5
$\begingroup$

The Continuum Hypothesis states that

$$2^{\aleph_0}=\aleph_1$$

And Cantor put it equivalently as:

"There is no uncountable subset $A$ of $\mathbb{R}$ such that $|A| <\mathbb{R}$."

Why are these two statements equivalent?

$\endgroup$
  • $\begingroup$ Do you see the implication in one of the directions? $\endgroup$ – Tobias Kildetoft Jun 22 '16 at 19:50
  • $\begingroup$ Mario Carneiro's "definition" of $\aleph_1$ isn't quite right. See my comment below his answer. $\qquad$ $\endgroup$ – Michael Hardy Jun 22 '16 at 22:27
  • $\begingroup$ THREE answers now say $\aleph_1$ is the least uncountable cardinal or that it's the next cardinal $\text{after }\aleph_0$. Could we recall the definition going back to Cantor, who introduced the notation: $$ \aleph_1 \text{ is the cardinality of the set of all countable ordinals.} $$ If the axiom of choice holds, then one can show as a corollary of this definition that it's the least uncountable cardinal. But the definition above remains the definition even if one drops the axiom of choice. $\endgroup$ – Michael Hardy Jun 22 '16 at 22:30
8
$\begingroup$

I am assuming you know that $|\Bbb R|=2^{\aleph_0}$, which can be proven by looking at binary expansions of numbers in $[0,1]$ (discounting countably many numbers with non-unique expansions).

The cardinal $\aleph_1$ is by definition the smallest cardinal larger than $\aleph_0$, meaning that there is no set $A$ such that $\aleph_0<|A|<\aleph_1$. Thus in particular, $2^{\aleph_0}\not<\aleph_1$.

In the other direction, if $\aleph_1<2^{\aleph_0}=|\Bbb R|$, then that means that there is an injection $f:\aleph_1\to\Bbb R$, and setting $A$ as the range of $f$, we have $A\subseteq\Bbb R$, and $\aleph_0<|A|=\aleph_1<|\Bbb R|$.

Thus assuming that there are no such sets $A$, we also have $\aleph_1\not<2^{\aleph_0}$, and assuming the axiom of choice this implies $\aleph_1=2^{\aleph_0}$.

$\endgroup$
  • $\begingroup$ While the binary argument is correct, it requires more than the naive approach, because there is a countable set of "mistakes". And one has to argue that a countable set does not matter here. $\endgroup$ – Asaf Karagila Jun 22 '16 at 19:59
  • $\begingroup$ @AsafKaragila Yes, since it was not the point of the post I didn't want to belabor it, but I rather prefer using ternary expansions to avoid this complication, basically proving that the cantor set is a subset of $\Bbb R$ of size $2^{\aleph_0}$ (with a natural bijection). $\endgroup$ – Mario Carneiro Jun 22 '16 at 20:02
  • $\begingroup$ While I understand this approach, with time I found it to be confusing. Because the naive approach does not work. It's fine not to delve into the details, but some caveat should be given. $\endgroup$ – Asaf Karagila Jun 22 '16 at 20:05
  • $\begingroup$ @Asaf (Caveat given.) It is a bit unfortunate that the binary expansion route is the most prevalent in introductory textbooks, because of this unnecessary complication. I remember reading the Cantor diagonal argument rendered in base 10 using $6$s and $7$s so as to avoid the nonuniqueness problem, but I don't think I've ever seen a direct proof of $|\Bbb R|=2^{\aleph_0}$ use this same trick. $\endgroup$ – Mario Carneiro Jun 22 '16 at 20:09
  • $\begingroup$ Now I can vote this up with clear conscience! $\endgroup$ – Asaf Karagila Jun 22 '16 at 20:10
4
$\begingroup$

Assuming the axiom of choice, every two cardinals are comparable. In particular either $\aleph_1\geq 2^{\aleph_0}$ or $\aleph_1\leq 2^{\aleph_0}$.

Since $\aleph_1$ is the least uncountable cardinal, and $2^{\aleph_0}$ is uncountable by the diagonal argument, it follows that it is necessarily the case that $\aleph_1\leq 2^{\aleph_0}$. So the continuum hypothesis is equivalent to saying that $2^{\aleph_0}=\aleph_1$, otherwise there is a set of size $\aleph_1$, pretty much by definition that $\aleph_1<2^{\aleph_0}$, which is intermediate between $\Bbb N$ and $\Bbb R$.

Do note, however, that the axiom of choice is essential here. It is possible that the axiom of choice fails, there are not intermediate cardinals between $\Bbb N$ and $\Bbb R$, but $\aleph_1\neq2^{\aleph_0}$ (in which case there are incomparable).

$\endgroup$
  • 1
    $\begingroup$ I'd ask what's wrong with the answer that a downvote is in order. But I don't think it has to do with the answer, as much as it has to do with the answerer. $\endgroup$ – Asaf Karagila Jun 24 '16 at 10:29
  • $\begingroup$ let's remedy that $\endgroup$ – user12802 Nov 22 '17 at 22:44
2
$\begingroup$

$\aleph_1$ is the least uncountable cardinal, and $|\mathbb{R}| = 2^{\aleph_0}$. To say that there is no uncountable set with cardinality less than that of $\mathbb{R}$ is precisely to say that $|\mathbb{R}|$ is the least uncountable cardinal; that is, $2^{\aleph_0} = \aleph_1$.

$\endgroup$
0
$\begingroup$

Several answers now say $\aleph_1$ is the least uncountable cardinal or that it's the next cardinal after $\aleph_0$. Could we recall the definition going back to Cantor, who introduced the notation:

$\aleph_1$ is the cardinality of the set of all countable ordinals.

If the axiom of choice holds, then one can show that all infinite cardinals are alephs, and those start with $\aleph_0$, $\aleph_1$, $\aleph_2$, etc., with nothing between alephs with consecutive indices.

Thus if there is nothing between $\aleph_0$ and $2^{\aleph_0}$, then $2^{\aleph_0}=\aleph_1$, but otherwise $2^{\aleph_0} > \aleph_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.