# Bolzano-Weierstrass Theorem implies Monotone Convergence.

Is my attempt at this problem correct?

Problem. Prove that the bolzano-weirstrass theorem implies the monotone convergence theorem. Do this without recourse to the axiom of completeness.

Proof. Let $(\alpha_n)$ be a increasing sequence that is bounded above, appealing to the Bolzano–Weierstrass Theorem yields a subsequence $(\alpha_{n_k})$ such that $(\alpha_{n_k})\to\beta$ for some $\beta\in\mathbf{R}$. We now show that $(\alpha_n)\to\beta$.

Let $\epsilon>0$, since $(\alpha_{n_k})\to\beta$, there exist a $n_K$ such that $|\alpha_{n_p}-\beta|<\epsilon/2,\forall p\ge K$. Now let $r\in\{n_K,n_K+1,n_K+2,\dots\}$, surely $n_K\leq r\leq n_{r}$ implying $|\alpha_{n_r}-\beta|<\epsilon$.

We now show that $|\alpha_r-\alpha_{n_r}|<\epsilon/2$, since $r\leq n_r$ and $(\alpha_n)$ is increasing it follows that $\alpha_r\leq \alpha_{n_r}<\alpha_{n_r}+\epsilon/2$. Now assume $\alpha_{n_r}-\epsilon/2\ge\alpha_r$ but then $\alpha_{n_r}>\alpha_r$ contradicting the fact that $(\alpha_n)$ is increasing thus $\alpha_{n_r}-\epsilon/2<\alpha_r$. In summary $\alpha_{n_r}-\epsilon/2<\alpha_r<\alpha_{n_r}+\epsilon/2$, equivalently $|\alpha_r-\alpha_{n_r}|<\epsilon/2$.

Thus $|\alpha_r-\beta| = |(\alpha_r-\alpha_{n_r})+(\alpha_{n_r}-\beta)|\leq |\alpha-\alpha_{n_r}|+|(\alpha_{n_r}-\beta|\leq\epsilon/2+\epsilon/2 = \epsilon$. An analogous argument can be constructed for the case where $(\alpha_n)$ is decreasing.

$\blacksquare$

Your argument is a bit mixed up about the indexing of the convergent subsequence given to you by the Bolzano-Weierstrass theorem: $(\alpha_{n_k})$ is a sequence indexed by $k$ and $(\alpha_{n_k}) \to \beta$ as $k \to \infty$. You also don't need to be taking estimates involving $\epsilon/2$ and you can take advantage of the monotonicity to avoid appealing to the triangle inequality.

Let's assume $(\alpha_n)$ is monotone non-decreasing. Given $\epsilon > 0$, the fact that $(\alpha_{n_k}) \to \beta$ says that there is a $K$ such that $\beta - \epsilon < \alpha_{n_k} \le \beta$ whenever $k > K$. But then taking $N = n_{K}$, we have (by monotonicity of $(a_n)$) that $\beta - \epsilon < \alpha_N = \alpha_{n_K} \le \alpha_r \le \beta$ whenever $r > N$. So $(\alpha_n) \to \beta$ if $(\alpha_n)$ is monotone non-decreasing.

Similarly for the case when $(\alpha_n)$ is monotone non-increasing.

(Apologies for being pedantic: I've written "non-decreasing"/"non-increasing" where you've written "increasing"/"decreasing" because we need to allow for sequences where we have $\alpha_n = \alpha_{n+1}$ for some $n$.)

• What do mean by "mixed up about indexing"? – Atif Farooq Jul 31 '18 at 23:25
• You write $\forall n_p \ge n_k$, where your should be talking about the indexes $p$ and $k$. That's like saying"for all $\sin(\theta)$" rather than talking about the value of $\sin(\theta)$ for all $\theta$. – Rob Arthan Jul 31 '18 at 23:30
• I have made some edits what would you say now about the argument above? – Atif Farooq Jul 31 '18 at 23:32
• You don't need the estimates involving $\epsilon/2$ or the triangle inequality. – Rob Arthan Jul 31 '18 at 23:33
• Thanks for the green tick! I was just writing the following: "think of the sequence as a train line with infinitely many stops: Bolzano-Weierstrass tells you that there is an express train that may miss some of the stops but has a limit point at a final destination. If there is a slow train in front of the express, the slow train is going to have the same limit point." – Rob Arthan Jul 31 '18 at 23:54