a question about the definition of limit point 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 :)
 
 A: 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 
$$\min\left\{\frac{\epsilon}2,1\right\}\in U\cap(A\setminus\{0\})=U\cap A\;.$$
A: Since your space seems to be limited to $\overline{\mathbb{R}}$, maybe this illustration will help:
Let $A = (0,1)$. The sequence $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots\}$ is in $A$.  And we know that $\lim\limits_{n\to\infty} \frac{1}{n} = 0$.  So $0$ is a limit point. It's a limit point of  $A$ since every neighborhood of $0$ will contain at least an element of $A$.  That is, you can always have an $n$ that is large enough so that $\frac{1}{n} \in A$ will be in that neighborhood.
However, no negative number will be a limit point of $A$.  For example, take $-0.000000001$.  You can have a neighborhood of that number that doesn't have any element of $A$ (because $0$ sits between $A$ and $-0.00000001$, and $0 \notin A$).  Intuitively, you can't come up with a sequence in $A$ whose limit is that negative number.  
So, in essence, the definition requires "every neighborhood" because if you can think of a neighborhood which doesn't have an element of $A$, that means that there is something "sitting" between $a$ and $A$ that isn't in $A$, and thus $a$ isn't a limit point of $A$.
Hope that helped.
