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Are there any examples of subspaces of $\ell^{2}$ and $\ell^{\infty}$ which are not closed?

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    $\begingroup$ The subspace generated by the FINITE linear combination of $(1,0,0,\cdots),(0,1,0,\cdots),(0,0,1,0,\cdots),\cdots$ in $l^2$ is not closed since for example $(1,\frac{1}{2},\frac{1}{3},\cdots)$ is a limit point of the subspace I defined but it is not lying in the subspace. $\endgroup$ – Li Li Dec 21 '16 at 3:10
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    $\begingroup$ Almost every infinite-dimensional subspace is not closed. $\endgroup$ – tomasz Dec 23 '16 at 11:25
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Take the subspace of $\ell^2$ (or $\ell^\infty$) consisting of sequences with finitely many non-zero coordinates.

Clearly, this is a subspace (sum of sequences still has at most # of nonzeros of first seq + # of nonzeros in second seq, scaling of a sequence has same non-zeros). However, you can take any sequence on $\ell^2$ (or $c_0 \subset \ell^\infty$, a closed subspace), $a$ and define the sequence of sequences $\{a_n\}$ where $a_n$ consists of only the first $n$ elements of $a$ and rest zeros, and see $a_n \to a$.

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  • $\begingroup$ The last statement is not true for $\ell^\infty$, only for $\ell^2$. The closure of the subspace of finite sequences in $\ell^\infty$ is $c_0$, not $\ell^\infty$. The subspace is still not closed in $\ell^\infty$ of course. $\endgroup$ – Tim B. Dec 24 '16 at 0:09
  • $\begingroup$ Oops, you're right. $\endgroup$ – Batman Dec 24 '16 at 15:24
  • $\begingroup$ @TimB. We can just take a more specific sequence, say $a_n=1/n$, right? $\endgroup$ – broncoAbierto Jul 8 '18 at 17:13
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Yes. For instance, the sequences which are eventually $0$.

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As an observation, every subspace generated by countably many linearly independent vectors (that is, a subspace of dimension $\aleph_0$) inside a Banach space is not closed, since a Banach space cannot have (algebraic) dimension $\aleph_0$. ( if it were closed, if would be itself a Banach space).

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For $1< x<2$ the space $l^x$ is a non-closed vector subspace of $l^2.$

Let $(A_j)_{j\in \mathbb N}$ be a strictly increasing positive real sequence converging to $1/x.$

For $i,j, \in \mathbb N$ let $B_{i,j}=0$ if $j>i,$ and $B_{i,j}=j^{-A_j}$ if $j\leq i.$ Let $v_i=(B_{i,j})_{j\in \mathbb N}.$

Since only finitely many co-ordinates of each $v_i$ are non-zero, we have $\{v_i:i\in \mathbb N\}\subset l^x.$

Let $u= (u_j)_{j\in \mathbb N}=(\;j^{-B_{j,j}}\;)_{j\in \mathbb R}=(\;j^{-A_j})_{j\in \mathbb N}.$

Take $d\in (0,1-x/2).$ Take $j_0$ such that $j\geq j_0\implies A_j>(1-d)/x.$ Then $j\geq j_0 \implies 2A_j>2(1-d)/x>1.$ So $$\sum_{j=j_0}^{\infty} u_j^2=\sum_{j=j_0}^{\infty}j^{-2A_j}\leq \sum_{j=j_0}^{\infty}j^{-2(1-d)/x}<\infty.$$ Therefore $u\in l^2.$

Now $u\not \in l^x$ because $u_j^x=j^{-xA_j}>j^{-1}$ so $$\sum_{j=1}^{\infty} u_j^x\geq \sum_{j=1}^{\infty} j^{-1}=\infty.$$

I will leave it to you to show that $\lim_{j\to \infty} \|u-v_j\|_2=0.$ So in $l^2,$ the vector $u$ belongs to $Cl(l^x)$ but not to $l^x.$

The fact that $l^x$ is a vector space for any $x>1$ is another story, which I am sure you can find on this site.

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  • $\begingroup$ Are you suggesting that $\ell^1$ is closed in $\ell^2$? Or, for that matter, $\ell^x$ for $x<1$? $\endgroup$ – tomasz Dec 23 '16 at 11:27
  • $\begingroup$ @tomasz .NO. Read my first sentence: "non-closed." And I only considered $1<x<2 $. $\endgroup$ – DanielWainfleet Dec 24 '16 at 0:01
  • $\begingroup$ What I meant is that you explicitly omitted $x=1$ as if it was any different, which may suggest that it actually is different, which (as far as I can tell), it isn't in this case, nor is $0<x<1$. $\endgroup$ – tomasz Dec 25 '16 at 0:36
  • $\begingroup$ @tomasz. The Q asked for some examples. Not for all possible instances $\endgroup$ – DanielWainfleet Dec 25 '16 at 1:27

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