Intuition for continuity in $\Bbb R^n$

Question

I'm trying to conceptually understand continuity in $\Bbb R^n$. I understand the one-dimensional case where for a function $f: \Bbb R \to \Bbb R$, arbitrarily small deviations of the input ($\delta$) will produce arbitrarily small deviations in the output ($\epsilon$). However, this picture is muddied when I start to consider functions of multiple variables. For example, what does a continuous function $g: \Bbb R^2 \to \Bbb R^2$ look like? Is it just piecewise continuous? For example, if I have a function $g(x,y) = (x+y, 2xy)$, would I just need to consider that an arbitrarily small change in $x$ and $y$, and show that each of these individually would produce an arbitrarily small change in the output?

More generally, is there a way that I can intuitively think about it? The picture for $\Bbb R$ is clear, but I think it is less clear for higher dimensions. Also, how should I picture any general function $f:\Bbb R^2 \to \Bbb R^2$?

Answer 1

It is instructive to reformulate the definition for continuity for functions $\mathbb{R} \rightarrow \mathbb{R}$. To this end, we first define some notation. Let $x \in \mathbb{R}^{n}$ be a real number and let $\epsilon \in \mathbb{R}_{>0}$ be a positive real number. Then we write \begin{equation*} B(x,\epsilon) = \{ y \in \mathbb{R}^{n}: \|x-y\| < \epsilon \} \end{equation*} for the open ball around $x$ with radius $\epsilon$.

In terms of open balls, the definition for continuity reads as follows.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Then $f$ is continuous in $x$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $y \in B(x, \delta)$ it follows that $f(y) \in B(f(x), \epsilon)$.

In $\mathbb{R}$, the condition $y \in B(x, \delta)$ means precisely that $|x-y|< \delta$ and the condition $f(y) \in B(f(x), \epsilon)$ means precisely that $|f(x) - f(y)| < \epsilon$.

The definition in terms of open balls generalizes readily to functions $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$.

The idea is as follows. If $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is continuous, then it should not make any sudden jumps, that is, points $x, y \in \mathbb{R}^{n}$ that are close to each other, should get sent to points $f(x),f(y)\in \mathbb{R}^{m}$ that are close to each other.

We formalize this idea by saying that the ball $B(x,\delta)$ around $x$, should get sent to a set of points contained in the ball $B(f(x), \epsilon)$.

Answer 2

The essential point is that we can measures distances in ${\mathbb R}^n$ as well as in ${\mathbb R}$: $$\|x-y\|:=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+\ldots+(x_n-y_n)^2}\ .$$ Continuity is explained in terms of such distances.

Consider a function $f:\>{\mathbb R}^n\to{\mathbb R}^m$ and assume $f(p)=q$. Then $f$ is continuous at $p$, if given any tolerance $\epsilon>0$ for the output error we can exhibit an allowance $\delta>0$ such that $\|x-p\|<\delta$ guarantees $\|f(x)-q\|<\epsilon$.

If, e.g., we know beforehand that $\|f(x)-f(y)\|\leq L\|x-y\|$ for any two points $x$ and $y$ in the domain of $f$ then we can just take $\delta={\epsilon\over L}$.