Let $f: \Bbb R^m \to \Bbb R^n$. Express $f(x)$ in terms of components: $$f(x)=(f_1(x), f_2(x), ... , f_n(x))$$
I need to prove that $f$ is continuous at $a$ if and only if each $f_i$ is continuous at $a$.
Here are the definitions for continuity I have:
Definition of Continuity at a Point: Let $f: U \in \Bbb R^m \to \Bbb R^n$, where $U$ is open. Then $f$ is continuous at $a \in U$ if $\lim_{x \to a} f(x)=f(a)$.
Alternative Definition: $f$ is continuous at $a \in U$ if and only if $\lim_{h \to 0} f(a+h)=f(a)$
And here's what I've got so far: $f$ is continuous at $a\iff$
$\lim_{h \to 0} f(a+h)=f(a) \iff$
$\lim_{h \to 0} (f_1(a+h), f_2(a+h), ... ,f_n(a+h))=(f_1(a), f_2(a), ..., f_n(a)) \iff$
$\lim_{h \to 0} (f_1(a+h), f_2(a+h), ... ,f_n(a+h)) = (\lim_{h \to 0} f_1(a+h), \ ...\ , \lim_{h \to 0} f_n(a+h))$.
How do I show that this last equation is true? Last year in class, we took this as a definition, but in the book I'm reading now it isn't.
Assume $f_i$ is continuous for $i \in [1,n] \cap \mathbb{N}$, that is, for every $\varepsilon >0$, there exists a $\delta_i > 0$ such that for all $x \in \mathbb{R}$, $|x-a|<\delta_i \Rightarrow |f_i(x)-f_i(a)|<\dfrac{\varepsilon^2}{n}$
Choose $\delta = \min \{\delta_1,...,\delta_n\}$
Let $\varepsilon>0$ be given. Let $a \in \mathbb{R}^m$ be given.
Then, for all $x \in \mathbb{R}^m$ with $\|x-a\| < \delta$, we have $$|f_i(x) - f_i(a)| \leq \|f(x) - f(a)\| =\sqrt{(f_1(a) - f_1(a))^2 + ... +(f_n(x)-f_n(a))^2 }< \varepsilon$$ Since $\varepsilon > 0$ was given arbitrarily, we know have: For all $a \in \mathbb{R}^m$, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that for all $x \in \mathbb{R}^m$ with $\|x-a\| < \delta$ implies $\|f(x) - f(a)\|< \varepsilon$