Is {-1,1}^N compact in $\ell_\infty$?

Question

I am reading A short Course on Banach Space Theory.

And it says that if ∑xn converges unconditionally in a Banach Space X, then the set of all vectors of the form ∑εnxn is a compact subset of X.

Its approach is to show f as in the picture is continuous

In which I am guessing it assume {-1,1}^N is compact in l∞.

Is there anyway to show if {-1,1}^N is compact in $\ell_\infty$?

Answer 1

In that context, $\{-1,1\}^{\mathbb N}$ is not a subspace of $\ell^\infty$. It is the set$$\overbrace{\{-1,1\}\times\{-1,1\}\times\cdots}^{\text{countable number of copies}}$$endowed with the product topology. And it is compact by Tychonoff's theorem.

Answer 2

It is possible to think of $\{-1,1\}^\mathbb{N}$ as a subset of $l^\infty$. That is the subset of all the sequences which only takes the values $-1,1$.

In this case the inherit topology on $\{-1,1\}^\mathbb{N}$ is the box topology which is discrete in this case. With this topology, the set $\{-1,1\}^\mathbb{N}$ is not compact.

However as noted by Jose, the author of the paper in the question does not think of $\{-1,1\}^\mathbb{N}$ as a subset of $l^\infty$. Instead he considers it with the product topology.