Let $\mathcal R = \{ r \in \mathbb Q^\mathbb N \mid r \text{ is a Cauchy sequence}\}$ and $\mathbf c_0 = \{ r \in \mathbb Q^\mathbb N \mid r \text{ is a null sequence}\}$. Then we construct the set $R$ of real numbers as the quotient ring of $\mathcal R$ by the ideal $\mathbf c_0$, that is, $R = \mathcal R/ \mathbf c_0$.
Theorem: Every increasing and bounded from above sequence $(x_k)_{k \in \mathbb N}$ in $R$ has a supremum $\sup \{x_k \mid k \in \mathbb N\}$.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
Lemma: Every increasing and bounded from above sequence in $\mathbb Q$ is a Cauchy sequence.
We denote by $\overline a$ the constant sequence $(a,a,\ldots,a)$.
WLOG, we assume that $(x_k)_{k \in \mathbb N}$ is strictly increasing. Each $x_k$ has the form $[r^k]$ where $r^k = (r^k_n)_{n \in \mathbb N} \in \mathcal R$. For all $k \in \mathbb N$, there exists $n_k, N_k \in \mathbb N$ such that $r^{k+1}_n - r^k_n \ge 1/N_k$ for all $n \ge n_k$. WLOG, we can assume that $(n_k)_{k \in \mathbb N}$ is strictly increasing.
Since $r^k$ and $r^{k+1}$ are both Cauchy sequences, there exists $m_k \ge n_k$ such that $$r^k_n - r^k_{m_k} < \frac{1}{4N_k} \quad \text{and} \quad r^{k+1}_{m_k} - r^{k+1}_{n} < \frac{1}{4N_k}, \quad n \ge m_k$$
Let $s_k := r^k_{m_k} + 1/(2N_k)$, we have $$s_k - r^k_n > \frac{1}{4N_k} \quad \text{and} \quad r^{k+1}_{n} - s_k > \frac{1}{4N_k}, \quad n \ge m_k$$
As such, $$x_k = [r^k] < [\overline{s_k}] = s_k [\overline 1] < [r^{k+1}] = x_{k+1}, \quad k \in \mathbb N \quad (\star)$$
Let $s := (s_k)_{k \in \mathbb N}$. By construction, the sequence $s$ is strictly increasing and bound from above. By Lemma, $s$ is a Cauchy sequence and thus $s \in \mathcal R$.
Next we prove that $[s] \in R$ is the supremum of $(x_k)_{k \in \mathbb N}$. We have $x_k < [\overline{s_k}] \le [s]$ for all $k \in \mathbb N$, so $[s]$ is an upper bound of $(x_k)_{k \in \mathbb N}$. Assume the contrary that there exists $\alpha \in R$ such that $x_k \le \alpha < [s]$ for all $k \in \mathbb N$. It follows from $(\star)$ that $$[\overline{s_k}] < x_{k+1} \le \alpha < [s], \quad k \in \mathbb N$$ which contradicts the fact that $s = (s_k)_{k \in \mathbb N}$. Therefore, $[s] = \sup \{x_k \mid k \in \mathbb N\}$.