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Let $S$ be a subset of topological space $X$. Let $S'$ be be the set of all limit points of $S$. Then, it can be shown that $$\bar S= S \,\cup\, S',$$ where $\bar S$ denotes the closure of $S$, the smallest closed set in $X$ containing $S$. Now, this is what I am accustomed to see as either the definition of $\bar S$ or as a proven theorem (e.g. in Munkres).

However, here the relationship between $\bar S$ and $S'$ is portrayed as if $\bar S = S'$ instead of $S' \subseteq \bar S.$

I can think of counterexamples to $\bar S = S'. $ For example, $S = \{\frac{1}{n}, n \in \mathbb{N} \}$, where $S' = \{0 \}.$

Am I simply misreading that post or missing something more fundamental? Thanks.

Neutrino
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2 Answers2

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That question isn't asking about the limit points of $S$. It's asking about all possible limits (in metric space $X$) of sequences of points in $S$.

Arthur
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  • Isn't the set of limits of all convergent sequences in $S$ the same as the set of limit points of $S$? In other words, if ${s_n} \subset S$ and $s_n\rightarrow s$, can't we say that for all such $s$, ${s} = S'$? – Neutrino Jun 11 '19 at 10:45
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    @Neutrino No. For $s$ to be a limit point, we require that the sequence converging to $s$ doesn't have $s$ as a term. The question you linked has no such requirement. – Arthur Jun 11 '19 at 10:53
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    No, $0$ is still a limit point of $[0,1)$. For instance, $\dfrac1n\to0$. –  Jun 11 '19 at 11:02
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    @Neutrino What Chris said above. There are sequences in $[0,1)$ which converge to $0$ but don't have $0$ as a term. Thus $0$ is a limit point. – Arthur Jun 11 '19 at 11:03
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This seems (to me) to boil down to whether one defines $S'$ as the set of all limits of sequences in $S$ (as i had been accustomed to doing) or, alternatively, as the set of all limits of nontrivial sequences in $S$ (as appears to be the custom on this site, and in various sources).

That is, if one includes "trivial" sequences, $x_n=x$ for $x\in S$, then one gets $S\subset S'$, and so $\bar S=S'$.