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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 :)

enter image description here

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I think it is depend on the way you define a neighbourhood. – Babak S. Sep 18 '12 at 8:40
As you've exhibited a neighbourhood of $a$ which does not contain at least one point of $A$, you've shown that $a$ is not an accumulation point of $A$. Also, your picture is in $\mathbb{R}^2$ but the definition you use is for a subset of $\mathbb{R}$. – Michael Albanese Sep 18 '12 at 8:43
up vote 5 down vote accepted

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

$$\min\left\{\frac{\epsilon}2,1\right\}\in U\cap(A\setminus\{0\})=U\cap A\;.$$

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M.Scott Thanks:) Can you recommend me a book, or have you a pdf which can you give me the link to read more? thanks. – Iuli Sep 18 '12 at 8:48
@Iuli: Any decent introductory topology text would do. Among English-language texts the one by Munkres is widely recommended. The one by Stephen Willard is a little more advanced but is quite readable and is available in an inexpensive edition from Dover. The text Topology Without Tears is available online; I’m not terribly fond of it, but that’s at least partly just a matter of taste; you should certainly take a look at it. – Brian M. Scott Sep 18 '12 at 8:55
Another question(excuse me) the limit points are the boundary points ? Am I right ? I cannot see how a point which is not in $A$ $(a \not \in A)$ can be limit point if is not a boundary point. Thanks :) – Iuli Sep 18 '12 at 12:01
about, continuous functions. "$\displaystyle (\forall) x_{n}, x_{n} \rightarrow a, x_{n} \in A $ then $\displaystyle f(x_{n}) \rightarrow f(a)$ ? Obviously this last proposition has sense if $a \in A$ is a accumulation point." So I asked you, if are there points $a \in A$ which are not limit(accumulation) point ? Thanks :) – Iuli Sep 18 '12 at 12:21
@Iuli: The limit points that are not in $A$ are all boundary points. Limit points that are in $A$ may be boundary points but need not be. The set of limit points of $A=(0,1]\cup\{2\}$ is $[0,1]$; the boundary points of $A$ are $0,1$, and $2$. Continuous functions can do anything at non-limit points. For instance, every function from $\Bbb Z$ to $\Bbb R$ is continuous, because $\Bbb Z$ has no limit points. – Brian M. Scott Sep 18 '12 at 19:01

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 $\lim_{n\to\infty} 1/n = 0$. So $0$ is a limit point. It's a limit point of $A$ since every neighbourhood 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 neighbourhood.

However, no negative number will not be a limit point of $A$. For example, take -0.000000001. You can have a neighbourhood 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 neighbourhood" because if you can think of a neighbourhood 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.

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