Skip to main content
MathOverflow

Return to Answer

fixed a typo
Source Link
Michael Hardy
Michael Hardy
  • 1
  • 12
  • 85
  • 126

For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$ m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases} $$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set of statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.)

For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$ m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases} $$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.)

For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$ m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases} $$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set of statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.)

Source Link
Michael Hardy
Michael Hardy
  • 1
  • 12
  • 85
  • 126

For some reason I like stating the ultraproduct proof of the compactness theorem in measure-theoretic language, as follows. A filter $F$ on the index set $I$ corresponds to a "finitely additive measure", like this: $$ m(A) = \begin{cases} 1 & \text{if }A \in F, \\ 0 & \text{if }I \setminus A \in F \\ \text{undefined} & \text{otherwise} \end{cases} $$ A filter is an ultrafilter if the "otherwise" case is vacuous. Then \Los's theorem says a first-order statement in the language of the model is true in the ultraproduct if and only if it's true in "almost all" of the factors. (Then as others noted above, let $I$ be the set of finite subsets of the set statements you're trying to satisfy; let $F$ be an ultrafilter that contains every co-finite set (one exists if you believe in Zorn's lemma); then for each $i \in I$ let the $i$th factor be a model that satisfies that finite set of statements; then prove by induction on the formation of first-order formulas that the ultraproduct satisfies all of the statements. As far as I know you have to include formulas with free variables in the proof to make the induction work. I'm skipping the details since others seem to have gone into those above.)