Let $(X,d)$ be a metric space, $E$ a subset of $X$ and $p$ a point of $X$.
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A neighborhood of $p$ is a subset of $X$ consisting of all points $q$ of $X$ with $d(p,q)<r$, some $r>0$.
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$p$ is a limit point of $E$ if every neighborhood of $p$ contains a point $q\neq p$ of $X$ which is a point of $E$
Definition 1 $E$ is said to be dense in $X$ if every point of $X$ is a limit point of $E$.
Definition 2 $E$ is said to be dense in $X$ if every point of $X$ is a limit point of $E$, or a point of $E$ (or both)
Is "..or a point of $E$, (or both)" really necessary in definition 2? In other words, are the two definitions defining the same class of subsets of $X$? I was thinking: If a point of $X$ is a point of $E$, then it is a limit point of $E$, unless $E$ has some isolated points, in that case I would have that $E$ is not dense into itself, according to definition 1. Is that the only circumstance where I need to specify "..or a point of $E$", or there are many cases of spaces $X$ and subspaces $E$ where the two definitions are actually not equivalent?
Best Answer
Consider any set $X \neq \emptyset$ with the discrete metric. Then $X$ is dense in itself but no point of $X$ is a limit point of $X$, so we must require ".. or a point of $E$" in the definition.
In other words, in my example definition (2) holds while definition (1) fails.
In fact, instead of asking for limit point as you defined, you can just define an adherent point of $E$ as a point $x \in X$ such that for every neighbood of $x$ contains a point of $E$. Then it is easy to see that a set $E$ is dense in $X$ iff every point of $X$ is an adherent point of $E$, so you don't need to distinguish cases. Note also that the set of adherent points of $E$ is exactly the closure of $E$, so basically the definition then becomes $X= \overline{E}$.