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I'm having trouble understanding why we need to add constants to a language in order to prove something when using the compactness theorem, in particular this:

Let L be a language with just a 2 place-symbol < for "less than". Show that there is no L-theory T such that the models of T are precisely the well-ordered sets.

The solution I have for this is: make a new language L' = L $\cup$ {$c_n | n \in \mathcal{N}$} and assume there is such theory T. Then T $\cup$ {$c_n > c_{n+1} | n \in \mathcal{N}$} is inconsistent so there is a finite subtheory that's inconsistent (by compactness theorem). Then there is a contradiction because this theory is true in $\mathcal{N}$.

Now my question is: why is it needed to add such new constants? I would not add them and say: $\phi_n = \exists x_1, \dots, x_n : x_1 >\dots > x_n$ and then take the theory T $\cup$ {$\phi_n | n \in \mathcal{N}$}, but this is apparently not correct.

xzeo
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1 Answers1

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We can't express the existence of an infinite chain by the formulas $\phi_n$ you defined.
Consider the linear posets $\{0,1,\dots, n\} $ put together as branches of a tree $(T, <) $ adjoined to a common root. This model satisfies each $\phi_n$ but it contains no infinite chain.

Berci
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  • Also, since the problem is about axiomatizing the well-ordered sets within the linearly ordered sets, it's worth noting that every infinite linear order satisfies ${\phi_n\mid n\in\mathcal{N}}$, so there is no contradiction. – Alex Kruckman Jun 17 '18 at 13:38