Math Stuff

Some Interesting Math I Have Come Across (Excerpts from Lectures and Otherwise)

The cardinality of \(\mathbb{R}\) is \(2^{\aleph_0}\)

This was a homework problem from Fall 2025 Math 131AH - Honors Real Analysis at UCLA. The rather cool aspect of this problem comes from the fact that the interval \(\left(0, 1\right)\) also has cardinality \(2^{\aleph_0}\). In particular, this is to say that the size of \(\mathbb{R}\) is the same as the size of the interval \(\left(0, 1\right)\).

It’s important that we don’t confuse size (and therefore, cardinality) with length (Lebesgue measure) because they describe two completely different properties of numbers. While the cardinality of \(\left(0, 1\right)\) equals that of \(\mathbb{R}\), the length of them differ. Indeed, if \(\ell\) describe the length of an interval, then \(\ell\left(\left(0, 1\right)\right) = 1\) while \(\ell\left(\mathbb{R}\right) = \infty\).

With this important note, let’s head to the proof.

Proof. We will use the fact that the interval \(\left(0, 1\right)\) has cardinality \(2^{\aleph_0}\). We want to use Schröder-Bernstein such that we can establish equipotence between \(\left(0, 1\right)\) and \(\mathbb{R}\) by showing an injection \(\phi:\left(0, 1\right) \longrightarrow \mathbb{R}\) and another injection \(\psi:\mathbb{R} \longrightarrow \left(0, 1\right)\).

Let \(\phi:\left(0, 1\right) \longrightarrow \mathbb{R}\) defined via the identity map \(\phi\left(x\right) = x\). Note \(\left(0, 1\right) \subseteq \mathbb{R}\).

As such, \[\phi\left(x\right) = \phi\left(y\right) \implies x = y ~~~~ \forall x, y \in \left(0, 1\right)\] Hence, \(\phi\) is injective.

Let \(\psi:\mathbb{R}\longrightarrow\left(0, 1\right)\) defined via the sigmoid function \[ \psi\left(x\right) = \frac{1}{1 + e^{-x}} \] As such, \[ \psi\left(x\right) = \psi\left(y\right) \implies \frac{1}{1 + e^{-x}} = \frac{1}{1 + e^{-y}} \iff 1 + e^{-y} = 1 + e^{-x} \iff e^{-y} = e^{-x} \iff y = x \] Hence, \(\psi\) is injective.

Because \(\phi:\left(0, 1\right) \longrightarrow \mathbb{R}\) injective and \(\psi:\mathbb{R}\longrightarrow\left(0, 1\right)\) injective, by Schröder-Bernstein, \(\left(0, 1\right) \sim \mathbb{R}\), that is, \(\mathbb{R}\) is equipotent to \(\left(0, 1\right)\). Because \(\left(0, 1\right)\) has cardinality \(2^{\aleph_0}\), we conclude that \(\mathbb{R}\) has cardinality \(2^{\aleph_0}\).

QED.

I hope you like this proof using quite a powerful theorem: Schröder-Bernstein. We can also skip Schröder-Bernstein and in fact construct a bijection which would require us to show surjectivity of a mapping we define. I find this simple and short proof quite elegant.

There’s another interesting proof showing that the set of irrational numbers \(\left(\mathbb{R} \backslash \mathbb{Q}\right)\) has the same cardinality as that of \(\mathbb{R}\) (and therefore, the same as \(\left(0, 1\right)\)). It also uses the Schröder-Bernstein theorem to establish equipotence but that too can be done so via direct bijection.