# Proof verification: Every bounded quasi-increasing sequence is convergent.

A sequence is quasi-increasing if: $$\forall \epsilon > 0 \quad \exists N \quad \forall n,m \quad \left( n> m \geq N \implies x_n > x_m - \epsilon \right)$$

My strategy is to show that every bounded quasi-increasing sequence is a Cauchy sequence, i.e: $$\forall \epsilon>0 \quad \exists N \quad \forall n,m \quad \left( n,m \geq N \implies | x_n - x_m | < \epsilon \right)$$

Since $$(x_n)$$ is quasi-increasing: $$\exists K \quad \forall n > m \geq K \quad \left( x_m - x_n < \epsilon/2 \right)$$

Let $$s_K = \operatorname{sup} \{ x_n : n \geq K\}$$. By definition of supremum: $$\exists N \geq K \quad x_N > s_K - \epsilon/2 \quad \text{or, equivalently:}\quad s_K - x_N < \epsilon /2 \tag{1}$$

I claim that the sequence is $$\epsilon$$-Cauchy beyond $$N$$. Consider any $$n,m \geq N$$; further, assume that $$n > m$$. Since $$n,m \geq N \geq K$$, the sequence is quasi-increasing for $$n,m$$, therefore: $$x_m - x_n < \epsilon /2 < \epsilon$$; this proves the negative side of the absolute difference. For the positive side:

\begin{align} &x_n - x_m \leq s_K - x_m & (\text{n > N \geq K \implies x_n \leq s_K})\\ &s_K - x_m < s_K - x_N + \epsilon /2& (\text{ m \geq N \geq K \implies x_N - x_m < \epsilon /2})\\ &s_K - x_N + \epsilon /2 < \epsilon /2 + \epsilon / 2& (\text{s_K - x_N < \epsilon / 2 from (1)}) \end{align}

For $$n < m$$, just switch $$n$$ and $$m$$. For $$n = m$$, the difference is $$0$$, so it always valid. The number $$N$$ depends on $$K$$ which depends on $$\epsilon$$. So there is always a valid $$N$$ for every $$\epsilon$$.

• Easier is to prove by contradiction Commented Aug 28, 2022 at 17:34
• Good. BTW you don't need \operatorname {sup}. Just type \sup. Also \inf . Commented Aug 28, 2022 at 21:34

I'll give an alternative approach: by contradiction. Suppose $$(x_n)_n$$ is quasi-increasing and bounded.
Suppose $$E:= \{ x_n: n\in\mathbb{N} \}\$$ has at least two limit points, $$\ l_1\$$ and $$\ l_2,\$$ with $$l_1
Let $$\varepsilon = \frac{l_2-l_1}{50},\$$ and find $$N$$ such that $$n>m>N\implies x_n > x_m - \varepsilon.$$
Since $$l_2$$ is a limit point of $$E,\ \exists\ k>N\$$ such that $$x_k > l_2 - \frac{l_2-l_1}{10},$$ i.e. $$x_k$$ is much closer to $$l_2$$ than to $$l_1.$$
But for all $$j\geq k,\ x_j > x_k - \varepsilon,\$$ i.e. $$\ x_j\$$ is sectioned away from $$l_1,\$$ and so $$l_1$$ is not a limit point of $$E,\$$ a contradiction. Therefore $$E$$ has at most one limit point.
Since $$E$$ is bounded, Bolzano-Weierstrass says $$E$$ must have a limit point. Since there is at most one limit point, there must be exactly one limit point: therefore $$E$$ is convergent and therefore Cauchy.