This is a short question concerning an application of the compactness theorem. If I understand correctly from my self-reading, it says that an infinite set of formula is satisfiable iff any finite subset of it is.

So suppose now that I have an infinite set of variables $x_1,..,x_\kappa,..$, and I have an infinite set of very simple linear inequalities which say $\sum_{i\in I} x_i>\sum_{j\in J} x_j$. Then suppose that I know each finite set of inequalities has a solution, does the compactness theorem tell me that there must be a solution for the infinite set?

My plan was to assign to each variable a constant, let the universe be the reals and allow for addition, scalar multiplication as functions. Then allow for simply the binary relation $>$. Then it seems to me that each inequality is expressible as a sentence, then the whole set of inequalities would be an infinite set of sentences which is satisfiable given that any finite subset would be.

Is there anything wrong with this approach? As I am very new to this part of math, I would be grateful if someone could point out some elementary mistakes or misunderstandings.


The compactness theorem will tell you that a particular set of formulas (here, equations) will be consistent. This means that you'll be able to find a solution for this set of formulas in an elementary extension of the reals. While the existence of this solution might help you, this solution won't belong to the reals in general.

For example, consider this set of equations : $S := \{ x > n \big| \ n \in \mathbb{N}\}$. $S$ is finitely satisfiable in $\mathbb{R}$, but no element of $\mathbb{R}$ will satisfy $S$. To find a realisation of $S$, you'll have to go into the nonstandard reals.

  • $\begingroup$ Fantastic, thank you very much. So in general, how can I tell if the elementary extension is distinct from the universe I began with? $\endgroup$ – En Hua Hu Nov 27 '19 at 13:43
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    $\begingroup$ That is, in general, a very difficult question. You might be interested by analysability, but I'm not sure this will help you for this particular question since the theory of $(\mathbb{R}, +, \cdot, <)$ isn't stable. $\endgroup$ – Olivier Roche Nov 27 '19 at 13:47

Not only does the compactness theorem not prove this as Olivier Roche explained, but the statement is false. Consider first the inequalities $x_2<x_1,x_3<x_1,$ and $x_2+x_3>x_1$, which together imply $x_1>0$. Now add on inequalities $x_n>x_1$ and $x_\omega>x_4+\dots+x_n$ for each $n>3$. This system of inequalities has no solution in $\mathbb{R}$, it implies $x_1>0$ and $x_\omega>nx_1$ for all $n\in\mathbb{N}$. However, it is easy to see any finite subset of them has a solution in $\mathbb{R}$.


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