Abbott Exercise 2.6.7 Assume the Bolzano-Weierstrass theorem and prove with it the Monotone Convergence Theorem without using the Archimedean Property.
I understand that I cannot use the fact that 1/n converges to zero, but can I use the least upper bound property in the proof or it presupposes the Archimedean Property? Thanks in advance.
 A: Assume that $(a_{n})$ is an increasing and bounded sequence. By Bolzano-Weierstrass Theorem, it has a convergent subsequence $(a_{n_{k}})$, say, $a_{n_{k}}\rightarrow l$ as $k\rightarrow\infty$.
Given $\epsilon>0$, there exists a $K$ such that $|a_{n_{k}}-l|<\epsilon$ for all $k\geq K$. For all $n\geq n_{K}$, then $a_{n}\geq a_{n_{K}}\geq l-\epsilon$. 
Now we prove that $\sup_{n\geq 1}a_{n}\leq l$, if this can be done, then $l-\epsilon<a_{n}\leq l$ for $n\geq n_{K}$, which shows that both $\sup_{n\geq 1}a_{n}=l$ and $a_{n}\rightarrow l$ as $n\rightarrow\infty$.
We see that for all $i$ with $n_{k}\leq i\leq n_{k+1}$, since $(a_{n})$ is increasing, then $a_{n_{k}}\leq a_{i}\leq a_{n_{k+1}}$, so $a_{1}\leq a_{2}\leq\cdots\leq a_{n_{k}}\leq a_{n_{k}+1}\leq\cdots\leq a_{n_{k+1}}\leq\cdots$ is then established. Since $a_{n_{k}}<l+\epsilon$ for all $k\geq K$, we get $a_{i}<l+\epsilon$ for all $i=1,2,...$ and hence $\sup_{n\geq 1}a_{n}\leq l+\epsilon$. Since this is true for all $\epsilon>0$, we get $\sup_{n\geq 1}a_{n}\leq l$, we are done.
Pretend that we don't assume the existence of $\sup_{n\geq 1}a_{n}$ at the first place. We still manage to prove that $l+\epsilon$ is an upper bound for $(a_{n})$ for each $\epsilon>0$. We claim that $l$ is an upper bound for $(a_{n})$. Assume that it were not, then there is an $n$ such that $a_{n}>l$. Take $\epsilon=(a_{n}-l)/2$, then by assumption $l+\epsilon$ is an upper bound for $(a_{n})$, in particular, $l+\epsilon>a_{n}$. But $l+\epsilon=l+(a_{n}-l)/2=(a_{n}+l)/2<a_{n}$, a contradiction.
So we establish the fact that $l$ is an upper bound for $(a_{n})$. But for any $\epsilon>0$, by the second paragraph, we can find an $n$ such that $a_{n}>l-\epsilon$, so by the definition of least upper bound, $l$ is really the least upper bound, and we have proved the existence of $\sup_{n\geq 1}a_{n}$. 
