The subspace of $\ell_2$ $M = \{(x_n) \in \ell_2 :\sum_{n=1}^\infty x_n = 0\}$
It is obvious that it is a linear set. But I don't know how to prove it is closed. I try to prove the complement is open, but it doesn't work. Can anyone help me?
 A: Actually, $M$ is a dense subspace of $\ell^2$ (in the definition of $M$, two things are given: the sum $\sum_{n=1}^{+\infty}x_n$ exists and is $0$). Take $x\in \ell^2$ such that $x\in M^\perp$, and let $e_n$ the sequence whose only non-zero element is the $n$-th, which is $1$. We have for $n\neq m$ that $e_n-e_m\in M$, hence $\langle x,e_n-e_m\rangle=0$. This gives $x_n=x_m$, and since $x\in\ell^2$, $x=0$.
A: The set $M$ is in fact not closed;
Consider the unit vector $e_1\in\ell_2$. For $n$ a positive integer, let
$$y_n=(1,\underbrace{\textstyle{-1\over n},{-1\over n},\ldots,{-1\over n}}_{n\text{-terms}},0,0,\ldots).$$
Then we have, for each $n$:
$$\Vert e_1-y_n\Vert_{\ell_2} = 1/\sqrt n;$$
whence, the sequence $(y_n)$ converges to $e_1$ in $\ell_2$.  
Since each $y_n\in M$ and $e_1\notin M$, it follows that $M$ is not a closed set.
(Similar constructions show that in fact every unit vector is in the closed linear span of $M$.  But then, since $M$ is a subspace of $\ell_2$, it follows that $M$ is dense in $\ell_2$.) 
A: This linear subset of $\ell _2$ is dense.
We will use a classic lemma which states:
Let $X$ be a normed space, and $M$ a linear subspace of $X$ if for every 
linear continuous functional for which $ f(M)= 0$ then $f \equiv 0$ then $M$ is dense.
Take a linear functional $f: \ell _2 \rightarrow \mathbb{R}$ then we know it has a representation
$$f(x)= \sum _{i=1}^{\infty }a_i x_i$$
Now we will take the image of some elements belonging to $M$ so to conclude $a_i = 0 \forall i\in \mathbb{N}$
Take firstly $b_1=(1,-1,0, \dots)$ so we get $f(b)=0 \Rightarrow a_1=a_2$
then take $b_2=(1,0,-1, \cdots )$ so we get $f(b)=0 \Rightarrow a_1=a_3$
so for $b_n$ we get $f(b_n)=0 \Rightarrow a_1=a_{n+1}$
and since $\lim _{n} a_n=0$ we get $a_1=0$.
and so from the previous steps $a_i=0 \forall i \in \mathbb{N}$. So $f\equiv 0$
and from the lemma we are done.
A: If you require that subspaces be closed then M is not a subspace of $\ell_{2}$. We construct a sequence of sequences as follows:
Define the first sequence by $x_1$=(1, -1, 0, ,0, ...), $x_2$=(1, -1, 1, -1, 0, 0, ...), etc.
Each of the $x_n$ is in M and in $\ell_2$ but the sequence converges to the Grandi series which does not even converge.
