Let $A$ be a subset, $A \subset \mathbb{R}$. A point $a \in \mathbb{\overline{R}}$ is a limit point(or accumulation point) of $A$ if every neighbourhood of $a$ contains at least one point of $A$ different from $a$ itself
I cannot unerstand this definition very well. For this I will draw a picture. I have a set $A$, and two neighbourhoods $V$ and $W$.
case I. For the neighbourhood $V$ our definition is verified because $V \cap A \neq \emptyset$
case II. neighbourhood $W$ is not ok because $A \cap W =\emptyset$.
Why in the definition is specified the word every? I can find at least a neighbourhood $U$ for that $U \cap A =\emptyset$.
Thanks 🙂
Best Answer
That point $a$ is not is not a limit point of $A$ precisely because it has a neighborhood, $W$, that does not contain any point of $A$ different from $a$ itself. (I’m assuming that you intended that $a$ belong to the set $A$, even though it’s detached from the rest of $A$.) Consider the set $A=(0,1]\cup\{2\}$. $1$ is a limit point of $A$, because every open set containing $1$ also contains other points of $A$. If $U$ is an open set containing $1$, then there is an $\epsilon>0$ such that $(1-\epsilon,1+\epsilon)\subseteq U$, and clearly
$$\max\left\{1-\frac{\epsilon}2,\frac12\right\}\in U\cap(A\setminus\{a\})\;.$$
$2$, on the other hand, is not a limit point of $A$, because the open set $(1,3)$ contains $2$ and no other point of $A$.
Finally, $0$ is a limit point of $A$, even though it does not belong to $A$: every open set $U$ containing $0$ contains an interval of the form $(-\epsilon,\epsilon)$, and
$$\min\left\{\frac{\epsilon}2,1\right\}\in U\cap(A\setminus\{0\})=U\cap A\;.$$