During studying set cardinality, I came across on this exercise:

$$2{}^{\aleph_0}2 = {}^{\aleph_0}2 + {}^{\aleph_0}$$

I understand that ${}^{\aleph_0}2$ is just $2^{\aleph_0}$ written in this way, at least previous text implies it. But I am really confused about what $^{\aleph_0}$ is supposed to be as there is no explanation for it whatsoever. There is a solution to the excercise, but it uses the same notation and therefore I do not understand it, even if it's quite short.

I think it's some typographical mistake, but having only brief background in cardinality, I can't figure out what the equation means. I bet it's either meant to be $2^{\aleph_0}$ or simply ${\aleph_0}$. Could you please find an explanation of this typo or post a meaningful version of the equation?

P.S. The textbook is in Slovak, so I won't post the solution here. Though I will translate it if it's necessary.

  • 4
    $\begingroup$ Looks like a typo to me. It's hard to guess what was intended without more context. Can you show some of the context? $\endgroup$ Jul 16, 2017 at 23:30
  • $\begingroup$ It could be related to infinity, $\aleph_0$ is sometimes used for the smallest infinite cardinal I think. Also, this seems to be a notation question I'll add that tag. youtube.com/watch?v=elvOZm0d4H0 and youtube.com/… talk about infinity, and youtube.com/watch?v=SrU9YDoXE88 talks about what's called aleph null. $\endgroup$
    – user451844
    Jul 16, 2017 at 23:44
  • 3
    $\begingroup$ I think the book's intention was $2\cdot {}^{\aleph_0}2={}^{\aleph_0}2+{}^{\aleph_0} 2. $ $\endgroup$ Jul 17, 2017 at 0:04
  • 1
    $\begingroup$ @RoddyMacPhee I highly doubt that's the issue. $\endgroup$ Jul 17, 2017 at 0:41
  • $\begingroup$ Something seems missing from the displayed equation. $\endgroup$
    – Asaf Karagila
    Jul 23, 2017 at 15:28

1 Answer 1


I have concluded that the stated equation really is a mistake in the textbook and that the intention was to write $2{}^{\aleph_0}2 = {}^{\aleph_0}2 + {}^{\aleph_0}2$ as Mr. Caicedo pointed out.


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