Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2019 at 18:47 | comment | added | Nick S | In line with a previous answer, the first claim also follows from the irrationality of $\zeta(3)$. | |
| Mar 5, 2012 at 0:56 | comment | added | Joel David Hamkins | Zsban, the usual definition of $\omega$ is that it is the least inductive set (containing $0$ and closed under successor $x\mapsto x\cup\{x\}$). The concept of finite is defined after this, since a set is finite if it is bijective with a proper initial segment of $\omega$. | |
| Mar 4, 2012 at 21:29 | comment | added | Zsbán Ambrus | Goldstern: avoiding ordinals is not the problem. The problem is how you define $ \omega $. The usual definition is that it's the least infinite ordinal. How do you avoid finiteness from that definition? | |
| Feb 21, 2012 at 20:07 | comment | added | Goldstern | @andres: "finite" does not have to use ordinals. A set $M$ is Tarski-finite iff every nonempty subset of $P(M)$ has a maximal element with respect to inclusion. Tarski-finite is equivalent to the usual notion of finite (in a weak version of ZF). | |
| Jan 1, 2012 at 5:35 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
added 3 characters in body
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| Nov 4, 2010 at 3:15 | comment | added | Andrés E. Caicedo | The issue with using Dedekind-finite is that Terry's trick no longer applies, which is too bad. | |
| Nov 4, 2010 at 2:26 | comment | added | Joel David Hamkins | Andres, you discovered an easier proof that finite sets can be well-ordered! (I don't think my argument is circular; it's just silly.) | |
| Nov 4, 2010 at 2:06 | comment | added | Andrés E. Caicedo | Joel, how is the third example not circular? Finite is defined in terms of the ordinals... Unless you mean Dedekind-finite, in which case there is something else to say (this whole thing is so silly!) | |
| Nov 3, 2010 at 22:17 | comment | added | Terry Tao | Well, what one could do here is first prove the claim using AC, and then use Godel's theorem that every statement in Peano Arithmetic that is provable in ZFC can be proven in ZF, which I think is very much in the "using a nuclear weapon to kill fleas" spirit of this exercise. | |
| Oct 20, 2010 at 16:28 | comment | added | Joel David Hamkins | The absurdity of the examples, to my way of thinking, is the idea that one should appeal to a big axiom such as AC to prove the completely trivial facts that every finite set has a well-order or that nonempty sets have members. Of course, AC is completely unnecessary here, and so this is using a nuclear weapon to kill fleas. | |
| Oct 20, 2010 at 11:01 | comment | added | Paul Siegel | Don't your second two examples actually imply the axiom of choice for finite collections of sets (and are therefore equivalent to it)? In any event, if the axiom of choice is the "nuke" in these examples then what is the approach that should be considered much simpler? My understanding is that the axiom of choice is only a big deal for infinite collections of sets. | |
| Oct 18, 2010 at 23:16 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |