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It is a very intuitive question, actually is very trivial... I have to show that $S\,'\!\subset \overline{S}$.

Using the definition of limit point ,we have :

$$\forall \epsilon>0, \big(]x- \epsilon ,x+\epsilon[\;\cap\; S\big)\setminus \{x\} \neq \emptyset$$

But, also, the definition of closure is

$$\forall \epsilon>0, \big(] x- \epsilon ,x+\epsilon [\;\cap\; S\big) \neq\emptyset$$

So, every limit point is also a closure point. Is that right?

How could I write it? That is being my problem: I know it, it's quite obvious by the definition, but I can't write it in a proper way...

I say that limit point is a subset of closure because "it is the closure minus a point"?

(I'm sorry my stupidity...)

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I am not used to notation $]a,b[$ for open interval, but I guess there are many people who use such notation. (IIRC such notation was used quite frequently in France, maybe still is.) –  Martin Sleziak Jun 27 '12 at 10:37
    
@MartinSleziak: It is still used (by almost everyone) in France. –  Najib Idrissi Jun 27 '12 at 13:05
    
I use it too.but in this notation i thought that would be too much parentheses...that´s it. –  Charlie Jun 27 '12 at 18:00
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2 Answers 2

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A point $x$ belongs to the closure $\bar S$ if each neighborhood of $x$ intersects $S$ in at least one point $y$. A point $x$ belongs to the set $S'$ if each neighborhood of $x$ intersects $S$ in at least one point $y$ which is different from $x$. Therefore obviously $S'\subset \bar S$.

How can we characterize possible points $x\in \bar S\setminus S'\,$? Every neighborhood of such a point intersects $S$ all right, but there is at least one neighborhood $U$ of $x$ for which $U\cap S$ contains no $y\ne x$; so this intersection is exactly $=\{x\}$. This is to say that $x\in S$, and $x$ is an isolated point of $S$.

Therefore your statement that $S'$ "is the closure minus a point" is not correct as it stands: The set $S'$ is the "closure of $S$ minus the isolated points of $S$".

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I think your explanation is pretty good, but if you want another idea I'd propose the following: if $\,x\notin\overline{S}\,$ then there exists $\,\epsilon >0\,\,s.t.\,\,(x-\epsilon\,,\,x+\epsilon)\cap S=\emptyset\Longrightarrow x\notin S'\,$ , by sheer definition of the derived set.

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