I'm having some trouble proving that BW implies MCT. Here's what I've done so far

If a bounded sequence $(a_n)$ is monotone, then the sequence is convergent.

Case 1: $(a_n)$ is always increasing.

Proof. By BW, there exists a subsequence $(a_{n_k})$ that converges to some limit $L$. Observe that $$|a_n-L| = |a_n-a_{n_k}+a_{n_k}-L|\leq |a_n-a_{n_k}|+|a_{n_k}-L|$$ I think the next step is to argue that you can make the left term as small as possible but I'm not sure how to show that.


You are close.

MCT : Every bounded monotone sequence is convergent.

BW: There exist a convergent subsequence $(a_{n_k})$.

Consider a monotonically increasing sequence.

Let $\epsilon >0$ be given.

There is a $k_0$ such that for $k \ge k_0$:

$|a_{n_k} -L|< \epsilon $.

Let $N:= n_{k_0}$.

For every $n \ge N$ there is a $k \ge k_0$ with

$n_k \le n \lt n_{k+1}$.

Since $(a_n)$ is mon. increasing we have

$a_{n_k} \le a_n \le a_{n_{k+1}} \le L$,


$|a_n-L| \le |a_{n_k} -L| \lt \epsilon$.

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  • $\begingroup$ Thank you very much! Can I modify my initial approach according to the above reply or is the way you did it the only way ? $\endgroup$ – Adam Mar 26 '18 at 22:41
  • $\begingroup$ Oh yeah I forgot that $a_{n_k}$ is also a Cauchy sequence. Thanks! $\endgroup$ – Adam Mar 26 '18 at 23:05
  • $\begingroup$ I'm not sure which line you're referring to. Basically what you did was set $N$ to the first $n_k$ term that satisfies the epsilon inequality and that guarantees that $a_n$ (for $n\geq N$) is between $(a_{n_k})$ and $(a_{n_{k+1}})$ since the difference is at least one, right ? $\endgroup$ – Adam Mar 26 '18 at 23:32

Note that $L>a_{n}$ for all n (why?). Since $a_{n_{k}}$ converges to $L$, $L-a_{n_{k}}$ is smaller than a given $\varepsilon$ if k is bigger than some Index $K$. But now since $a_{n}$ is increasing, $L-a_{n}$ must also be smaller than $\varepsilon$ for $n>n_{K}$.

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  • $\begingroup$ Since $(a_n)$ is monotone, if $L$ was less than any element then it would be less than all the infinitely many elements after it but that would contradict the fact that it's the limit. Thanks for the explanation! $\endgroup$ – Adam Mar 26 '18 at 22:03

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