Let $(x_n \in \mathbb{R})_{n \in \mathbb{N}}$ be a cauchy sequence of reals.
Each number in the sequence is an equivalence class of Cauchy sequences of rational numbers. That is, $x_n = \left[x_{n, m} \in \mathbb{Q} \right]_{m \in \mathbb{N}}$ under the usual equivalence relation between Cauchy sequences.
The equivalence class may contain several Cauchy sequences which are "equivalent" to each other. Any one of such Cauchy sequences $(x_{n, m} \in \mathbb{Q})_{m \in \mathbb{N}}$ can be a "representative" for the real number $x_n$.
Is it possible to find a representative $(x_{n, m} \in \mathbb{Q})_{m \in \mathbb{N}}$ such that $\lim x_{n,n} \neq \lim x_n$?
Okay, here is my try for redemption. Consider: $$ \begin{aligned} x_1 &:= 1,12345 \dots , \\ x_2 &:= 1,02345 \dots, \\ x_3 & := 1,00345 \dots, \\ \dots \end{aligned} $$ Then $x_n \to 1$. Now consider the following cauchy sequences $$ \begin{aligned} x_{nm} := \begin{cases} 1,\underbrace{0\dots0}_{n-1}n \,(n+1) \dots m\, 0 \dots & n < m \\ 0 & \text{otherwise} \end{cases} \end{aligned} $$ So for example we have $x_{35} = 1,0034500\dots$ . So $x_{nm} \to x_n$, right? But in particular $x_{nn} = 0$ and so $x_{nn} \to 0$.
It seems to me this works?