Cfrgtkky

Let $X$ be any metric space. Let $f_1:Xto R$ and $F_2:Xto R $ be
defined. Then

$f:Xto R^2$

$f(x)= (f_1(x),f_2(x))$ is continuous on X

$iff $

$f_1$ and $f_2$ are continuous.

Instead of proving this via $epsilon -delta$ definition, can't we straight away use the sequential criterion to prove this?

$f:Xto R^2$

$f(x)= (f_1(x),f_2(x))$ is continuous on X

$iff$

$forall langle x_n rangle in X $ such that $langle x_n rangle to ain X$
we have $f(x_n)= (f_1(x_n),f_2(x_n)) to (f_1(a),f_2(a)) ; ; forall ain X$

$iff$

$f_1(x_n) to f1(a)$ and $f_2(x_n) to f_2(a)$

$iff$

Both $f_1$ and $f_2 $ are continuous on X.

Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$

If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.

Such a proof is indeed valid as soon as you know (and have proved) the following facts:

1. 
   $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).


2. A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.

Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):

Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)

Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)

But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.