Schauder bases of subspaces of the sequence space $\ell^p(\mathbb{N})$ Consider the canonical Schauder basis $\{e_i:i\in \mathbb{N}\}$ for $\ell^p(\mathbb{N})$, where $e_i(j)=\delta_{ij}$. Let $M$ be a subspace of $\ell^p(\mathbb{N})$. Is it right that $\{e_i:i\in \mathbb{N}\}\cap M$ is a basis for $M$?
For example, if we take the subspaces $$M_1=\{(\cdots,0, 0, x_1,x_2,x_3,0,0,\cdots)\}$$ or $$M_2=\{\{x_i\}:x_i=0 \text{ if } i \text{ is even }  \}$$
Then the answer is clearly yes. However, what about the general cases? 
If the answer is no, is there any condition to be added such that the question has a positive answer?
 A: Note The theorem below is actually trivial without the lemma. The lemma seems interesting as well; this will be rewritten as a three part tfae soon.
Theorem Suppose $1\le p<\infty$ and $X$ is a subspace of $\ell^p$. Then $X$ has some set of $e_j$ for a basis if and only if $X$ is monotone, meaning that if $x\in X$ and $|y(n)|\le|x(n)|$ for all $n$ then $y\in X$.
First a lemma, inspired by the fact that if $X$ is any of various function spaces on the circle then a subspace $Y$ of $X$ is spanned by the characters it contains if and only if it is rotation-invariant:
Lemma Suppose $1\le p<\infty$. A subspace $X$ of $\ell^p$ has some set of $e_j$ as a Schauder basis if and only if $X$ is invariant under multiplication by the sequence $(e^{ijt})$ for every $t\in\Bbb R$.
(Meaning that if $x=((x(1),x(2),\dots)\in X$ then $y=(e^{it}x(1),e^{2it}x(2)\dots)\in X$.)
Proof If $X$ is spanned by some set of $e_j$ then it is clear that $X$ is invariant under those multiplication operators. Suppose that $X$ is invariant, and let $x\in X$. It follows that $y\in X$, if $y$ is defined by $$y(j)=\frac1{2\pi}\int_0^{2\pi}e^{-int}e^{ijt}x(j)\,dt.$$But $y(n)=x(n)$ and $y(j)=0$ for $j\ne n$, which is to say that $y=x(n)e_n$.
So if $x\in X$ then $e_n\in X$ for every $n$ with $x(n)\ne0$. QED
Proof of the theorem If $X$ has a Schauder basis consisting of a set of $e_j$ it is clear that $X$ is monotone. If $X$ is monotone then $X$ is invariant under multiplication by $(e^{ijt})$.

(
Proof of the theorem without the lemma If $X$ has a Schauder basis consisting of a set of $e_j$ it is clear that $X$ is monotone. If $X$ is monotone and $x\in X$ then it is clear that $e_j\in X$ for every $j$ with $x(j)\ne 0$.)
A: The answer is no. A subspace of $\ell_p$ need not have a basis at all. However you don't need that. Take the subspace spanned by $e_i+e_{i+1}$ for $i=2,4,6,\ldots$.
