I want to prove this statement,

If $x_n$ is an increasing sequence and if some subsequence of it converges, then $x_n$ also converges.

My proof is that suppose $x_{n(k)}$ is the subsequence that converges. Since $x_n$ is an increasing sequence, we only need to show that it is bounded; then by the Monotone Bounded Theorem, $x_n$ converges.
Since $x_{n(k)}$ converges, then it is bounded say by $M$.
Suppose $x_n^*$ is any arbitrary term of the sequence $x_n$, then since the $x_n$ is increasing, $ x_n^* \leq x_{n(k)} $ for some $ k \in \mathbb{N}$. Since the subsequence is bounded, then $x_n^* \leq M$ as well. Thus $x_n$ is bounded and the sequence converges.

Is my proof correct? Thank you.

  • 2
    $\begingroup$ Yes, you proof is correct. $\endgroup$ – user258700 Jul 27 '16 at 2:23
  • 1
    $\begingroup$ The idea to use the monotone convergence theorem is right. But your write-up is not very clear and your notatoin is really not good. $\endgroup$ – Gregory Grant Jul 27 '16 at 2:23

A self-contained proof (no Monotone Bounded Theorem).

Assume that $\{x_{n_k}\}_{k\geq 1}$ is the converging subsequence and let $L=\lim_{k\to \infty}x_{n_k}$. Then for all $\varepsilon>0$ there is a $K$ such that $|x_{n_k}-L|<\varepsilon$ for all $k\geq K$.

Now, for any $n\geq n_K$, there is a $J\geq K$ such that $n\leq n_J$. Since $\{x_{n}\}_{n\geq 1}$ is increasing, it follows that $$x_{n_K}\leq x_{n} \leq x_{n_J}\leq L$$ and $$|x_{n}-L|=L-x_{n}\leq L-x_{n_K}=|x_{n_K}-L|<\varepsilon.$$ Hence $\lim_{n\to \infty}x_{n}=L$ by definition of limit.


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