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I have a question, why isn't an alternative definition of a $p \in X$ being a limit point of a set $E$ : $\forall r>0 N_r(p) \cap E \neq \emptyset$, why does it have to be the punctured neighborhood? Note that $E$ is a subset of $X$ and $X$ is a metric space. Note that the definition of a limit point of a set E is a point $p \in X$ such that every neighborhood contains a point $q$ such that $q\neq p$ and $ q\in E$.

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  • $\begingroup$ Then the limit point is same as a point of the closure of the set. $x\in Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty $\endgroup$
    – IrbidMath
    Jan 14, 2019 at 3:40
  • $\begingroup$ Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure. $\endgroup$
    – M A Pelto
    Jan 14, 2019 at 4:16

1 Answer 1

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The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).

We typically say $p \in \text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.

To summarize:

  1. $p$ is a limit point of $E$ means there is a sequence of points $\{p_n\}_{n=1}^\infty$ with $p_n \in E$ and $p_n \neq p$ for every $n \in \mathbb N$ and such that $\lim_{n \to \infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
  2. $p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p \cap E=\emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.

I hope this helps to clarify things.

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