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?
 A: 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 $\aleph_1=2^{\aleph_0}$.
A: 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).
A: $\aleph_1$ 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, $2^{\aleph_0} = \aleph_1$.
A: 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 $2^{\aleph_0} > \aleph_1$.
