## 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 and and all that:

A function is continuous at if and only if for each there is a so that implies .

where the bars denote the norm in one or the other of the spaces or as depends on context. Again, the idea is that if we pick a metric ball around , we can find some metric ball around 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 using a very different subbase: preimages of neighborhoods in 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 is continuous at if and only if for each with all there is a so that implies for all .

That is, if we pick a small enough metric ball around its image will fit within the “box” which extends in the th direction a distance on each side from the point .

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 , 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 so that is the th component of . For each of these real-valued functions, I have a definition of continuity:

A function is continuous at if and only if for each there is a so that implies .

So each is continuous if I can pick a that works with a given . And if *all* the are continuous, I can pick the smallest of the and use it as a that works for each component. But then I can wrap the up into a vector and use the I’ve picked to satisfy the box definition of continuity for itself! Conversely, if is continuous by the box definition, then I must be able to use that the for a given vector to verify the continuity of each for the given .

The upshot is that a function from a metric space (generalize this yourself to other metric spaces than ) to is continuous if and only if each of the component functions is continuous.

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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 |