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I have a handwavy argument for this. Is it sound enough or if not, how to make it more mathematically rigorous? Thanks.

My Argument: The closure of the subset $l_c^\infty(\mathbb{N}; \mathbb{R})$ consists of all sequences in $l_c^\infty(\mathbb{N}; \mathbb{R})$ together with all limit sequences of the sequences in $l_c^\infty(\mathbb{N}; \mathbb{R})$.

If $x \in l_c^\infty(\mathbb{N}; \mathbb{R})$, then there exists a $k^* \in \mathbb{N}$ such that for all $k > k^*$, we have $x^k = 0$. Call such a sequence $x_{k^*}$ meaning this sequence has at most $k^* + 1$ nonzero terms. But for every such finite sequence, the sequence $x_{k^*+1}$ is also a finite sequence. Hence, the limit of all finite sequences are actually infinite sequences. Therefore, the closure of $l_c^\infty(\mathbb{N}; \mathbb{R})$ is the entire $l^\infty(\mathbb{N}; \mathbb{R})$ set.

rims
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1 Answers1

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If you are refferring to the closure (w.r.t. sup norm) of the set of sequence $(a_n)$ such that $a_n=0$ for $n$ sufficiently large then the closure is not all of $\ell^{\infty}$. The sequence $(1,1,\cdots )$ cannot be approximated in $\ell^{\infty}$ norm by a finitely non-zero sequence.

It is the space of all sequences that converge to $0$. If $(x_n)$ belongs to this closure and $\epsilon >0$ then there exists a finitely nonzero sequence $(a_n)$ with $|a_n-x_n| <\epsilon $ for all $n$ , so $|x_n| <\epsilon $ for all $n$ sufficiently large. Hence $x_n \to 0$. I leave it to you to verify that any sequence converging to $0$ can be approximated by a finitely non-zero sequence.

Kavi Rama Murthy
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  • If $x$ belongs to this closure, then it is a limit point of the set $l_c^\infty$. Then, for every $\epsilon > 0$, there exists a sequence $(a_n \in l_c^\infty)_{n \in \mathbb{N}}$ such that $\sup_k|a_n^k - x^k| < \epsilon$ for n large enough. What I am trying to say is that the limit point $x$ will be approximated by a sequence of sequences in $l_c^\infty$ – rims Sep 23 '19 at 00:00
  • If $(1,1,...)$ is in the closure then there must be one finitely non-zero sequence $(a_n)$ with $|a_n-1|<\frac 1 2 $ for all $n$. Do you agree? Do you see a contradcition from this? – Kavi Rama Murthy Sep 23 '19 at 00:03
  • I do agree that that the closure is the set of all sequences that converge to zero. The point I am making is about what does one mean by "every sequence converging to zero is approximated by a sequence of points in $l_c^\infty$". The points are themselves sequences so we have a sequence of sequences approximating $x$. See https://math.stackexchange.com/questions/3365635/question-about-l-infty-as-a-complete-metric-space/3365649#3365649 – rims Sep 23 '19 at 00:08
  • If you don't understand, then it is fine. You have answered my question anyway; I just have a different notion of what it means to be a limit point of $l_c^\infty$. – rims Sep 23 '19 at 00:09
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    @Black In a metric space $x$ belongs to closure of $A$ iff for every $\epsilon >0$ there exists $a \in A$ such that $d(a,x) <\epilosn$. It is not necessary that you write down a sequence in $A$ converging to $x$. – Kavi Rama Murthy Sep 23 '19 at 00:13
  • I think there is an error in your second paragraph first couple of lines. If $(x_n)$ belongs to the closure then for some finitely nonzero sequence ($a_n$), we have $d(a,x) = \sup_n|x_n - a_n| < \epsilon$ for every $\epsilon > 0$. This would imply $\sup_n |x_n| < \epsilon$ for large enough $n$, thus we have $x_n \to 0$. The conclusion remains the same but I think we should have the notion of a sequence object "approximating" the sequence object, instead of coordinates approximating the coordinates which is a weaker and insufficient condition for what it means to be a "limit point". – rims Sep 26 '19 at 02:11
  • @Black Thank you for your comment. – Kavi Rama Murthy Sep 26 '19 at 05:22