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I am going through the proof of the Question:

Let $A$ be a subset of a (point set) topological space $(X,T)$. When will a point $p$ not be a limit point of $A$.

Proof: $p$ is not a limit point of $A$ if there exists an open set $G \in T$ such that $p \in G$ and $(G \setminus \{p\}) \cap A = \phi$

or equivalently $p \in G$ and $G \cap A = \phi$ or $G \cap A = \{p\}$.

I did not understand why $G \cap \{p\}'$ is avoided here, since $(G \setminus \{p\}) \cap A = G \cap \{p\}' \cap A = \phi $ which will be true if $G \cap \{p\}' = \phi$(where $\{p\}' = X \setminus \{p\}$).

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I suppose $\{p\}' = X\setminus \{p \}$. What do you mean by $G \cap \{p\}'$ "is true"? –  Henno Brandsma Jun 26 '13 at 4:03
    
What is your definition of limit point of $A$? What do you mean by $\{p\}'$? –  Brian M. Scott Jun 26 '13 at 4:04
    
@BrianM.Scott: I think he means the complement of $\{ p \}$. But of course, there'll be a lot of cases, where $G \cap \{ p \}^c \neq \emptyset$, but $G \cap \{ p \}^c \cap A$ is. –  user49685 Jun 26 '13 at 4:14

2 Answers 2

up vote 3 down vote accepted

Henno has explained the result itself, but it occurs to me that you may be asking why it’s written

$\qquad\qquad\qquad\qquad\qquad\qquad p\in G$ and $(G\setminus\{p\})\cap A=\varnothing$

instead of in the equivalent form

$\qquad\qquad\qquad\qquad\qquad\qquad p\in G$ and $G\cap\{p\}'\cap A=\varnothing$,

since, as you point out, $(G\setminus\{p\})\cap A=G\cap\{p\}'\cap A$.

It could be written either way; it really makes no difference as far as the meaning is concerned. It could, in fact, just as well be written $G\cap A\subseteq\{p\}$, since

$\qquad\quad G\cap A\subseteq\{p\}$ if and only if $(G\setminus\{p\})\cap A=\varnothing$ if and only if $G\cap\{p\}'\cap A=\varnothing$.

These are three equivalent ways of expressing the same fact; which one you choose is a matter of taste. I usually use the third. Many people prefer to avoid the second, because the notion of complement is a bit tricky: it requires specifying the universe in which the complement is to be taken. Here that’s the whole space, so we have to understand that $\{p\}'$ is really just an abbreviation for $X\setminus\{p\}$. And once we write that out explicitly, making the expression $G\cap(X\setminus\{p\})\cap A$, we might as well simplify it to $(G\setminus\{p\})\cap A$ anyway.

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$p$ is a limit point of $A$ iff every open set $G$ with $p \in G$ (every open neighbourhood of $p$) intersects $A$ in a point different from $p$.

So not being a limit point means that there exists some $G$ with $p \in G$, such that $G \cap A$ contains no point of $A$, except possibly $p$, which would be unavoidable if $p \in A$. So either $p \in A$ and there is some $G$ open in $X$ such that $G \cap A = \{p\}$, which says that $p$ is an isolated point of $A$, or $p \notin G$ and there is some open $G$ that contains $p$ and $G \cap A = \emptyset$, so this says that $p \in G \subset X \setminus A$ or equivalently, $p$ is an interior point of the complement of $A$.

So the non-limit points are the isolated points of $A$ (if any) and the interior of the complement of $A$. It's not a matter of "avoiding $G \cap \{p\}'$".

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