Continuity of a function $f : \mathbb R^n \to \mathbb R^k$ So preparing for the end-term of the semester, strolling trough exercises in our huge book (Th. Rassias Calculus 2) and in the internet, I found one more theoritical one, that as always bothers me (because it's theoritical !!).
Show that in the function $f : \mathbb R^n \to \mathbb R^k$ , $f=(f_1,\dots,f_k)$ is continuous at $\vec x_o$, if and only if the function $f_i$ is continuous at $\vec x_o$ for $i = [1,2,\dots,n]$.
Now, it's pretty obvious that since the "big" function $f$ is formed by the "smaller" $f_i$'s , all the sub-functions must be continuous, but that's not a mathematical proof of course. Also, I could say that for $f$ to be continuous, its limit must be defined, which means that $\lim_{x \to x_o}f = f_o$, which would mean that : $\lim_{x \to x_o}f = (\lim_{x \to x_o}f_1^o,\dots,\lim_{x \to x_o}f_k^o) = f_o = (f_1^o,\dots,f_k^o)$. Now, by definition that means that each of the sub-functions must be continuous, which means that they are continuous if and only if $\forall e>0$ $\exists d>0$ : $\forall x \in \mathbb R^n$ with distance $d(x,x_o)<d \Rightarrow d(f(x),f(x_{io})<d$. Is that a kind of correct claim ? I am pretty sure it's not as mathematical it should be. Any tips-hints-corrections or help over this will be really appreciated !
 A: What I would say is this:
For a function $F : \mathbb{R}^n \to \mathbb{R}^m$ to be continuous at a point $\vec a$ ($\vec a \in \mathbb{R}^n$) then $\forall \epsilon > 0, \exists \delta > 0 : ||\vec x - \vec a || < \delta \Rightarrow || F(\vec x) - F( \vec a) || < \epsilon$.
Now $|| F(\vec x) - F( \vec a) || = \sqrt{\left( F_1(\vec x)-F_1( \vec a) \right)^2+ \ldots}$ which means that $|x_1-a_1|<\delta_1 \leq \delta \Rightarrow |F_1(\vec x)-F_1(\vec a)| < \epsilon_1 \leq \epsilon$ which implies that $F_1$ must be continuous in a ball of center $\vec a$. Then you can easily generalise for the other component functions.
From a student also preparing for his final, I would say your answer is correct, all I did was use the actual definition of what it means to be continuous at a point. Also, for someone else more comfortable with this, is my reasoning (and @charalampos) correct?
A: $(\Leftarrow)$ Suppose $f$ is continuous, you could use the fact that composition of continuous functions is continuous to prove that all of the $f_i\text{'s}$ are continuous as well..
Define the $i\text{-th}$ coordinate projection function $$\pi_i:\mathbb{R}^k \rightarrow \mathbb{R}\\(x_1,\ldots,x_k)\mapsto x_i,\quad i\in\{1,\ldots,k\}$$
which is linear and hence continuous. Then,
$$f_i=\pi_i \circ f,\quad i\in\{1,\ldots,k\} $$
are continuous as compositions of continuous functions.
$(\Rightarrow)$ For the other direction, continuity of the $f_i\text{'s}$ implies,
$$\forall \epsilon>0. \exists \delta>0:(\Vert x-x_0\rVert<\delta)\implies\lVert(f_i(x)-f_i(x_0)\rVert<\epsilon$$
Taking the supremum norm in $\mathbb{R}^k$, defined as $\lVert y\rVert_{\infty}=\sup_k\{\lvert y_k\rvert\}, y\in \mathbb{R}^k$, we see that point-wise convergence of the $f_i\text{'s}$ suggests convergence in $\mathbb{R}^k$.
