# Equivalent form of continum hypothesis

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?

• Do you see the implication in one of the directions? – Tobias Kildetoft Jun 22 '16 at 19:50
• Mario Carneiro's "definition" of $\aleph_1$ isn't quite right. See my comment below his answer. $\qquad$ – Michael Hardy Jun 22 '16 at 22:27
• 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. – Michael Hardy Jun 22 '16 at 22:30

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

• 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. – Asaf Karagila Jun 22 '16 at 19:59
• @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). – Mario Carneiro Jun 22 '16 at 20:02
• 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. – Asaf Karagila Jun 22 '16 at 20:05
• @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. – Mario Carneiro Jun 22 '16 at 20:09
• Now I can vote this up with clear conscience! – Asaf Karagila Jun 22 '16 at 20:10

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).

• 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. – Asaf Karagila Jun 24 '16 at 10:29
• let's remedy that – user12802 Nov 22 '17 at 22:44

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

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 $2^{\aleph_0} > \aleph_1$.