dense subspace of $c_0( \mathbb N)$

Prove that

$$Y= \left\{ x=(x_n)_{n \in\mathbb{N}} \in c_{0}(\mathbb N )~ \Bigg | ~\sum_{n=1}^{\infty} x_n = 0 \right\}$$

is a dense linear subspace of $c_0( \mathbb N)$.

where $\displaystyle{c_0( \mathbb N) = \left\{ x=(x_n)_{n \in\mathbb{N}} \in \mathbb R ^{\mathbb N} : \lim_{n \to \infty} x_n =0 \right\}}$

I cannot prove that it is dense.

Any help?

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For denseness, you can show each unit vector is in the closure of $Y$. For example, to show $(1,0,\ldots)$ is in the closure of $Y$, consider vectors of the form $(1,\underbrace{-1/n,\ldots ,-1/n}_{n-\text{terms}},0,\ldots)$. – David Mitra Oct 22 '12 at 13:15
O.K I can show that each $e_n$ is in the closure of $Y$ but I can't see how I can get density. – passenger Oct 22 '12 at 13:26
@DavidMitra: Yes you are right! I think the most simple solution! Thank you for your time! – passenger Oct 22 '12 at 14:44
Sorry, there was a "typo" in my previous comment (now deleted). I meant to say just use the fact that the linear span of the set of unit vectors is dense in $c_0$. So the closure of $Y$ contains the closure of the linear span of the set of unit vectors, and hence is all of $c_0$. – David Mitra Oct 22 '12 at 15:53
Yes i understand that! Thank's again! – passenger Oct 23 '12 at 12:28

The elegant proof is the following. Consider linear functional $$f:c_{00}(\mathbb{N})\to\mathbb{R}:x\mapsto\sum\limits_{n=1}^\infty x_n$$ Then

1. Show that $f$ is unbounded and $\mathrm{Ker}(f)\subset Y$.
2. Show that that kernel of each unbounded functional is dense in the domain space.
3. Recall that $c_{00}(\mathbb{N})$ is dense in $c_0(\mathbb{N})$.
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I'm confused. Consider $x = (1, 0 , 0 \dots) \in c_0$. If $Y$ is dense in $c_0$ then there is a sequence $y_k = (y_n)_k \in Y$ such that $\|y_k - x\| = \max_n |y_{kn} - x_n| \to 0$ as $k \to \infty$. What would be such a sequence? – Rudy the Reindeer Oct 22 '12 at 13:00
The problem is that $f$ is not well defined on the whole $c_0$ (take $x_n=n^{-1}$). – Davide Giraudo Oct 22 '12 at 13:01
@MattN. Consider $y_k=(1-k^{-1},k^{-1}2^{-1},k^{-1}2^{-2},k^{-1}2^{-3},\ldots)$ – Norbert Oct 22 '12 at 13:03
@DavideGiraudo You are right. I'll try to salvage my proof see edits. – Norbert Oct 22 '12 at 13:04
@Norbert: Can you prove step 2 ? – passenger Oct 22 '12 at 13:35

The fact that $Y$ is a subspace is quite clear. To see density, we can use a corollary of Hahn-Banach theorem: we just need to show that each linear continuous functional on $c_0(\Bbb N)$ which vanished on $Y$ vanishes on the whole space.

Let $f$ such a functional. We have $f(e_n-e_m)=0$ if $m\neq n$, where $e_n$ is the sequence whose $n$-th term is $1$, the others $0$. So $f(e_k)=:K$. As $\left\lVert\sum_{k=0}^ne_k\right\rVert_{\infty}=1$, we show have $nK\leq \lVert f\rVert$ and $K=0$.

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Can you explain a little more the last line of your solution? – passenger Oct 22 '12 at 13:09
$f(e_k)$ as the same value $K$. Then I use $l(\sum_{1\leq k\leq n} e_k)=nK, and it's supposed to be$\leq$the norm of$f$. – Davide Giraudo Oct 22 '12 at 13:20 Taking on Norbert's answer: since the linear functional$\,f\,$is obviously not the zero functional, we know$\,\ker f\,$is a maximal subspace of$\,c_0(\Bbb N)\,$, from which it follows that $$c_0(\Bbb N)=\operatorname{Span}\{\ker f\,,\,v\,\}\;\;,\;\forall\;v\notin\ker f$$ Perhaps this now will make it simpler to find the solution ( hint: a subset$\,A\,$of a topological space$\,X\,$is dense in it iff$\,A\cap Y\neq\emptyset\,$for every non-empty open subset$\,Y\subset X\,\$ )

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