Limit point intuition

Question

Quoting Rudin,

"A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q\not=p : q \in E$."

This would imply that the points in an open ball would all be limit points, since for any $p$ in $E$ there are $q$ such that $d(p,q) < r$ for all $q \in E$. So E is also a neighborhood of the open ball.

Is my intuition correct? What can be improved about this statement?

If our underlying space is say $\mathbb{R}^n$, sure. However, if our underlying set is a subspace of $\mathbb{R}^n$, the result is not necessarily true. — André Nicolas, Aug 23 '13 at 19:55
Like discrete metric spaces, where no point is a limit point and each subset is open (and closed). — njguliyev, Aug 23 '13 at 20:00
@Jossie: Consider $\Bbb Z$ as a subspace of $\Bbb R,$ for example. If you look at the open ball about $0$ of radius $1$ in that space, then the only point in it is $0$. It turns out that no set has a limit point in that space. — Cameron Buie, Aug 23 '13 at 20:01
Basically, every topology has $\Bbb R$ as its domain? That is, all topologies are continuous (or have continuous cardinality)? @CameronBuie — Don Larynx, Aug 23 '13 at 20:11
One can imbue any set with a topology. Some are nicer and more familiar than others--like metric spaces. Some can get really weird. — Cameron Buie, Aug 23 '13 at 20:15
@Jossie: Cameron is using the subspace topology for $\mathbb{Z}$ as a subset of $\mathbb{R}$. — augurar, Aug 24 '13 at 01:41

score 6 · Accepted Answer · answered Aug 24 '13 at 01:36

6

You can think of the set of limit points $L(S)$ of a set $S$ as all points which are "close to" $S$. In the example of an open ball in $\mathbb{R}^n$, the limit points are all points of the open ball, plus all points lying on the boundary, since every punctured neighborhood of such points will intersect the set.

Note, however, that if $S$ is some set and $L(S)$ is the set of limit points, then it is not always true that $S \subseteq L(S)$. For example, in $\mathbb{R}$ under the ordinary topology, the set of integers has no limit points. (An element of a set which is not a limit point of the set is called an isolated point, which provides a good intuitive way of thinking about such points.)

answered Aug 24 '13 at 01:36

augurar

1,856

When you say "close to" $S$, it seems you're talking about the closure points of the set $S$. Though there must a difference between closure points and limit points, right? – johnny09 Apr 20 '19 at 19:07
2

@johnny09 They're closely related. In fact, one way to define the closure of a set is the union of the set and its limit points. The difference is precisely the points of $P \in S$ which are "isolated" from the rest of $S$ -- i.e there exists a neighborhood of $P$ that does not contain any point of $S$ except $P$ itself. – augurar Feb 04 '20 at 01:03

score 1 · Answer 2 · answered Aug 23 '13 at 20:20

1

The following is likely to be not relevant, since the space being discussed is undoubtedly $\mathbb{R}^n$ for some $n$. But it is too long for a comment.

Suppose that our space $S$ consists of the points $0$, and $\frac{1}{k}$, where $k$ ranges over the positive integers. Use the ordinary distance function $|x-y|$.

Then $0$ is a limit point of $S$, since every interval about $0$ contains a point of $S$ different from $0$. However, if we pick any $\frac{1}{k}$ in this interval, then there is an interval about $\frac{1}{k}$ that contains no point of $S$ other than $\frac{1}{k}$.

answered Aug 23 '13 at 20:20

André Nicolas

507,029

So the set is perfect? – Don Larynx Aug 23 '13 at 20:36
The derived set of $S$ is not equal to $S$, so no. – André Nicolas Aug 23 '13 at 21:02

Limit point intuition

2 Answers2