1. ## Limit Points

Definition: Suppose that S is a set of real numbers and that x is any real number. We say that the number x is a limit point of the set S if for every number $\delta>0$ we have $(x-\delta, x+\delta)\cap S$ \ $\{x\}$ is not equal to $\emptyset$. The set of limit points of a given set S is written as L(S).

Problem: Given any set S of real numbers, prove that the set L(S) must be closed.

2. Let x1,x2,... be a convergent sequence of L(S) and let x be its limit. We must prove that x belongs to L(S). This means proving that x is a limit point of S. Fix epsilon>0. Take x_n such that | x_n - x | < epsilon/2. Since x_n is a limit point of S, there is a y in S such that | y - x_n | < epsilon/2. Hence y is in the epsilon neighborhood of x (apply triangle inequality). Since epsilon was arbitrary, this proves that any neighborhood of x has a point of S, hence x is a limit point of S. Therefore x is in L(S) and L(S) is closed.

3. Originally Posted by Slazenger3
Definition: Suppose that S is a set of real numbers and that x is any real number. We say that the number x is a limit point of the set S if for every number $\delta>0$ we have $(x-\delta, x+\delta)\cap S$ \ $\{x\}$ is not equal to $\emptyset$. The set of limit points of a given set S is written as L(S).

Problem: Given any set S of real numbers, prove that the set L(S) must be closed.
An alternative way of thinking about it is this:

Let $s$ be a limit point of $L(s)$ then for every neighborhood $N$ of $s$ there is a point $s'$ of $L(s)$, but $N$ is also a neighborhood $s'$ and so it must contain infinitely points of $S$ in particular it must contain a point of $S$ different from $s$...so