Answer to Question #116679 in Real Analysis for Supun Sampath

Question

If {xn} does not converge to L, show that there exist  > 0 and a subsequence {xnk
}
of {xn} such that |xnk − L| ≥  for each k ∈ N.

Answer 1

Suppose (x_n) does not converge to L.

So $\exists\isin$ > 0 such that for each integer N there is an integer n = n(N) ≥ N

|x_n − L| ≥ 0.

For N = 1 we obtain n₁ = n(1) ≥ 1

such that |x_n1 − L| ≥ 0.

let N = n_k + 1 to obtain n_k+1 = n(N) ≥ n_k + 1

such that |x_nk+1 − L| ≥ 0.

SO, there is a subsequence (x_nk for k = 1, 2, · · ·

such that |x_nk − L| ≥ 0 ∀ k = 1, 2, ·