The multiscale coefficient space is separable and has vanishing projection tails

Let $(x^{(m)})$ be Cauchy in $M_0(w)$. Each coordinate is Cauchy because $\mathrm{\mid }{x}_{lk}^{(m)}-x_{lk}^{(n)}\mathrm{\mid }\le w_l\mathrm{\parallel }{x}^{(m)}-x^{(n)}{\mathrm{\parallel }}_{M(w)}$. In the Haar case also $\mathrm{\mid }{x}_{\ast }^{(m)}-x_{\ast }^{(n)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(n)}{\mathrm{\parallel }}_{M(w)}$. Let $x_{lk}$ and, when present, $x_{\ast }$ be the scalar limits. The uniform Cauchy bound passes to every finite collection of coordinates and then to the supremum, so $x^{(m)}\to x$ in the weighted supremum norm. To see $x\in M_0(w)$, fix $\epsilon >0$, choose $m$ with $\mathrm{\parallel }x-x^{(m)}{\mathrm{\parallel }}_{M(w)}<\epsilon$, then choose $J$ so the normalized tail of $x^{(m)}$ is below $\epsilon$; for $l>J$ the normalized tail of $x$ is at most $2\epsilon$. Thus $M_0(w)$ is complete.

Finite-support sequences with rational coordinates are countable and dense: first choose $J$ to make the tail of any $x\in M_0(w)$ small, then approximate its finitely many coordinates through level $J$, including the scaling coordinate when present, by rationals. Hence $M_0(w)$ is separable. The formula for the norm shows $P_J$ is a contraction, and the defining tail limit gives $\mathrm{\parallel }x-P_{J}x{\mathrm{\parallel }}_{M(w)}\to 0$.

If $x\in {\mathrm{\ell }}^2$, then $w_l\ge 1$ implies $w_l^{-1}{\max }_k\mathrm{\mid }{x}_{lk}\mathrm{\mid }\le ({\sum }_k\mathrm{\mid }{x}_{lk}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. The right side tends to zero because its squares are level blocks of a convergent nonnegative series. Its supremum and the absolute scaling coordinate, when present, are at most $\mathrm{\parallel }x{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}$, so $x\in M_0(w)$.