# Continuous Motion?

In the new issue of Analysis that Brian just mentioned, there’s an article by Hud Hudson called “How to Part Ways Smoothly”. He describes two point-objects that are colocated at every time before 100, but then are at a different location at that time, although both move continuously. The way this works is that both rotate around a clock face doubling their speed after each rotation, so that they go around infinitely many times before 100, and that at 100 one of them is at the 3 on the clock face and the other is at the 9.

I don’t understand why he is justified to claim that “neither character ever moves discontinuously”. It’s true as he says that no matter how small an interval you look at before time 100, there are time-slices of each character that are arbitrarily close to their destination. However, I normally think of continuity as being characterized by a different set of quantifiers – for every spatial distance, there is some temporal duration such that all time-slices within that duration are within that distance of the destination.

Because their trajectories satisfy Hudson’s definition of continuity but not mine, the space-time trajectories are said to be “connected, but not path-connected”, and curves like this are standard counterexamples in topology. But is there any reason why metaphysicians might adopt Hudson’s account of continuous motion and not mine? If not, then an example like his could be constructed whereby a particle traces out successive approximations to a Peano space-filling curve with constantly doubling speed, so that in the limit every point in space could be the result of “continuous” motion.

## 8 Replies to “Continuous Motion?”

1. Kenny,

Were there meant to be links in this post? I get bits of text which look like links, but they aren’t live.

2. Timothy Bays says:

As Kenny notes, the definition of continuity that Hud is using is very odd. If I’m reading the definition right, then it would make some pretty standard examples of discontinuous functions come out continuous. I would bet, for instance, that over half of the calculus textbooks in the US give the following as an example of a discontinuous function:

f(0) = 0

f(x) = sin(1/x) for x≠0

It looks, however, like this fits Hud’s definition of a continuous function: no mater how small an interval around 0 you look at, there will be points in that interval whose image is arbitrarily close to 0 (indeed, there will be infinitely many points whose image is 0).

3. In fact, the function that Timothy gives just is the projection of Hud Hudson\‘s function onto the y-coordinate (well, except that his is sin(1/2x) instead of sin(1/x), and that time goes from 0 to 100 instead of -1 to 0).

4. Timothy Bays says:

After thinking about it some more, Hud’s definition looks even stranger than it did at first. Consider a standard example of a totally discontinuous function—-i.e., a function that is discontinuous at every point:

f(x) = 0 for x rational
f(x) = 1 for x irrational

The graph of this function isn’t even connected, but it looks like the function satisfies Hud’s criterion for continuity (since for any point on the graph, there are other points which are arbitrarily close to it).

5. That’s interesting – I suppose that graph is at least locally connected, but you’re right that under any intuition that looks discontinuous.

I think I can make it look even worse – consider the standard example of a function that’s continuous at every irrational and discontinuous at the rationals, which is 0 at the irrationals, and 1/q at the point p/q (in lowest terms). Instead of letting it be 1/q, let it be f(q), where f is some enumeration of the rationals. In every open interval, this function attains all but finitely many rational values, so the graph is dense in the plane. Clearly, this function is nowhere continuous – but it also satisfies his criterion.

6. Oops – I’ve made the mistake before of thinking that the rationals (or irrationals) were connected. They’re actually both totally disconnected. For the same reason, the graph of my function is also disconnected (just consider any irrational y value, and the entire graph is contained in the open half plane above that value, and the open half plane below it).

I guess his criterion just says there are no isolated points, and doesn’t really say anything about connectedness.

7. Hud Hudson says:

Hi all,

Sorry not to have posted sooner — I have been out of town and away from email. I think Kenny and Tim are simply right on this point (and that I wasn’t). Many thanks to them for clearly and fairly showing why.