for the difference between N and R there is a discrete operation, R is the powerset of N, or anything equivalent you like. There is some kind of discontinuity (and thus, I guess, cardinality). But when we use ordinals, it is not clear to me what operation or property makes countable ordinals before $\omega_1$ to become uncountable at $\omega_1$. I cannot see any obvious "change" or operation that makes this possible. Am I asking some stupid question? (Yes, I know)

  • $\begingroup$ Your question is getting upvotes, so I'm guessing it's not stupid ;-) $\endgroup$ – Kevin Mar 23 '13 at 7:46

The only change is that $\omega_1$ has uncountably many predecessors, while each $\alpha<\omega_1$ has only countably many predecessors. In character it’s the same kind of change that occurs at $\omega$: $\omega$ has infinitely many predecessors, while each $n<\omega$ has only finitely many predecessors.

  • $\begingroup$ I promise not to ask questions about ordinals for a long time! $\endgroup$ – Wolphram jonny Mar 23 '13 at 6:00
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    $\begingroup$ you have no idea how great is to have your wisdom available from anywhere in the world! (and for free) $\endgroup$ – Wolphram jonny Mar 23 '13 at 6:01
  • $\begingroup$ @julian: :-) (And yes, MSE is a wonderful resource.) $\endgroup$ – Brian M. Scott Mar 23 '13 at 6:06

Perhaps another way to see Brian's answer is the following:

  • Consider $\omega$. This is what we get when we gather all of the finite ordinals together, and well-order them in the natural way. This well-ordering cannot have finite order-type, since otherwise it would be the largest finite ordinals, and every proper initial segment of this order is finite, so it is the first infinite ordinal. Thus we have the least infinite ordinals.

  • Similarly, $\omega_1$ is what we obtain when we gather all of the countable ordinals together and well-order them in the natural way. Again, the order-type of this well-ordering cannot be countable, since otherwise it would be the largest countable ordinal, and all proper initial segments of this ordering are countable. Thus we have the least uncountable ordinal.

  • $\begingroup$ @ Arthur: so, the set of countable limit ordinals between any arbitrarliy high countable limit ordinal and $\omega_1$ is uncountable, right? because the set before it is countable, and the union of two countable sets is countable? $\endgroup$ – Wolphram jonny Mar 24 '13 at 18:11
  • $\begingroup$ @Julian: That's correct. $\endgroup$ – user642796 Mar 24 '13 at 19:33

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