Nonstandard Natural Numbers

Question

Let $n \in \mathbb{N}$ and consider the set of nonstandard natural numbers $^*\mathbb{N}$ in sense of nun standard analysis.

I want to show that for each $m \in (^*\mathbb{N}) \backslash \mathbb{N} $ we have $n <^* m$ where $<^*$ is the extension of $<$ to nonstandard natural numbers.

My ideas:

I tried to consider the set $A:= \{a \in \mathbb{N} \vert a < n\}$ and it's extension $^*A$ in order leading it to a contradiction or maybe directly by showing $^*A \subset \mathbb{N}$ using transfer principle.

But till now I don't get how to construct it.

Do you mean $n<^* m$ for all $n\in\Bbb N$? – Berci Apr 30 '18 at 16:38 — Berci, Apr 30 '18 at 16:38
Yes, of course. Thank you. – user267839 Apr 30 '18 at 16:41 — user267839, Apr 30 '18 at 16:41

score 5 · Accepted Answer · answered Apr 30 '18 at 16:37

For a fixed natural number $n$, consider the first order formula $$\forall x:(x=0)\lor(x=1)\lor\dots\lor(x=n) \lor (x>n) $$ This is valid in $\Bbb N$, so it also holds in ${}^*\Bbb N$.
In particular, if the given $m\in{}^*\Bbb N$ is not in $\Bbb N$, it follows that $m>n$.

Nonstandard Natural Numbers

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