A little help with point set topology Can anyone give me an EXAMPLE of adherent point that is not an accumulation point or just opposite.
 A: A point $x$ is an adherent point of a set $A$ if every open set containing $x$ contains a point of $A$; Thus, if $x \in A$, $x$ is automatically an adherent point of $A$. However, $x$ is an accumulation point of $A$ if every open set containing $x$ contains a point of $A$ other than $x$. Thus, to find an example of a set $A$ with an adherent point $x$ that isn’t an accumulation point of $A$, you want a set $A$ that contains some point $x$ that is contained in an open set that otherwise misses $A$ completely.
A simple example in the real line is to let $A = [0,1] \cup \{3\}$ and $x=3$. $3 \in A$, so $3$ is an adherent point of $A$, but the open interval $(2,4)$ is an open set containing $3$ that does not contain any other point of $A$: $(2,4) \cap A = \{3\}$.
It should be clear from the definitions that every accumulation point of a set is also an adherent point: if every open set containing $x$ contains a point of $A$ different from $x$, then every open set containing $x$ certainly contains a point of $A$!
A: In the real numbers, letting $A=\{0\}$, $0$ is an adherent point of $A$ since every neighborhood of $0$ contains at least one point of $A$, namely $0$; but $0$ is not a limit point of $A$, since no neighborhood of $0$ contains points of $A$ different from $0$.
A: Well you can see the below link for an answer.


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*http://www.physicsforums.com/archive/index.php/t-473352.html
