0
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.