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Let $V = \{ v \in \ell^2 \setminus \lvert \sum_{n=1}^\infty v_n \rvert < \infty \}$ be the subspace of $\ell^2$ over $\Bbb C$ of summable complex sequences

Let $U=\overline{V}$ and be $u \in U-V$

I would like to know if $u$ is summable $\lvert \sum_{n=1}^\infty u_n \rvert < \infty$

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    I changed \space \backslash \space to \setminus. That is standard usage. – 2017-01-20
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    ok thanks @MichaelHardy – 2017-01-20
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    @DuchampGérardH.E. $u$ is not a generic element of $\ell^2$ – 2017-01-20
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    Doesn't "summable" mean $\sum_{n=1}^\infty |u_n | < \infty$ rather than $\left| \sum_{n=1}^\infty u_n \right|<\infty\text{?}\qquad$ – 2017-01-20
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    @DuchampGérardH.E. i don't understand because in your example $u,v \in V$ – 2017-01-20
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    @MichaelHardy summable i mean just i wrote – 2017-01-20
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    Unless I missed something, $U=\bar{U}=V$ (these are vector subspaces). Then $U-V=U$ and this is trivial, no ? even with the stronger requirement $\sum_{n=1}^\infty |u_n | < \infty$. – 2017-01-20
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    @DuchampGérardH.E. not as you say but $U=\overline{V}$ – 2017-01-20
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    @DuchampGérardH.E. : Vector subspaces in infinite-dimensional spaces are not always closed. – 2017-01-20
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    @MateyMath For any vector subspace $U-U=U$ – 2017-01-20
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    @MichaelHardy It seems one does not need closure. – 2017-01-20
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    @DuchampGérardH.E. $U-U=\emptyset$ – 2017-01-20
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    @MateyMath In this context "minus" is the [Minkowski operation](https://en.wikipedia.org/wiki/Minkowski_addition) ($-$ is different from $\setminus$) – 2017-01-20
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    It seems that $U-V$ was intended to mean what is sometimes denoted by $U\setminus V$, i.e. it contains everything in $U$ that's not in $V$, whereas Duchamp Gérard H. E. may be construing it as $\{x-y : x\in U\ \&\ y\in V\}. \qquad$ – 2017-01-20
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    @DuchampGérardH.E. i meant the compement of $U$ – 2017-01-20
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    yes @MichaelHardy it is so – 2017-01-20
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    @MateyMath : $U\setminus V$ is the complement of $V$ relative to $U$, not the "complement of $U$". $\qquad$ – 2017-01-20
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    By closure, do you mean the $\ell^2$ closure? – 2017-01-20
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    sorry you are right @MichaelHardy – 2017-01-20
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    the completion i meant – 2017-01-20
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    Perhaps the OP search for an example such that $v_m(n)=1/n$ for $1\leq n\leq m$, $v_m(n)=0$ for $n\geq m+1$, and $v(n)=1/n$ for $n\geq 1$; then $v_m\in V$ for all $m$, $v_m\to v$ (in $l^2$) but $v\not \in V$ ? – 2017-01-20
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    thanks @Kelenner but i think my question is the contrary i should konw if $v \in U$ not if $v \in V$ – 2017-01-20
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    If $u \in U \setminus V$ it **cannot** be summable by definition, since $V$ contains all the summable sequences. – 2017-01-20
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    ok @copper.hat you're right, so my question becames is $V$ closed? – 2017-01-20
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    @Matey Math But, $U\setminus V=\emptyset$ – 2017-01-20

2 Answers 2

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since the zero sequence is in V, if every $u\in U-V$ is summable (in your sense) then $\bar{V}=V$. And reciprocally.

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    thanks @basti so $V$ is closed – 2017-01-20
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It seems that, in this context "minus" is the Minkowski operation ($-$ is different from $\setminus$). Then, unless I missed something, $U=\bar{U}=V$ (these are vector subspaces), one has $U−V=U$ (otherwise $U\setminus V=\emptyset$ and this should be meaningless). All this holds with the stronger requirement $\sum_{n\geq 1}|u_n|<+\infty$ (and $U=\ell^2\cap \ell^1=\ell^1$ in this case).