Mathematicians seem to be reluctant to say the Hypothesis of the continuum is true, many even arguing against. Ever since 2021, when it was proven that the cardinality of the continuum is at least ${\aleph_2}$. We also know that the cardinality of the natural numbers is ${\aleph_0}$.

But now strictly intuitively speaking, wouldn't it be very logical that the cardinality of the real numbers is $2^{\aleph_0} $? Let me explain: Every positive (for the sake of simplicity) real number can be written as the fractional part + integer part. And since the length of the integer part is a lot smaller than the factionary part in a real number, we may just consider that cardinality of the fractional part. The next step would be that we write the fractional part in binary.

Something like this:


But there are ${\aleph_0}$ digits in such a number (since its never-ending - even the rational numbers have a never ending either of repeating digits, or of 0s). And since there are only 2 digits that we can use in binary - 0 and 1 -, this means that for a sequence of n digits we can create exactly $2^{n}$.

And since there are ${\aleph_0}$ digits in a real number, we may conclude that there are $2^{\aleph_0} $ real numbers.

Please note that I'm just an amateur in math, and enjoying to play around with concepts. I have most likely made a mistake, and I'm eagerly waiting to understand what that mistake is and why it cannot be the way I said. So take it with a grain of sault.

Thanks alot!

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    $\begingroup$ It was not "proved" that the continuum hypothesis is false. I suppose that you mean that article in Quanta about the Asperó–Schindler paper. That is definitely not what they prove there. $\endgroup$
    – Asaf Karagila
    Commented Jan 27, 2022 at 13:32
  • $\begingroup$ @AsafKaragila Yes. That's exactly what I mean :D $\endgroup$ Commented Jan 27, 2022 at 13:36
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    $\begingroup$ For convenience for the non-expert and the ignorant (e.g., me), here is a link to the Quanta article quantamagazine.org/… $\endgroup$
    – peter a g
    Commented Jan 27, 2022 at 13:57

1 Answer 1


The cardinality of the continuum is $2^{\aleph_0}$, the argument is roughly as you describe, although there are some subtleties to be dealt with (e.g. some numbers would have two distinct binary expansions, but we can correct for that).

The problem is that assuming only the axioms of $\sf ZFC$, we are unable to determine whether or not $2^{\aleph_0}=\aleph_1$ or $2^{\aleph_0}=\aleph_2$, or maybe it is actually much larger.

The only thing we can prove that that the cofinality of $2^{\aleph_0}$ is uncountable. This is a technical concept, but in this case we can state it simply: if $F\colon\Bbb{R\to N}$ is any function, then there is some $n\in\Bbb N$ such that $\{r\in R\mid F(r)=n\}$ has the same cardinality as the continuum. This mean that $2^{\aleph_0}\neq\aleph_\omega$, where $\aleph_\omega$ is the first infinite cardinal which has infinitely many infinite cardinals below it.

But we know that this is the only limitation. We can have the continuum as large or as small as we want as long as it satisfies the above requirement.

As a side note, nobody proved recently that $2^{\aleph_0}\geq\aleph_2$. The work of Asperó and Schindler show that assuming a technical axiom called $\sf MM^{++}$, which implies $2^{\aleph_0}=\aleph_2$, we can prove an even more technical axiom known as $(*)$.

But $\sf MM^{++}$ itself is not provable. You can assume it, or you can assume its negation. And whether or not there is an absolute mathematical truth, so far as a mathematical society we seem more keen to assume that it is something like $\sf ZFC$, and so we do not have enough information to determine whether or not $\sf MM^{++}$, $(*)$ or $\sf CH$ are true.

  • $\begingroup$ I see. But how can a number have two distinct binary expansions? Could you give me an example? And perhaps even an explication? It seems fascinating to me. $\endgroup$ Commented Jan 27, 2022 at 13:52
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    $\begingroup$ $0.011111\ldots=0.10000\dots$ $\endgroup$
    – Asaf Karagila
    Commented Jan 27, 2022 at 13:52
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    $\begingroup$ Oh yeah. Obviously $\endgroup$ Commented Jan 27, 2022 at 13:53
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    $\begingroup$ $\aleph_1$ is the number of countable well-ordered sets (up to order isomorphism). $\aleph_2$ is the number of well-ordered sets (up to isomorphism) which have cardinality $\aleph_1$ or less. These can be represented concretely as a set of equivalence classes of orders of subsets of $\Bbb N$, or more directly as the von Neumann ordinals corresponding to the $\aleph$ number. $\endgroup$
    – Asaf Karagila
    Commented Jan 27, 2022 at 13:57
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    $\begingroup$ @Daron: That is just because you've spent a lot more time learning about $\Bbb R$. Getting down to brass tacks, in the language of set theory, $\omega_1$ is arguably simpler to write down than $\Bbb R$. $\endgroup$
    – Asaf Karagila
    Commented Jan 27, 2022 at 22:04

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