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 $\epsilon$.

  • $\begingroup$ Easier is to prove by contradiction $\endgroup$ Aug 28, 2022 at 17:34
  • 1
    $\begingroup$ Good. BTW you don't need \operatorname {sup}. Just type \sup. Also \inf . $\endgroup$ Aug 28, 2022 at 21:34

1 Answer 1


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 $l_1<l_2.$

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.


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