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 (xn) that satisfies 0 ≤ xn ≤ 1/n. Note that any such sequence belongs to the Hilbert space l2, so the Hilbert cube inherits a metric from there.
Since l2 is not locally compact, no point has a compact neighbourhood, so one might expect that all of the compact subsets are finite-dimensional. 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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