hilbert cube?

Question

prove that a hilbert cube is closed in L2.

need your help.

L2 is hilbert space

Answer 1

Because a Hilbert cube is a topological product of the intervals [0,1/n], for all natural numbers n, it consists of all sequences (x_n) such that 0 <= x_n <= 1/n. Any such sequence belongs to l2 (not L2).

Suppose that (y_n) is in l2, but not in the cube. Then, at least one element of y_k of (y_n) lies outside the interval [0,1/k]. No matter how close y_n is to any sequence (x_n) that belongs to the cube, we can always find a ball

B(y_n,e) = { (z_n) | ||y_n-z_n|| < e }

that does not intersect the cube, so the complement of the cube is open, therefore the cube is closed.

This requires more details to be rigorous but it's a blueprint.