Proof of theorem 4.10 in Walter Rudin Analysis I have managed to prove the theorem on one hand, the other is stated as follows:
let $(X, d_x)$ be a metric space and $f_i: X \rightarrow R$ be continuous for $i = 1, 2, ..., n$. to prove that:
$f = (f_1, ..., f_n): X \rightarrow R^n$
$x \mapsto f(x) = (f_1(x), ..., f_n(x))$
is continuous.
My attempt is as follows:
$f_i$ is continuous for every I, therefore, $\forall \varepsilon \gt 0, \exists \delta \gt 0$ such that $|x - y| \lt \delta $ implies $|f_i(x) - f_i(y)| \lt \varepsilon$. This is true $\forall i$.
therefore, $|\vec f(x) - \vec f(y)| \lt \varepsilon ~~ \forall \varepsilon \gt 0 ~~~ s.t~~ |x-y| \lt \delta$
therefore $\vec f$ is continuous.
is this proof correct? any help is appreciated. 
 A: Fixed $\epsilon >0$ and $x$. For each $f_i$ there exists $\delta(i, x, \epsilon)$, i.e. depends on $i$ such that 
\begin{align}
|f_i(x)-f_i(y)|<\frac{\epsilon}{\sqrt{n}}
\end{align}
provided
\begin{align}
d(x, y)<\delta(i, x, \epsilon).
\end{align}
Since $i = 1, \ldots, n$ is finite, then we can define
\begin{align}
\delta(x, \epsilon) = \min_{1 \leq i \leq n}\delta(i, x, \epsilon)
\end{align}
such that
\begin{align}
\|f(x)-f(y)\|=\sqrt{ \sum^n_{i=1}|f_i(x)-f_i(y)|^2} < \sqrt{n\times\frac{\epsilon^2}{n}}= \epsilon
\end{align}
whenever
\begin{align}
d(x, y) <\delta. 
\end{align}
A: Depends on what your metric for $R^n$ is. 
If you're using $\sup$ distance ($d(f(x),f(y))\triangleq \sup_i d_x(f_i(x),f_i(y))$) then yeah this is fine, since $\sup \{f_1(x),\dots,f_n(x)\}$ is attained somewhere in the set.
If you're using some other distance, say $d(f(x),f(y))\triangleq \sum_{i=1}^n d_x(f_i(x),f_i(y))$ then no this is not fine in general: the first line of your proof doesn't imply the second line: with $n=2$ you could have a situation where $|f_i(x)-f_i(y)|=2\varepsilon/3$ for $i=1,2$. Then $d(f(x),f(y))=2\cdot(2\varepsilon/3) > \varepsilon.$
A: Let $p$ be an element of $X$.  
If $p$ is an isolated point of $X$, then there exists $\delta_0 > 0$ such that $\{x \in X | d_X(x, p) < \delta_0\} = \{p\}.$
So, for an arbitrary positive real number $\epsilon$, $d_X(x, p) < \delta_0 \implies \sqrt{\sum^k_{i=1}|f_i(x)-f_i(p)|^2}=\sqrt{\sum^k_{i=1}|f_i(p)-f_i(p)|^2}=0 < \epsilon.$
So, $\mathbf{f}(x)$ is continuous at $p$.  
Suppose that $p$ is an element of $X$ and a limit point of $X$.
$f_1(x), f_2(x), \cdots, f_k(x)$ are continuous at $p$.
The constant functions $f_1(p), f_2(p), \cdots, f_k(p)$ are continuous at $p$.
By Theorem 4.9 on p.87,
$f_1(x)-f_1(p), f_2(x)-f_2(p), \cdots, f_k(x)-f_k(p)$ are continuous at $p$.
By Theorem 4.9 on p.87,
$(f_1(x)-f_1(p))^2, (f_2(x)-f_2(p))^2, \cdots, (f_k(x)-f_k(p))^2$ are continuous at $p$.
By Theorem 4.9 on p.87,
$\sum^k_{i=1}(f_i(x)-f_i(p))^2$ is continuous at $p$.
By Theorem 4.6 on p.86,
$\lim_{x \to p} \sum^k_{i=1}(f_i(x)-f_i(p))^2 = \sum^k_{i=1}(f_i(p)-f_i(p))^2 = 0$.
So, for an arbitrary positive real number $\epsilon$, $0 < d_X(x, p) < \delta_0 \implies |\sum^k_{i=1}(f_i(x)-f_i(p))^2-0|=\sum^k_{i=1}(f_i(x)-f_i(p))^2 < \epsilon^2.$
So, for an arbitrary positive real number $\epsilon$, $0 < d_X(x, p) < \delta_0 \implies \sqrt{\sum^k_{i=1}(f_i(x)-f_i(p))^2} < \epsilon.$
Because $\sqrt{\sum^k_{i=1}(f_i(x)-f_i(p))^2} = 0 < \epsilon$ when $x = p$, $\mathbf{f}(x)$ is continuous at $p$.  
