Completeness and Topology of $\ell ^p,0<p<1$

Question

We can not define a norm in $\ell^p,0<p<1,$never the less,
we can define $\ell^p$ ,$0<p<1$, as a metric topological vector space
with the following metric:

$\qquad \qquad \bbox[5px,border:2px solid red] {ρ(x,y):=\sum_{n=1}^{\infty} \lvert x(n)-y(n)\rvert^p }$

Notice that:

$\qquad \qquad ρ(x+z,y+z)=ρ(x,y). $

$\qquad \qquad$ $\ell^p$ are linear subspaces of $\ell^1 , \forall p\in (0,1).$
Could such a space be complete? separable? or even compact?

My approach for completness:

1. Let $x_n$ be a Cauchy sequence in $\ell^p$ and $ε>0$,
   Then, $ρ(x_n,x_m)=\sum_{k=1}^{\infty} \lvert x_n(k)-y_m(k)\rvert^p\leε $
   That implies that for a fixed $k$ $x_n(k)$ is Cauchy, since $\mathbb R$ is complete,
   $\exists$ $x(k)$ s.t. $x_n(k) \rightarrow x(k) \in \ell^p $ since $x_n$ is Cauchy in $\ell^p.$
2. $x_n$ converges toward $x$ in $\ell^p$: Let $ε > 0$. For $d\in \mathbb N$ and $n,m$ sufficiently large:

$\qquad \sum_{k=1}^d |x_n(k)-x_m(k)|^p \le ρ(x_n,x_m)\leε,$

Letting $m → \infty $ and then $d → \infty$ allows to conclude ($ρ(x_n,x)\le ε)$
Hence we have complteness.
Is it correct?