Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

2 Answers 2

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

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

share|improve this answer
    
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
1  
@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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.