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Cincinnatus
Nov9-05, 12:41 AM
So, what exactly is "cubelike" about the hilbert cube?

I think I am having trouble "visualizing" it. Is it just called that because it it homeomorphic to I^inf. ?

Cincinnatus
Nov9-05, 12:48 AM
I realize there are different ways of defining the hilbert cube, so this question probably doesn't make much sense.

My class defined it to be the subset of l^2 space given by 0<x_n<1/n (actually less than or equal to).

fourier jr
Nov9-05, 01:52 AM
the hilbert cube is the product [0,1]^{\mathbb{N}} with the product topology. if you take the product of just 3 of them it looks like a cube, hence the name. some people like to define it as [0,1] x [0,1/2] x [0,1/3] x ... x [0, 1/n] x ... just because it's easier to work with, but it doesn't really matter since all closed intervals are homeomorphic to [0,1]

fourier jr
Nov9-05, 02:01 AM
actually a cube is a product of any closed intervals. someone working in a hilbert space would rather use [0,1] x [0,1/2] x [0,1/3] x ... x [0, 1/n] x ... as the definition since it is isometric, rather than just homeomorphic, to a subspace of itself.