7. “finite” sequences are dense

problem

Show that the set of all sequence $x∈𝒞$ with finitely many $1$s is countable and dense.

Let $𝒮$ denote the set of all strings with finitely many $1$s. This set is countable because it is a countable union of countable sets (define $𝒮_k$ the set of strings with no $1$s after position $k$; then $𝒮=⋃𝒮_k$).

It is dense: fix any $x∈𝒞$ and $ϵ>0$. Take $k∈ℕ$ so that $1\/2^(k+1)<ϵ$. Then define $y∈𝒮$ as the string agreeing with $x$ on the first $k$ coordinates, and zeros everywhere else. Notice then that $d(x, y) = ∑_(i=0)^∞ abs(x(i)-y(i)) / 2^(i+1) = ∑_(i=k+1)^∞ abs(x(i)-y(i)) / 2^(i+1) ≤ ∑_(i=k+1)^∞ 1 / 2^(i+1) = 1 / 2^(k+1) < ϵ.$