Thread: compactness

The Hilbert cube Q is the subset of l∞ given by {a : N→R││a(i)│≤2-i}.
Show that Q is a compact subspace of l∞ (with the norm ∥a∥=supn∈N│a(n)│).


can anyone please help me to do this question ?? ,,

Originally Posted by jin_nzzang

The Hilbert cube Q is the subset of l∞ given by {a : N→R││a(i)│≤2-i}.
Show that Q is a compact subspace of l∞ (with the norm ∥a∥=sup_{n∈N}│a(n)│).

Let denote the unit ball of . Then consists of all sequences of real numbers such that for all n. Give the topology that it has as the product of an infinite number of copies of [–1,1]. In this topology, a sequence converges to if it converges coordinatewise. That is to say, if for each coordinate n.

The map x_n)\mapsto(2^{-n}x_n)" alt="fx_n)\mapsto(2^{-n}x_n)" /> takes onto Q. The idea of the proof is to show that f is continuous from (with the above product topology) to Q (with the topology from the norm). By Tychonoff's theorem, any product of compact spaces is compact. So is compact, and hence Q, as a continuous image of a compact space, is also compact.

To see that f is continuous, suppose that in , and let . Choose N so that . Then for all n>N. But (by the definition of the product topology in ) there exists K such that for all k>K and all n with 1≤n≤N. Therefore , so that in Q.

[That is essentially the argument given here to show that the Hilbert cube is compact for the topology that comes from the -norm, except that the Wikipedia page completely ducks the central part of the proof, which is to show that the map f is continuous.]