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$M = \{(x_n) \in \ell_2 :\sum_{n=1}^\infty x_n = 0\}$

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?

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What is your definition of subspace? It's not usually required that they're closed. For example that would make it hard to talk about dense linear subspaces (e.g. $C_c(\mathbb R) \subset L_1(\mathbb R)$). –  kahen Oct 14 '12 at 21:25
    
@kahen In the book, subspace should be linear and closed. –  Joe Berg Oct 14 '12 at 21:26
    
Are you sure this is true? –  abatkai Oct 14 '12 at 21:36
    
@abatkai I don't if it is closed. If it isn't,can you give me a counterexample? –  Joe Berg Oct 14 '12 at 21:38
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... and furthermore the domain of that functional is not the whole space ... –  GEdgar Oct 14 '12 at 23:23

4 Answers 4

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 $x\in\ell^2$, $x=0$.

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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.

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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.

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Are you sure the sequence $\{x_n\}$ is convergent in $\ell^2$? –  Davide Giraudo Oct 14 '12 at 22:07
    
I realized it wasn't after posting. My mistake. –  Some Math Guy Oct 14 '12 at 22:13

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 $\ell_2$.)

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