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Can anyone give me an EXAMPLE of adherent point that is not an accumulation point or just opposite.

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    *posted from mobile .. please feel free to edit2011-07-17

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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 $A$!

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    Mathematics and mathematicians are pretty hard to understand ... you should have put it there more explicitly ... anyway thanks for answer.2011-07-17
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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 $0$.

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Well you can see the below link for an answer.

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    @Chandru: His name appears properly on the paper, as ‘Bühler’, with u-umlaut; ‘Buehler’ is the proper substitute spelling when umlauts are unavailable.2011-07-17