Show that (a_n) must converges to a.?

Question

Assume (a_n) is a bounded sequence with the property that every convergent subsequence of (a_n) converges to the same limit a element of the real numbers. Show that (a_n) must converges to a.

Answer 1

Suppose the contrary: a_n does not converge to a. Then ∃ε>0: ∀N∈ℕ, ∃n>N: |a_n-a|>ε. Then define a subsequence (a_(k_m)) as follows: let a_(k_1) be the first element of (a_n) whose distance from a is greater than ε, and let a_(k_(m+1)) be the first element of (a_n) that occurs after a_(k_m) and whose distance from a is greater than ε (it is possible, because ∀N∈ℕ, ∃n>N: |a_n-a|>ε, so in particular ∃n>k_m such that |a_n-a|>ε). Now, no subsequence of (a_(k_m)) can converge to a, because all of the elements in (a_(k_m)) are bounded away from a. However, since (a_n) is a bounded sequence, so is (a_(k_m)), so per the Bolzano-Weierstrass theorem, (a_(k_m)) has a convergent subsequence, which is then a convergent subsequence of (a_n) and thus per the stipulated property, converges to a -- a contradiction. Therefore, our initial assumption that (a_n) did not converge to a must be wrong, so (a_n) converges to a. Q.E.D.