Show that if a set of real numbers E has at least one point

Show that if a set of real numbers E has at least one point of accumulation, then for every > 0 there exist points x, y E so that 0 < |x-y| <.

Solution

A function f : E R, where E R, is continuous at y E if for every neighborhood A of f(y) there is a neighborhood B of y such that x E B implies that f(x) A. As we know that y belongs to E and y is an accumulation point of E, then continuity of f at y is equal to,

lim(xy) f(x) = f(y).

Now if y is an isolated point of E, then the continuity condition holds. Therefore we can say that for small > 0, the only point x E with |x y| < is x = y, and also 0 = |f(x) f(y)| < , where >0.