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I think the canonical example is that WO means there's a well ordering of the reals. We have to accept it as true, but what the hell would that look like? It's
by v64 9y ago
I think the canonical example is that WO means there's a well ordering of the reals. We have to accept it as true, but what the hell would that look like? It's generally assumed that such a well ordering wouldn't be constructible, so it does seem like a little much to ask for.
Edit: For those curious why this is problematic, a well ordering on a set means that for any non-empty subset of the set, there's a consistent way to define the least element of the set.
An easy example is the natural numbers starting from 0 going up. The well ordering on the naturals is the usual <, and 0 is the lowest of the entire set.
This doesn't work this way for the integers. Because the integers go to negative infinity, we have to order them in a different way (but we know it's possible because the well ordering principle says so). It turns out the way to do it is to define the order on the integers this way: 0 1 -1 2 -2 3 -3 etc. Now the integers are ordered in such a way that we have a well defined least element, 0.
Now, WO says the real numbers have a well ordering as well. What would that look like? For instance, < fails us because the open interval (0,1) has no least element. The well ordering has to hold for any non-empty subset, and there are ways to create very strange and pathological subsets of R. To think that the entirety of R can be ordered in such a way like the integers is asking for a lot. But the well ordering principle states that even if we can't figure out the order, we can assume that one exists and use it in proofs. But it's weird because we're pretty much resigned to never knowing how it'd actually look.
- deleted 9y ago[deleted]
- monochromatic 9y agoGreat explanation.
- danbruc 9y agoI guess the following won't work then. 0 / 0.1, 0.2, 0.3...0.9 / 0.01, 0.02, 0.03...0.09, 0.11, 0.12, 0.13...0.19, 0.21...0.99 / 0.001, 0.002, 0.003...0.009, 0.011, 0.012... And when ever you list x, you also list -x, 1/x, and -1/x. Except for 0 of course. And you have to throw in +1 and -1 somewhere. Somehow it feels like you should cover all the reals eventually but that can't be true because that would imply there are only countable infinite many reals. It actually looks more like a convoluted version of the standard way to enumerate the rationals by visiting diagonals in order of increasing distance from the origin on a two dimensional integer lattice.
- giomasce 9y agoIf you could provide some way to construct the WO we could ditch AC. I guess it is not happening soon...
- Retra 9y agoYeah, that's obviously just an ordering of the rationals.
- danbruc 9y agoI don't think it's that obvious, actually I think it's not actually an ordering of the rationals. If this were an ordering of the rationals, then it better included rationals with an infinite number of non-zero decimal digits, for example one third. But if this would include one third, then I am unable to see why it wouldn't also include square root two. Why would this be able to include repeating decimal expansions but not non-repeating ones?
- Retra 9y agoYou're right, I misread what you wrote. You're actually only covering a small portion of the rationals. Either way, to order the reals, you have to include the non-computable ones, which cannot be written down in any systematic way. This also means that ordering and equality relations cannot be systematically defined either.
- deleted 9y ago[deleted]