is a accumulation point if for every there exists a with and
Choose a .
For every you can choose a with such that:
Now as
I'll leave the other implication to you.
Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.
I'm not really sure how to start. I know since it's an if and only if statement we have to prove it both ways.
You did not tell us what sort of top-space we are considering.
I will assume that you are working in a metric space (so the statement is true).
If then is a ball.
Using the definition of accumulation point we get:
.
.
Now you have a way to generate your sequence by induction.
Let be a sequence in such that and . Now, let be arbitrary. Since there exists an such that and since and we see there exists a point of distinct from in this ball. The conclusion follows from the arbitrariness of the open ball.
Conversely, choosing any arbitrary point in finishes the job. There is no need for them to be all distinct.