Let {a_n} be a sequence and let set S={a_n|n?N}. If s0 is an accumulation point

Question

Let {a_n} be a sequence and let set S={a_n|n?N}. If s0 is an accumulation point of S,prove that there exists a subsequence {a_n} that converges to s0

Explanation / Answer

If x is an accumulation point of a set S, then there exists a sequence, in S that converges to x. Proof: Let e > 0 If x is an accumulation point of S, then the intersection of a ball with any arbitrary radius, and a center of x, with S is non-empty. In particular, we may take the radius to be 1/n, where n is any positive interger. That is B_(1/n)(x) n S ? Ø (B_(1/n)(x) is "ball of radius 1/n and center x). But since the sequence 1/n converges to zero, we can find an N such that for all n > N, 1/n N, B_(1/n) (x) is inside B_e (x) So B_e (x) n S ? Ø Since we can do this infinitely many times for k > N, we have constructed a subsequence that converges to x. why?: For each k > N, we can find an element that is both in S and in B_e(x) Since it is in S, it is of the form a_n_k, and since it is in B_e(x) it follows that |a_n_k - x| N) Thus a_n_k converges to x

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