I'm having some trouble understanding the following proof that every real Cauchy sequence is bounded:
Proof:
Let $(a_n)_{n\in \mathbb N}$ be a Cauchy sequence and suppose, for the sake of contradiction, that $(a_n)_{n\in \mathbb N}$ is not bounded. Then, there exists $\sigma\in\ ^*\mathbb N_\infty$ such that $a_\sigma\not\in\ ^*\mathbb R_\flat$. Consider the set $$\mathcal{N}=\{n\in\ ^*\mathbb N:|a_\sigma - a_n|<1\}$$ then this set is an internal set. Because $(a_n)_{n\in \mathbb N}$ is a Cauchy sequence, then $$\forall\sigma,\nu\in\ ^*\mathbb N_\infty:a_\sigma\approx a_\nu$$ This allows us to conclude that $^*\mathbb N_\infty\subseteq\cal N$. Because $\cal N$ is internal and contains infinite elements arbitrarily small, then it must have some finite element $n_1 \in \mathbb N$, but that would imply that $a_{n_1}\not\in \ ^*\mathbb R_\flat$, which is a contradiction, thus the sequence must be bounded.
I don't understand why the set $\cal N$ is internal. The definition the book I'm reading gave was:
A subset $A\subseteq\ ^*\mathbb R$ if there exists a sequence of subsets $(A_n)_n$ of $\mathbb R$ such that, for any number $x\equiv [(x_n)_n]$ one has that: $$x\in A\iff\{n\in \mathbb N:x_n\in A_n\}\in \cal U$$ where $\cal U$ is the ultrafilter used to construct the hyper-reals.
But I can't find any sequence of sets that satisfy this property for the set $\cal N$. Why is this set internal?
Disclaimer: The book I'm using to learn nonstandard analysis is not very standard, so if any of this notation is not common, let me know.
Did the book not prove yet that every Cauchy sequence in the metric space $(\mathbb{R},d_{Eu})$ converges and that every convergent sequence in said space is bounded? Try this direct proof then in lieu of your question.
Let $(a_n)$ be a Cauchy sequence. Then for any $\epsilon>0$ there exists a natural number $H(\epsilon)$ such that for $m,n\geq H(\epsilon)$, $|a_n-a_m|<\epsilon$. Take $\epsilon=1$ and consider $H(1)$. For all $n\geq H(1)$, $|a_n-a_{H(1)}|<1$ (Why?). Applying the triangle inequality gives us $|a_n|<|a_{H(1)}|+1$. Since $H(1)$ is a natural number there are a finite amount of terms in $(a_n)$ that come before $a_{H(1)}$. Knowing this, we let $M=max\{|a_1|+1,|a_2|+1,...,|a_{H(1)}|+1\}$ and it will be guaranteed that the absolute value of any term of $(a_n)$ will be less than $M$. Hence, $(a_n)$ is bounded.
This should be easier to follow. Perhaps someone else can better examine the book's proof but this is the proof I was taught.