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Shizuo Kakutani's fixed point theorem

Shizuo Kakutani's fixed point theorem. I'm confused about the 3-D example...

By Jason Kottke    Aug 20, 2004 at 09:55 am

There are 8 reader comments

jkottke    Aug 20 2004    9:58AM

Last example. Consider a cupful of coffee. Each molecule is at some point in 3-dimensional space. Stir. At least one molecule ends up in the same place as it began.

Here's my problem with this. Let's say the cup has 2 molecules. Stirring switches their positions...Molecule A is now where Molecule B was and vice versa. Therefore the statement "at least one molecule ends up in the same place as it began" is incorrect. Isn't it? I can see how the paper example is correct, but this one I just can't visualize. Anyone up on their fixed point theory?

Graham    Aug 20 2004    10:19AM

I can't seem to get to the linked site, but here's my best guess. The vital thing is the continuity, which you've thrown out in your version -- there aren't just two molecules, there is a full solid of them. Actually, talking about molecules at all kind of finesses the point, since there are only going to be a finite (though very large) discrete bunch of them. Fixed-point theorems are about continua, and aren't going to make any sense for finite sets.

Graham    Aug 20 2004    10:21AM

Here's another fixed-point theorem, with a nice 2-d example. Visualizing an analogous 3-d example isn't easy.

jkottke    Aug 20 2004    10:25AM

Fixed-point theorems are about continua, and aren't going to make any sense for finite sets.

Ok, that makes sense. The other two examples talked about points, but molecules aren't points. I could buy that in a full cup of coffee there's a high probability that the same molecule would occupy the same space after stirring, but definitely? Didn't seem right.

L.N. Hammer    Aug 20 2004    11:53AM

And the article's been revised to reflect that molecules was being confusing. It's a continuum thing.

---L.

jf    Aug 20 2004    2:28PM

There is also a strict requirement on the 'stirring'. For simplicity, consider a right-cylindrical coffee cup. Now 'stir' by slicing the coffee into slabs and shuffle them. Clearly under this mapping no point is mapped to itself. This is really a property of continuous maps of continuous sets.

vacapinta    Aug 20 2004    11:20PM

The whole coffee example has so many restrictions that it robs these theorems of their true wow potential.
My favorite example of the intermediate value theorem was that you can easily show that there are two antipodal points on earth that have the same temperature right now - in fact, there is a whole band of them.
This is true for any continuously varying quantity you impose on the surface of the Earth (e.g. barometric pressure, elevation etc)

Maciej Ceglowski    Aug 23 2004    6:49AM

Here's my problem with this. Let's say the cup has 2 molecules

Sounds like the coffee they serve at the Fairmount/Hutchinson croissanterie in Montreal. Great croissants, but you might want to sneak in an insulated hip flask from a less stingy cafe down the street. Fair warning.

This thread is closed to new comments. Thanks to everyone who responded.

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