Here's the question:

Assume $a_n$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit $a \in \mathbb{R}$. Show that $(a_n)$ must converge to $a$.

Here's my proof for it: Assume $(a_n)$ converges to some $b\in R$ with $b\ne a$. Then its subsequences must converge to $b$. However, this leads to a contradiction since all of its subsequences converge to $a$. Thus, $b=a$ and the sequence must converge to $a$. $\square$

Is my proof correct? Apparently, my textbook uses the definition of a sequence not converging to some $x$ and then uses the Bolzano-Weirstrass Theorem to prove it.

  • 7
    $\begingroup$ No, it's not correct. The negation of "$a_n$ converges to $a$" is not "$a_n$ converges to some $b \neq a$". Divergence is also a possibility. $\endgroup$ – MathematicsStudent1122 Dec 23 '17 at 14:02

Hint: Since the sequence is bounded, $\{a_n\}\subseteq [-R, R]$ for some positive real number $R$. Assume for contradiction that $(a_n)$ does not converge to $a$. Then $\exists \varepsilon >0$ such that $\forall N$, $\exists n \geq N$ such that $|a_n- a| \geq \epsilon$. Then we can pick a subsequence (why?) $\{a_{n_k}\} \subseteq [-R, R] \setminus [a- \varepsilon, a+\varepsilon]$. What can you say about the behavior of this subsequence given that it is bounded?

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