Topologically, the Hilbert cube may be defined as the product of countably infinitely many copies of the unit interval [0,1]. That is, it is the cube of countably infinite dimension. As a product of compact Hausdorff spaces, it is itself a compact Hausdorff space as a result of the Tychonoff theorem.
It's sometimes convenient to think of the Hilbert cube as a metric space, indeed as a specific subset of a Hilbert space with countably infinite dimension. For these purposes, it's best not to think of it as a product of copies of [0,1], but instead as [0,1] × [0,1/2] × [0,1/3] × ···; for topological properties, this makes no difference. That is, an element of the Hilbert cube is an infinite sequence (x_{n}) that satisfies 0 ≤ x_{n} ≤ 1/n. Note that any such sequence belongs to the Hilbert space l_{2}, so the Hilbert cube inherits a metric from there.
Since l_{2} is not locally compact, no point has a compact neighbourhood, so one might expect that all of the compact subsets are finitedimensional. The Hilbert cube shows that this is not the case. But the Hilbert cube fails to be a neighbourhood of any point p because its side becomes smaller and smaller in each dimension, so that an open ball[?] around p of any fixed radius e > 0 must go outside the cube in some dimension.
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