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The paragraph leaves out an important constraint: it’s only for functions that map some compact convex set to itself. So, the theory doesn’t apply to the siblin
by fshbbdssbbgdd 5y ago
The paragraph leaves out an important constraint: it’s only for functions that map some compact convex set to itself. So, the theory doesn’t apply to the sibling commenter’s example of f(x) = x + 1 because it doesn’t map a compact set to itself.
One example of a compact convex set is a circle. Imagine some continuous function that maps a circle to a circle. An easy example is rotation. For rotation, the center stays in the same place. Now imagine some more complicated function that might do rotation, stretching, spiral distortions, whatever you can imagine. But it has to be continuous, meaning you can’t go off one edge and come back on the other side (among other things). Try to draw, or even just imagine a transformation like this and you’ll find that there’s always some point that stays in the same place.
For a 3D object like a glass it’s hard to see the same thing intuitively, but Brouwer’s theorem says that it holds (and it also holds for higher-dimensional sets).
- kevinventullo 5y agoMinor wording nit: rather than the circle, I would say “consider the disc”. Usually in math “circle” means just the boundary whereas “disc” means the circle together with its interior.
- OrderlyTiamat 5y agothank you, that did clear it up for me.
- antman 5y agoWhat about a circle with 60cm diameter that I move one meter to the right? Something is missing in my understanding of the definition.
- legolas2412 5y agoIt has to map the same set onto itself.
- enriquto 5y ago> For a 3D object like a glass A glass is not convex so this theorem does not apply. You'd need other fixed-point theorems for that. If the glass has a handle the result is trivially false (consider rotating a doughnut, this does not have any fixed point).
- FabHK 5y agoI think GP meant the water in the glass (as the article).