Note that without replacement, you can’t construct the von Neumann ordinal omega*2, or any higher ones, so certainly not omega_1. Of course, this doesn’t prevent uncountable well-ordered sets (obviously these follow from choice, though I guess you’re taking that out as well), but you need replacement to show that every well-ordered set is isomorphic to a von Neumann ordinal.
So I don’t think that this should prevent the construction of an order of type omega_1, even if it can’t be realized as a von Neumann ordinal. Of course losing canonical representatives means you have to talk about equivalence classes, but if all we want to do is talk about omega_1, it suffices to consider well-orderings of subsets of N, so that the equivalence classes in question will in fact be sets. Maybe there’s some other technical obstacle I’m missing here (like it somehow wouldn’t be the first uncountable ordinal despite being the right order?) -- this isn’t really my area and I haven’t bothered to work through it, I can try that later—but I wouldn’t expect one.
Maybe there’s some other technical obstacle I’m missing here
There’s not. The Hartog’s number construction gives us the set H(N) of all isomorphism classes of well-orders on subsets of any fixed countably infinite set, and we can prove that H(N) is uncountable and every proper initial segment of H(N) is countable, using power set and separation (but only bounded separation) but not replacement. I verified this just now by looking at Wikipedia’s article on Hartog’s number and checking through the proof myself.
The next step (step 4 in Wikipedia, ETA: which can be saved for the end, although WP did not do so) is to replace the elements of H(N) with von Neumann ordinals, but this is really beside the point. You already have a representation of the least uncountable ordinal, and this step is just making it canonical in a certain way.
Note that without replacement, you can’t construct the von Neumann ordinal omega*2, or any higher ones, so certainly not omega_1. Of course, this doesn’t prevent uncountable well-ordered sets (obviously these follow from choice, though I guess you’re taking that out as well), but you need replacement to show that every well-ordered set is isomorphic to a von Neumann ordinal.
So I don’t think that this should prevent the construction of an order of type omega_1, even if it can’t be realized as a von Neumann ordinal. Of course losing canonical representatives means you have to talk about equivalence classes, but if all we want to do is talk about omega_1, it suffices to consider well-orderings of subsets of N, so that the equivalence classes in question will in fact be sets. Maybe there’s some other technical obstacle I’m missing here (like it somehow wouldn’t be the first uncountable ordinal despite being the right order?) -- this isn’t really my area and I haven’t bothered to work through it, I can try that later—but I wouldn’t expect one.
There’s not. The Hartog’s number construction gives us the set H(N) of all isomorphism classes of well-orders on subsets of any fixed countably infinite set, and we can prove that H(N) is uncountable and every proper initial segment of H(N) is countable, using power set and separation (but only bounded separation) but not replacement. I verified this just now by looking at Wikipedia’s article on Hartog’s number and checking through the proof myself.
The next step (step 4 in Wikipedia, ETA: which can be saved for the end, although WP did not do so) is to replace the elements of H(N) with von Neumann ordinals, but this is really beside the point. You already have a representation of the least uncountable ordinal, and this step is just making it canonical in a certain way.
Heh, I’d forgotten how simple Hartogs number was in general.