Timeline for How can we define non-finitely axiomatizable extensions of set theories?
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11 events
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| Jun 28 at 22:28 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, the matter of non-finite axiomatizability is technical. I think that $T$ is consistent. My point was really about trying to prove the consistency of a theory by compactness. This cannot be done when $T$ is finitely axiomatizable. So, I thought one can prove the consistency of $T$ by proving the consistency of an extension of it that is not finitely axiomatizable through using the compactness feature. So, I didn't want that extension to have high consistency difference from the original theory. | |
| Jun 28 at 22:18 | comment | added | Michael Hardy |
@NoahSchweber : The reason one font looks very good and another looks terrible is just a large number of little things like these. Donald Knuth invented TeX, from which LaTeX and MathJax evolved, and he brought his prodigious intelligence to it, i.e. he thought about things like this. I wonder whether the asymmetry in $$ a+...+z $$ makes this appear differently to you. And if you write a+...+z in LaTeX (as opposed to MathJax) you get something like this: $$ a+\text{...} + z $$
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| Jun 28 at 20:57 | comment | added | Joel David Hamkins | How can it be reasonable to strenuously reject the iterated consistency statements, which anyone who believes in the truth of your theory would surely also believe, but accept something at the level of Con(T)? I don't get the philosophical attitude toward the theory that is being expressed here. Do you think T is true or not? But also, since the Lindenbaum algebra is dense, you can always find strengthenings as low as you like, which are not finitely axiomatizable. Between $T$ and any extension of $T$ there will be a nonfinitely axiomatizable theory. As close to $T$ as you want. | |
| Jun 28 at 20:47 | comment | added | Noah Schweber | @MichaelHardy OK? For what its worth, of your three examples the first two are equi-good to me and the third is the only one which strikes me as wrong. I think these sorts of typographical preferences are going to vary between people, and aren't really worth much energy. | |
| Jun 28 at 20:41 | comment | added | Michael Hardy | $\ldots\,$and the third line also differs from the first two. | |
| Jun 28 at 20:41 | comment | added | Michael Hardy | $$ \begin{align} & T+Con(T)+Con(T+Con(T))+... \\ {} \\ & T+\operatorname{Con}(T)+\operatorname{Con}(T+\operatorname{Con}(T))+\ldots \\ {} \\ & T+\operatorname{Con}(T)+\operatorname{Con}(T+\operatorname{Con}(T))+\cdots \end{align} $$ @NoahSchweber You will have noticed that the formatting of the three dots in the first line above looks conspicuously different from what you see in the second line above. Do you know that what is in the first line will strike some people the way a spelling error does? And $Con$ in the first like differs from $\operatorname{Con}$ in the second. And${}\,\ldots$ | |
| Jun 28 at 20:07 | vote | accept | Zuhair Al-Johar | ||
| Jun 28 at 19:19 | answer | added | Noah Schweber | timeline score: 4 | |
| Jun 28 at 19:16 | comment | added | Zuhair Al-Johar | No! That's too much stronger. Something like $T+Con(T)$ or something near this vicinity. | |
| Jun 28 at 19:13 | comment | added | Noah Schweber | What does "slightly stronger than $T$" mean? Would $$T+Con(T)+Con(T+Con(T))+...$$ work? (Note that for any "reasonable" theory $T$, the infinite consistency extension $T+Con(T)+...$ is never finitely axiomatizable by the second incompleteness theorem.) Of course this trick assumes that $T$ isn't too unsound; if you want one which always works, with more effort we can find a family of Rosser-type sentences $(\sigma_i^T)_{i\in\omega}$ which are uniformly independent of each other over $T$ (= no finite set implies one not in the set), and consider $T\cup\{\sigma_i^T: i\in\omega\}$. | |
| Jun 28 at 18:25 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |