# The Unapologetic Mathematician

## Multivariable Continuity

Now that we have the topology of higher-dimensional real spaces in hand, we can discuss continuous functions between them. Since these are metric spaces we have our usual definition with $\epsilon$ and $\delta$ and all that:

A function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is continuous at $x$ if and only if for each $\epsilon>0$ there is a $\delta>0$ so that $\lVert y-x\rVert<\delta$ implies $\lVert f(y)-f(x)\rVert<\epsilon$.

where the bars denote the norm in one or the other of the spaces $\mathbb{R}^m$ or $\mathbb{R}^n$ as depends on context. Again, the idea is that if we pick a metric ball around $f(x)\in\mathbb{R}^n$, we can find some metric ball around $x\in\mathbb{R}^m$ whose image is contained in the first ball.

The reason why this works, of course, is that metric balls provide a neighborhood base for our topology. But remember that last time we came up with an equivalent topology on $\mathbb{R}^n$ using a very different subbase: preimages of neighborhoods in $\mathbb{R}$ under projections. Intersections of these pre-images furnish an alternative neighborhood base. Let’s see what happens if we write down the definition of continuity in these terms:

A function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is continuous at $x$ if and only if for each $\epsilon=\left(\epsilon_1,\dots,\epsilon_n\right)$ with all $\epsilon_i>0$ there is a $\delta>0$ so that $\lVert y-x\rVert<\delta$ implies $\lvert\pi_i(f(y))-\pi_i(f(x))\rvert<\epsilon_i$ for all $i$.

That is, if we pick a small enough metric ball around $x\in\mathbb{R}^m$ its image will fit within the “box” which extends in the $i$th direction a distance $\epsilon_i$ on each side from the point $f(x)$.

At first blush, this might be a different notion of continuity, but it really isn’t. From what we did last yesterday we know that both the boxes and the balls provide equivalent topologies on the space $\mathbb{R}^n$, and so they much give equivalent notions of continuity. In a standard multivariable calculus course, we essentially reconstruct this using handwaving about how if we can fit the image of a ball into any box we can choose a box that fits into a selected metric ball, and vice versa.

But why do we care about this equivalent statement? Because now I can define a bunch of functions $f_i=\pi_i\circ f$ so that $f_i(x)$ is the $i$th component of $f(x)$. For each of these real-valued functions, I have a definition of continuity:

A function $f:\mathbb{R}^m\rightarrow\mathbb{R}$ is continuous at $x$ if and only if for each $\epsilon>0$ there is a $\delta>0$ so that $\lVert y-x\rVert<\delta$ implies $\lvert f(y)-f(x)\rvert<\epsilon$.

So each $f_i$ is continuous if I can pick a $\delta_i$ that works with a given $\epsilon_i$. And if all the $f_i$ are continuous, I can pick the smallest of the $\delta_i$ and use it as a $\delta$ that works for each component. But then I can wrap the $\epsilon_i$ up into a vector $\epsilon=\left(\epsilon_1,\dots,\epsilon_n\right)$ and use the $\delta$ I’ve picked to satisfy the box definition of continuity for $f$ itself! Conversely, if $f$ is continuous by the box definition, then I must be able to use that the $\delta$ for a given vector $\epsilon$ to verify the continuity of each $f_i$ for the given $\epsilon_i$.

The upshot is that a function $f$ from a metric space (generalize this yourself to other metric spaces than $\mathbb{R}^m$) to $\mathbb{R}^n$ is continuous if and only if each of the component functions $f_i$ is continuous.

September 16, 2009 - Posted by | Point-Set Topology, Topology

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6. Handwaving is not necessary – it is easy to prove the equivalence of the two using the Cauchy-Scwarz inequality, which is also easy/quick to prove.

Comment by Brian Burns | October 8, 2012 | Reply