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Working in $l_p : 1

$\rho(x,0)=\big|\sum_{n=1}^{\infty} |x_n|^p\big|^{\frac{1}{p}}$

$S=\{\{x_n\}_{n=1}^{\infty} : |x_n| \leq \frac{1}{n}\}$

I'm wondering if this "set of sequences" has no limit points. This would mean that the closure is the empty set which implies S is closed. Is there maybe another way of showing this set is closed?

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    What topology are you using on sets? – 2017-02-14
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    I edited the question to include that. – 2017-02-14
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    You are using $n$ for two different indices here. Do you mean $S= \{\{x_n\}_{n=1}^{\infty} : |x_n|\leq \frac{1}{n}, n\in\mathbb{N}\}$ ? – 2017-02-14
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    I wrote it just as I saw it in the notes. I'm assuming that is what is meant. – 2017-02-14
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    So, you're talking about a set of sequences, I wouldn't call it a set of sets when the $l_p$ norm is being used. – 2017-02-14
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    Just fixed that. Thanks for the correction. – 2017-02-14
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    @user153126 Perhaps, just drop the $n\in\mathbb{N}$? – 2017-02-14
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    Why would you believe that this set has no limit points, I would think that it has limit points. Also the closure would not be empty because the closure contains the set itself. – 2017-02-14
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    @Masacroso that isn't true. The question appears to be about $\ell_p$ spaces with $1$(\sum_{n=1}^{\infty} |x_n|^p)^{1/p}$ does converge for every sequence if $|x_n|\leq\frac{1}{n}$. – 2017-02-14
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    So since $p>1$ I can argue that the series is a convergent p-series and so after some work I can show $\rho(x,0)<\epsilon$ which implies 0 is a limit point of S so S is closed. I guess I just felt like I needed to start with an arbitrary limit point and show that that limit point is in S. – 2017-02-14
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    @user153126 you are totally right... I missread something, obviously if $|x_n|<1/n$ then these series converges absolutely, so the norm $\|{\cdot}\|_p$ is well defined for these sequences. – 2017-02-14
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    @Kevin I mean I gave an example doing essentially that as an answer. – 2017-02-14

2 Answers 2

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For any set A, A is a subset of the set of limit points of A (also called the closure of A). Therefore the only set with no limit points is the empty set.

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Let $m\geq 1$ then the sequences $y_m = (\frac{1}{m}, 0, 0,\dotsc)$ all reside in $S$ as $\frac{1}{m} \leq \frac{1}{1}$ and $0<\frac{1}{n}$ for $n>1$. We can easily show that $y_m$ is a Cauchy sequence with respect to $m$ and that $y_m \to (0,0,0,\dotsc) \in S$. Hence $(0,0,0,\dotsc)$ is a limit point of $S$.