Consider the set of natural numbers $\mathbb{N}=\{0,1,2,3,\ldots\}$. There is a countably $\aleph_0$

Claim: $\aleph_0 \neq \aleph_1$

Proof: We

Consider some set $X$ and its power set $2^X$. Let the cardinality of $|X|=\aleph_0$

Define $f$ to be some set function, which associates a subset of $X$ to each element $x \in X$. \[ x \to f(x). \] To show $\aleph_0 \neq \aleph_1$, we have to show that there is no bijection between the set and its powerset (since $\aleph_1$ is identified as the cardinality of the powerset of a set of cardinality $\aleph_0$).

We give a proof by contradiction. Assume that a bijection $f$ exists, and consider the subset of $X$ which consists of all elements which are not contained in their image under $f$. \[ Y = \{ x \in X \ | \ x \not \in f(x) \}, \qquad \qquad (1) \] which means that: \[ \forall y \in Y, y \not \in f(y). \qquad \qquad (2) \]

Now if $f$ is a bijection, then there must be some $\tilde{y} \in X$ that is the pre-image of $Y$, such that: \[ f(\tilde{y}) = Y. \]

Now we ask the question: “is $\tilde{y} \in Y$? ”
If yes, then we get a contradiction with $(2)$. If no, then this means $\tilde{y} \not in f(\tilde{y})$ and from the definition of the set $Y$ in $(1)$, $\tilde{y}$ should have been included in $Y$. So we have a contradiction, and $f$ must not exists.

On the other hand, we have the trivial injection $g$ which assigns to each element of a set the set with the single element: \[ g(x) = \{ x \}, \forall x \in X. \]

Therefore we have shown that there is an injection from $X$ to its powerset, but there is no bijection, hence the cardinality of the set $X$ must differ from the cardinality of its powerset. By definition, we call this next cardinality $\aleph_1$.

There is a bijection between the reals $(0,1)$ and subsets of the Naturals. Let $r=0.r_1r_2r_3r_4$ be some real numbers written in binary, where $r_j \in \{0,1\}$. Now consider the set $S_r \subset \mathbb{N}$, which corresponds to the number $r$. We define $S$ to consist of all numbers $j$ such that $r_j=1$.

Note that there is also a bijection between the interval $(0,1) \subset \mathbb{R}$ and the whole $\mathbb{R}$.

I want to thank Paul Raymond-Robichaud for showing me these results, concisely and cleanly.