Continuity of function with "internal" functions Suppose we have some $h:\mathbb{R}\to\mathbb{R}^2$ where $h(x)=(f(x),g(x))$ where $f,g:\mathbb{R}\to\mathbb{R}$. $h$ is continuous $\iff f,g$ are continuous. I'm not sure exactly what these sorts of functions are called, as they're not exactly multivariable functions.
I've been trying to prove this with the epsilon-delta method but it's a little tricky because I'm not sure exactly what epsilon is in the context of $\mathbb{R}^2$. Would this be an epsilon ball still, not a square? So $\epsilon = x^2+y^2$ for some $x,y$?
Moreover (this isn't the main question), I believe with an inductive argument we can prove that $\phi:\mathbb{R}\to\mathbb{R}^n$ where $\phi(x)=(\lambda_1(x),\dots,\lambda_n(x))$ and each $\lambda_i(x):\mathbb{R}\to\mathbb{R}$, $\phi$ is continuous $\iff$ each $\lambda_i$ is continuous.
Would appreciate any pointers in the right direction.
Best,
Adam
 A: $f$ and $g$ are often called coordinate functions. $h$ is then called a curve.
Continuity for $h$ in some $x \in \mathbb{R}$ means that for every $\varepsilon > 0$ there is some $\delta > 0$ such that if $y \in \mathbb{R}$ and $\lvert x -y \rvert < \delta$ then $\lVert h(x) - h(y) \rVert := \sqrt{(f(x) - f(y))^2 + (g(x) - g(y))^2} < \varepsilon$.
Now let $f$ and $g$ be continuous in $x$ and $\varepsilon > 0$. Choose $\delta > 0$ such that $\lvert y - x \rvert < \delta$ implies
$$
\lvert f(x) - f(y) \rvert < \frac{\varepsilon^2}{2} \text{ and } \lvert g(x) - g(y) \rvert < \frac{\varepsilon^2}{2}.
$$
(You will get a different $\delta$ from $f$ and $g$, just take their $\min$.)
Then
$$
\sqrt{(f(x) - f(y))^2 + (g(x) - g(y))^2} < \sqrt{\frac{\varepsilon^2}{2} + \frac{\varepsilon^2}{2} }= \varepsilon.
$$
If $h$ is continuous in $x$, then note that for $\lvert x - y \rvert < \delta$ we have
$$
\lvert g(x) - g(y) \rvert = \sqrt{\lvert g(x) - g(y) \rvert^2} < \sqrt{\lvert g(x) - g(y) \rvert^2 + \lvert f(x) - f(y) \rvert^2} < \varepsilon.
$$
The continuity of $f$ you prove the same way.
