this post was submitted on 23 Apr 2026
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[–] someacnt@sh.itjust.works 19 points 3 weeks ago* (last edited 3 weeks ago) (2 children)

I am not a topologist, but I can try..

A space (shape) is contractible if you can "contract" (shrink) it to a point without cutting, pinching or punching through holes. For example, a mattress is contractible, since you can shrink it to the center - each point can follow the line to the center, continuously. Meanwhile, a doughnut, a circle or a hollow sphere are not contractible, you can never remove the inner "hole" to shrink to a point without cutting.

In general, any dimensional sphere is not contractible... Until it is - infinite dimensional sphere is contractible. Somehow, it loses the "hollow space" inside.

[–] zipsglacier@lemmy.world 17 points 3 weeks ago (1 children)

For each finite dimension n (1, or 2, or 3, etc...), the sphere in dimension n can't be contracted because of that empty n-dimensional space it surrounds. But that same sphere is the "equator" of the sphere in the next higher dimension, n+1. There, the n-dimensional equator can contract along one of the hemispheres, to a pole. But then that whole (n+1)-dimensional sphere still isn't contractible, because of the (n+1)-dimensional space it surrounds.

BUT the (n+1)-dimensional sphere can contract along one of the hemispheres in the (n+2)-dimensional sphere. And so on.

For any particular finite dimension n, there is an n-dimensional obstruction to contracting the sphere in that dimension. But if you go all the way to infinitely-many dimensions, there is no obstruction that ever stops contractibility of the infinite-dimensional sphere.

[–] ComicalMayhem@lemmy.world 4 points 3 weeks ago (1 children)

Sphere refers to the surface area, correct? Ball would be the volume inside?