The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith 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 ith 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

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