# Thread: Accumulation point proof. Need assistance.

1. ## Accumulation point proof. Need assistance.

Problem: Prove that a real number s_0 is an accumulation point of a set S if and only if there exists some sequence {a_n} in S such that a_n =/= s_0 for every n in N and lim n-> INF a_n = s_0.

I understand the "<=" implication. I am struggling with the "=>" implication. The definition of accumulation point I am using is "A point s_0 is an accumulation point of a set S if for every epsilon > 0, there exists a number t in S such that 0 < |t - s_0| < epsilon."

So far I no that t is in (s_0 - e, s_0 + e) but I am having difficulty showing there is a sequence {an} in (s_0 - e, s_0 + e) s.t. lim is s_0. I think that since t is in S and {an} is a sequence of numbers in S, that t should be in {an}. So, I have been finding sequences such that an = t/n. I don't know if a_n lies in the interval after a certain point or if s_0 even converges.

I've seen proofs using epsilon = 1/n and stating there exists a sequence of numbers in (s_0 - 1/n, s_0 + 1/n) but no one explains why this is true. I'm missing that little piece (may be obvious to other people). What is the reason that lets us simply say "there exists a sequence of numbers {an} that lies in the interval (s_0 - 1/n, s_0 + 1/n)? I thought we need to show there was a sequence of numbers not simply say there is.

2. ## Re: Accumulation point proof. Need assistance.

You certainly know "in any interval of the real numbers there exist at least one real number", do you not? For any n, there exist a real number in the interval (s0- 1/n, s0+ 1/n). Let a_n be such a number. (There are, in fact, an infinite number. Choose any one.)

3. ## Re: Accumulation point proof. Need assistance.

Originally Posted by HallsofIvy
You certainly know "in any interval of the real numbers there exist at least one real number", do you not? For any n, there exist a real number in the interval (s0- 1/n, s0+ 1/n). Let a_n be such a number. (There are, in fact, an infinite number. Choose any one.)
So we are defining our sequence {an} as

a1 taken to be any point =/= s_0 in the interval (s_0 - 1, s_0 + 1).
a2 taken to be any point =/= s_0 in the interval (s_0 - 1/2, s_0 + 1/2).
a3 taken to be any point =/= s_0 in the interval (s_0 - 1/3, s_0 + 1/3).
... a_n is taken to be any point =/= s_0 in the interval (s_0 - 1/n, s_0 + 1/n).

Okay this makes sense now. So then this implies 0 < |a_n - s_0| < 1/n.

Proof: Let epsilon > 0. Choose n* > 1/epsilon. For n > n* we have that |a_n - s_0| < 1/n < 1/n* < epsilon?

I think I understand now. This t in the definition of accumulation point isn't fixed. It depends on n. Since s_0 was an accumulation point and e = 1/n we know that the "t" in (s_0 - 1, s_0 + 1) is different than the "t" in (s_0 - 1/2, s_0 + 1/2). Then I can just choose a_n to be these t_n. Is that correct?