A point \(x \in \mathbb{R}\) is a limit point of the set \(A\) if for every \(\epsilon > 0,\) \[((x - \epsilon, x + \epsilon) \cap A) - \{x\} \neq \emptyset\] where \((x - \epsilon, x + \epsilon)\) is the interval \(\{y \in \mathbb{R} : x - \epsilon < y < x + \epsilon\}.\)

The set of all limit points of \(A\) is written \(A'.\)

For example, the limit point of the set \(A = \{\frac{1}{n} : n \geq 1\}\) is the point 0. Every interval around 0 contains some point of \(A.\)

The limit points of the set \(B = (1, 2]\) is the set \(B' = [1, 2].\)

The set \(C = \{1, 2, 3\}\) has no limit points, so \(C' = \emptyset.\) The reason \(1\) is not a limit point of \(C\) is that \[\left(\left(\frac{1}{2}, \frac{3}{2}\right) \cap A\right) - \{1\} = \emptyset\] In the above interval, \(\epsilon = \frac{1}{2}.\)

Element \(\in\)

Empty set \(\emptyset\)

Intersection \(\cap\)

Set difference \(-\)

The following is the reason the points are called "limit points".

Proposition: A point \(x \in \mathbb{R}\) is a limit point of the set \(A\) if and only if there exists a sequence of points \((a_n : 1 \leq n \leq \infty)\) such that

• \(\lim_{n \rightarrow \infty} a_n = x,\) and
• \(a_n \neq a_m\) if \(n \neq m.\)

Proof: Since the statement is an "if and only if statement", we need to prove both directions.

First, suppose \(x\) is a limit point of \(A.\) By definition of a limit point, there exists a point \(a_1 \in A\) such that \(a_1 \neq x\) and \(a_1 \in (x - 1, x + 1).\) Since \(a \in (x - 1, x + 1),\) \(|x - a_1| < 1.\)

Let \(\epsilon_1 = \frac{|a_1 - x|}{2} \wedge \frac{1}{2}.\) Since \(a_1 \neq x,\) \(\epsilon > 0.\) Using the definition of a limit point again, there exists a point \(a_2 \in A\) such that \(a_2 \neq x\) and \(a_2 \in (x - \epsilon_1, x + \epsilon _1).\) Since \(a_2 \in (x - \epsilon_1, x + \epsilon_1),\) \(a_2 \neq a_1\) and \(|x - a_2| < \frac{1}{2}.\) Next let \(\epsilon_2 = \frac{|a_2 - x|}{2} \wedge \frac{1}{3}.\)

This process can be iterated to generate a sequence \(a_n\) such that \(a_n \neq x,\) the \(a_n\)'s are distinct, and \(|x - a_n| < \frac{1}{n}\) for all \(n.\) The sequence \(a_n : n \geq 0\) is a sequence of distinct point that converge to \(x.\)

To show the other direction, assume that \((a_n : 1 \leq n \leq \infty)\) is a sequence described in the proposition. By definition of \(\lim_{n \rightarrow \infty} a_n = x,\) if \(\epsilon > 0\) there exists \(N\) such that if \(n \geq N,\) \(|x - a_N| < \epsilon.\) Since \(a_n \neq a_m\) if \(n \neq m,\) there must exist \(a_M\) with \(M \geq N\) such that \(a_M \neq x.\) Then \(a_M \in (x - \epsilon, x + \epsilon) \cap A) - \{x\}.\) Since \(\epsilon > 0\) was arbitrary, for any \(\epsilon > 0\) the set \((x - \epsilon, x + \epsilon) \cap A) - \{x\}\) is non-empty.

A discrete set is a set \(A\) such that \(A' = \emptyset.\)

The set \(\{1, 2, 3, \dots\}\) is a discrete set in \(\mathbb{R}\). Any finite subset of \(\mathbb{R}\) is discrete.

The set \(S = \left\{\frac{1}{n} : n \geq 1\right\}\) is not a discrete set, because \(S' = \{0\} \neq \emptyset.\)