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# Suppose x is an accumulation point of {a_n: n in J}. Show that there is a subsequence of {a_n} that converges?

Suppose x is an accumulation point of {a_n: n in J}. Show that there is a subsequence of {a_n} that converges to x.

### 1 Answer

- Mr. BrightsideLv 510 years agoFavorite Answer
I'll write S instead of {a_n: n in J}.

If x is an accumulation point of a set S, then there exists a sequence, in S that converges to x.

Proof:

Let ε > 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) ∩ S ≠ ∅

(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 < ε

So for all n > N, B_(1/n) (x) is inside B_ε (x)

So B_ε (x) ∩ S ≠ ∅

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_ε(x)

Since it is in S, it is of the form a_n_k, and since it is in B_ε(x) it follows that |a_n_k - x| < ε (for k >N)

Thus a_n_k converges to x

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