Apply the compactness theorem to show that there are nonstandard models [closed]

Apply the compactness theorem to show that there are nonstandard models of complete arithmetic (the set of all arithmetic truths). Apply it likewise to show there are nonstandard models of analysis, which contain "infinitesimal" real numbers.

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closed as not a real question by Andres Caicedo, Austin Mohr, Asaf Karagila, LVK, WilliamAug 20 '12 at 17:37

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What have you tried? –  Austin Mohr Apr 30 '12 at 22:39

Here is a detailed sketch for half of what you are asking. Let $L$ be the usual language of Peano Arithmetic, with constant symbol $0$, unary function symbol $s$, and binary function symbols $p$ (for plus) and $t$ (for times). Add a constant symbol $B$ to $L$, to form the language $L'$.
Let $T$ be the following set of axioms: (ii) all sentences of the language $L'$ that are true in the natural numbers, under the usual interpretation of $0$, $s$, $p$, and $t$; (ii) the sentences $\lnot(B=0)$, $\lnot(B=s(0))$, $\lnot(B=s(s(0)))$, $\lnot(B=s(s(s(0))))$, and so on forever.
Note that every finite subset of $T$ has a model, indeed has the natural numbers as a model, if $B$ is interpreted suitably. So by the Compactness Theorem the theory $T$ has a model $M$. That model cannot be the standard model, since the interpretation of $B$ in $M$ cannot be equal to the interpretation of $0$, nor of $s(0)$, nor of $s(s(0))$, nor of $s(s(s(0)))$, and so on.
A similar argument will produce a non-standard model of analysis. It is best to use a very large language $L$ as a starting point. Throw in a constant symbol for every real number, a function symbol of the appropriate arity for every function $f:\mathbb{R}^n\to \mathbb{R}$, and a predicate symbol of the appropriate arity for every relation, that is, for every subset of $\mathbb{R}^n$. Add a constant symbol $\epsilon$ to make a new language $L'$. Let $T$ be the theory whose axioms are all sentences of $L'$ true in the reals, together with an infinite number of special axioms that involve $\epsilon$. Continue.