# Accumulation point proof. Need assistance.

• Oct 12th 2013, 12:13 PM
mathguy25
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.

• Oct 12th 2013, 02:52 PM
HallsofIvy
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.)
• Oct 12th 2013, 07:41 PM
mathguy25
Re: Accumulation point proof. Need assistance.
Quote:

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?