Assume that each coordinate function $f_j$ is continuous at $a$ in the sense of Continuity at a Point. Let $\varepsilon>0$. For each $j\in\{1,\dots,m\}$, continuity of $f_j$ at $a$ with tolerance $\varepsilon/m$ yields a number $\delta_j>0$ such that whenever $x\in E$ and

$$\sum_{i=1}^n (x_i-a_i)^2<\delta_j^2,$$

one has

$$|f_j(x)-f_j(a)|<\frac{\varepsilon}{m}.$$

Set

$$\delta=\min\{\delta_1,\dots,\delta_m\}.$$

If $x\in E$ and

$$\sum_{i=1}^n (x_i-a_i)^2<\delta^2,$$

then in particular

$$\sum_{i=1}^n (x_i-a_i)^2<\delta_j^2 \quad\text{for every } j\in\{1,\dots,m\},$$

so

$$|f_j(x)-f_j(a)|<\frac{\varepsilon}{m} \quad\text{for all } j\in\{1,\dots,m\}.$$

Therefore,

$$\sum_{j=1}^m \bigl(f_j(x)-f_j(a)\bigr)^2 < \sum_{j=1}^m \left(\frac{\varepsilon}{m}\right)^2 = \frac{\varepsilon^2}{m} \le \varepsilon^2,$$

because $m\in\mathbb{N}$ implies $m\ge 1$. Thus $f$ is continuous at $a$ in the sense of Continuity at a Point for Maps Between Euclidean Spaces.