Rick Santorum and rest of Intelligent Design Crowd Get Ahead of Themselves.

With the conventional definition of "well ordered" (every non-empty subset has a least member) then the reals are not well ordered. I'm sure you know the proof as well as I do.

But I seem to have stupidly failed to grasp your point. Why is this an important question? The important question, to me, is rather the question whether the conventional definition of "well ordered" makes sense. Which has very little to do with Platonism.

It's very important for Platonism. Do the "reals" exist independently of set theory? If so, which set theory applies, with or without the axiom of choice or the continuum hypothesis? Both ways are consistent. The Platonist position generally has been (at least pre-1963) that either the axiom of choice is true or false and this is to be determined. It's possible to assume that both sets of "reals" are possible (as well as many more types of "reals".) In that case, one can ask which set of "reals" applies to physics.

Sorry for the long delay, it's been a busy week.

In no particular order: Yes, the "axiom of choice" must be either true or false. So, which is it? My position is simple (may be too simple). If the AoT is true, then the reals are well ordered. But the reals are evidently not well ordered, since the subset denoted by (0,1] has no least member.

The proof of that is a pretty strong one, namely that if you offer me a candidate least member, say r, then I can offer a better candidate, namely r/2. This is the exact same strategy as we use in the proof that there is no largest prime, which no mathematician has ever believed fallacious.

Hence, everything "proved" by the AoT is dubious, including your previous example (by Vidali wasn't it?) of a set that can be shown to have measure both zero and non-zero.

As another digression, do I believe (or suspect) that there are incomparable transfinite cardinals, ie t1, t2 such that it is not the case that t1>t2 or t1=t2 or t1<t2? Again, it's not something I've ever considered relevant to Platonism as such, but I would neither be surprised if two such cardinals existed nor devastated if they did not.

It's actually inner measure zero and outer measure one. This makes the set non-measurable. I prefer (in the context of probability theory) to accept the axiom of countable choice but not the unrestricted axiom of choice. This allows all sets to be measurable; that is, all Lesbegue sets are Borel sets. It's only my choice. As one can add the AoC to set theory or not, I just use whichever version works best for the problem at hand. Thus, my preferences seem the same as yours. The difference is that I do not consider any of these versions more real than the other.

Another amusing problem is that each of these possitilities can be translated into a problem in elementary number theory. So the truth of trichotomy of transfinite sets, for example, is equivalent to some elementary number theory statement; which statement is probably something like: "there exist no solutions to the following equation: blah, blah, blah..."

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