Under ZFC, the real numbers can be well-ordered. So, there is some ordinal number whose cardinality is that of the continuum. Is there a standard notation for this number?

For example, the first infinite ordinal is usually denoted $\omega$, and the first uncountable ordinal is usually denoted $\omega_1$. But unless we appeal the continuum hypothesis (or something just as presumptuous), $\omega_1$ may not have cardinality of the continuum.

Here is a related question, which is just my curiosity at work: Do we know whether there is such thing as a least and/or greatest ordinal with cardinality of the continuum?


P.S. One last recreational-math question: Are there any other well-known measures of a number's size besides ordinals and cardinals?

  • $\begingroup$ Please don't ask unrelated questions on the same post. $\endgroup$ – Asaf Karagila Oct 1 '18 at 23:34
  • $\begingroup$ (And before you run to post it on a separate question, please clarify what you mean. Ordinals only measure order types, specifically of well-ordered sets. They require some structure with constraints. You can talk about arbitrary order types, but it still requires some order to it. If you start adding requirements, requiring a set is a Borel set on $[0,1]$ gives you a good measure of it, etc.) $\endgroup$ – Asaf Karagila Oct 1 '18 at 23:36
  • $\begingroup$ I admit, it is difficult to explain what I'm asking in my postscript. Perhaps a more specific example would help illustrate what I'm after. Let's take the set of naturals, and the set of even naturals. Most laypersons would assume that the set of naturals is in some sense "larger" than the set of even naturals. Mathematicians might be tempted to counter that this is false, as the naturals and the evens have identical cardinality. But frankly, I'm not altogether certain that the laypersons are incorrect here. Are cardinals really the only way to measure the size of a set? $\endgroup$ – Ben W Oct 1 '18 at 23:45
  • $\begingroup$ Ben, I've written about this before: math.stackexchange.com/questions/40309/cardinality-density math.stackexchange.com/questions/125412/… math.stackexchange.com/questions/168258/… math.stackexchange.com/questions/242057/… I hope these satisfy you. If not, your "P.S." question should really be removed and expanded into a new question on its own. $\endgroup$ – Asaf Karagila Oct 1 '18 at 23:48
  • $\begingroup$ Yes, I had thought about natural density, but it is highly specific to subsets of naturals. I'm not aware of general measures of density for sets of numbers---and yes, I realize that last sentence is a bit garbled, but hopefully you take my meaning. $\endgroup$ – Ben W Oct 2 '18 at 0:00

Not really. If one assumes the axiom of choice, as one usually does, then one has that $2^{\aleph_0}$ is in fact an ordinal, an initial ordinal. So writing $\alpha<2^{\aleph_0}$ has a very clear meaning. To that end, using $\frak c$ is also quite common as denoting the cardinality of the continuum. So it is not unusual to see $\alpha<\frak c$ in set theoretical papers.

So yes, there is such a least ordinal, this is the initial ordinal (by the definition of an initial ordinal). There is no maximal ordinal, just like there is no "largest countable ordinal". If $X$ is an infinite set, there is no largest ordinal equipotent to $X$. In fact, there are exactly $|X|^+$ ordinals which are equipotent to $X$. Just like there are $\aleph_1$ countably infinite ordinals, and $\aleph_2$ ordinals of size $\aleph_1$, and so on.

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  • $\begingroup$ So, if I understand you correctly, in the literature, c doubles as a label for some unspecified order type of the reals, yes? $\endgroup$ – Ben W Oct 1 '18 at 23:41
  • $\begingroup$ It is the cardinality of the reals. If cardinality also means the suitable initial ordinals, yes. But $\frak c$ is also used sometimes in contexts where the reals cannot be well-ordered, which is why I am somewhat reluctant to say "yes" outright. But again, if you were to write $\alpha<\frak c$, I think that most readers should be able to infer what you meant. $\endgroup$ – Asaf Karagila Oct 1 '18 at 23:42
  • $\begingroup$ Not an unspecified order type, but specifically the least ordinal which can be mapped bijectively with the reals. $\endgroup$ – Ned Oct 1 '18 at 23:45
  • $\begingroup$ So, there is such a least ordinal? $\endgroup$ – Ben W Oct 1 '18 at 23:46
  • $\begingroup$ @Ben: I've added a bit on that. I missed that part of your question. Sorry! $\endgroup$ – Asaf Karagila Oct 1 '18 at 23:49

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