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 numberwe have
\
is not equal to
. 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.
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.
An alternative way of thinking about it is this: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
Problem: Given any set S of real numbers, prove that the set L(S) must be closed.
Letbe a limit point of
then for every neighborhood
of
of
in particular it must contain a point of
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