Timeline for Lindenbaum algebras and models
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 24, 2011 at 15:29 | comment | added | Emil Jeřábek | How could it be far away from Boolean algebras? Every Lindenbaum algebra is a Boolean algebra. | |
| May 24, 2011 at 15:26 | comment | added | Hans-Peter Stricker | Are there examples really "far away from Boolean algebras", i.e. not at first sight "Boolean algebra with some extra conditions"? I mean, just some "innocent" theory and - alas! - its L-algebra turns out to be a model of it! | |
| May 24, 2011 at 15:13 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
one more fairly common example
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| May 24, 2011 at 15:13 | comment | added | Emil Jeřábek | You mean $T$ such that the Lindenbaum algebra of $T$ is a model of $T$ itself? Clearly, the theory of Boolean algebras does the job. Some of its proper extensions also work. For example, if $T$ is the theory of nontrivial Boolean algebras of size at most $2^n$, then its Lindenbaum algebra has indeed size $2^n$. | |
| May 24, 2011 at 15:03 | comment | added | Hans-Peter Stricker | Let me add: "a non-trivial algebraic theory". | |
| May 24, 2011 at 15:02 | comment | added | Hans-Peter Stricker | Is there an algebraic theory which its Lindenbaum algebra is a model of? (I am not quite sure whether this is correct English;-) | |
| May 24, 2011 at 14:47 | comment | added | Emil Jeřábek | Which question is the side question? | |
| May 24, 2011 at 14:41 | comment | added | Hans-Peter Stricker | @Emil: I take "there are not that many different Boolean algebras" as the essence of your answer. Thank you! (What about my side question? Do you have an example?) | |
| May 24, 2011 at 14:29 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |