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In this link they show how the compactness theorem fails in english:

(first page with the starbucs example):

https://ocw.mit.edu/courses/linguistics-and-philosophy/24-241-logic-i-fall-2005/readings/chp09.pdf

My question is: Will it fail as well when we replace 'there are finitely many starbucks' with 'there are more girls than boys' in formal language? what are the other propositions I should add to the list (or set) instead of the other propositions mentioned there (there is at least X starbucks etc.), in order to show that every subset is compossible and yet the compactness theorem fails.

Thank you.

  • I do not quite follow what the point is, but formally you can do something like x is greater than y, x is less than 3/2y, x is less than 4/3y, x is less than 5/4y, ... , for continuous x and y. This is conistent but ω-inconsistent. It does not quite work with x boys and y girls because at some point the difference is becomes less than 1, which would mean it is zero. – Conifold Feb 2 '17 at 0:44
  • very interesting solution. So let me see if I understand it correctly. Because this infinite series tends to 1, the compactness theorem fails? – Akira Feb 2 '17 at 1:18
  • More or less. If x/y is greater than 1, then it is greater than 1 by some 1/n (in standard analysis, no infinitesimals), and then it is not less than 1+1/n=(n+1)/n. But n can be arbitrarily large, so compactness fails. It is parallel to the usual calculus non-compactness of (1,2] here: we have an infinite cover of (1,2] by open intervals, but it has no finite subcover. – Conifold Feb 2 '17 at 1:35

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