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Borealis
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Understanding proof that $\mathbb{R}$ is Cauchy complete

Context: The real numbers were constructed using Cauchy sequences of rational numbers, where every real corresponds to the equivalence class of a rational Cauchy sequence. The fact that $\mathbb{R}$ is Cauchy complete will be used later to prove the Least Upper Bound Property, so we can't use that here.

The following proof was provided for $\mathbb{R}$ being Cauchy complete. Note that $\mathbb{Q}$ being dense in $\mathbb{R}$ has already been proven and the definition of absolute value mentioned in the proof is this one:

Definition. Let $x\in\mathbb{R}$. The absolute value function $|\cdot |:\mathbb{R}\to\mathbb{R}_{\geq 0}$ is defined by \begin{equation*} |x| := \begin{cases} x, & x\geq 0 \\ -x, & x<0 \end{cases} \end{equation*} Furthermore, when considering $x$ as the equivalence class of a Cauchy sequence $\left(x_n\right)$, we may define the absolute value function by \begin{equation*} \left|\left[\left(x_n\right)\right]\right| := \lim_{n\to\infty} \left|x_n\right|. \end{equation*}

Now the proof.

Proof: To prove that $\mathbb{R}$ is Cauchy complete, we wish to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{R}$, $\exists x\in\mathbb{R}$ such that $\lim_{n\to\infty} x_n = x$. Because $\mathbb{Q}$ is dense in $\mathbb{R}$, it is sufficient to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{Q}$, $\exists x\in\mathbb{R}$ such that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*} Let the desired $x$ be $x=\left[\left(x_1,x_2,\ldots\right)\right]$. We may observe that $x\in\mathbb{R}$ since $\left(x_n\right)$ is Cauchy sequence of rational numbers (assumed above). Then, by our original statement, we wish to show that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = \left[\left(x_1,x_2,\ldots\right)\right]. \end{equation*} By definition, this is equivalent to \begin{equation*} \lim_{n\to\infty} \left|\left[\left(x_n,x_n,\ldots\right)\right] - \left[\left(x_1,x_2,\ldots\right)\right]\right| = 0. \end{equation*} Then by the definition of absolute value: \begin{equation*} \lim_{n\to\infty} \left|\left[\left(x_n,x_n,\ldots\right)\right] - \left[\left(x_1,x_2,\ldots\right)\right]\right| = \lim_{n\to\infty} \lim_{m\to\infty} \left|x_n - x_m\right|. \end{equation*} Finally, the limit on the right is $0$ because $\left(x_n\right)$ is Cauchy, which is equivalent to what we originally wanted to show: \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*}

My problem: First of all, I'm not clear on the notation. Is $\left(x_n,x_n,\ldots\right)$ an infinite sequence where each term is the $n^{\text{th}}$ term of $\left(x_n\right)$?

Second, if that's the case, why is it sufficient to show that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = \left[\left(x_n\right)\right]. \end{equation*} (Where $[(x_n)]=[(x_1,x_2,\ldots)]$). Isn't that kind of... self-evident?

Third, I don't understand the premise. Why does $\mathbb{Q}$ being dense in $\mathbb{R}$ $\implies$ it is sufficient to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{Q}$, $\exists x\in\mathbb{R}$ such that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*} Seeing as how $(x_n)\in\mathbb{Q}$, $[(x_n,x_n,\ldots)]$ is basically just a rational number in $\mathbb{R}$. So, we're saying that the limit of a rational number is equal to a real?

Note: I've looked at this question, but the proofs are completely different, especially since the proof in the above question uses the least upper bound property.

Any help would be much appreciated.

Borealis
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