. Any Two Nondegenerate Closed Intervals Have the Same Cardinality

Glomzzz
2026-10-26
About

A geometric picture and an explicit affine bijection show why any two nondegenerate closed intervals have the same cardinality.

A longer segment has greater length, but does it have more points? For cardinality, we compare sets by pairing their elements, not by measuring their lengths. Two sets have the same cardinality if there is a bijection between them: every point in either set has exactly one partner in the other.

Let $𝑎, 𝑏, 𝑐, 𝑑 \in ℝ$ , with $𝑎 < 𝑏$ and $𝑐 < 𝑑$ . The goal is to pair the points of $[𝑎, 𝑏]$ and $[𝑐, 𝑑]$ without leaving any point unpaired.

Geometric idea: one ray, one pair

Draw the intervals as two parallel segments, as in the picture. The lines through the corresponding endpoints meet at a point $𝑂$ outside both segments. Given a point $𝑥$ on the upper segment, draw the ray from $𝑂$ through $𝑥$ . Its unique intersection with the lower segment is the partner of $𝑥$ , which we call $𝑓 (𝑥)$ .

This pairing works in both directions:

Each point $𝑥$ on the upper segment determines one ray and hence exactly one point $𝑓 (𝑥)$ on the lower segment.
Two distinct points on the upper segment lie on different rays, so they cannot have the same partner on the lower segment.
Every point on the lower segment lies on a ray that also meets the upper segment, so no point is left unpaired.

The picture shows a shorter upper segment. If it is longer, interchange the roles of the segments in the picture; if their lengths are equal, simply translate one segment onto the other.

By similar triangles, paired points are the same fraction of the way from the left endpoint to the right endpoint:

$\frac{𝑥 - 𝑎}{𝑏 - 𝑎} = \frac{𝑓 (𝑥) - 𝑐}{𝑑 - 𝑐} .$

For example, the point two-fifths of the way along one segment is paired with the point two-fifths of the way along the other. Writing this relative position as $𝑡 \in [0, 1]$ , we have

$𝑥 = 𝑎 + 𝑡 (𝑏 - 𝑎), 𝑓 (𝑥) = 𝑐 + 𝑡 (𝑑 - 𝑐) .$

Solving for $𝑡$ gives the formula used in the proof.

Statement and proof

Theorem Any Two Nondegenerate Closed Intervals Have the Same Cardinality [closed-intervals-have-the-same-cardinality]

2026-10-26
Glomzzz

Let $𝑎, 𝑏, 𝑐, 𝑑 \in ℝ$ , with $𝑎 < 𝑏$ and $𝑐 < 𝑑$ . Then the closed intervals $[𝑎, 𝑏]$ and $[𝑐, 𝑑]$ have the same cardinality.

Proof.

Define $𝑓 : [𝑎, 𝑏] \to [𝑐, 𝑑]$ by

$𝑓 (𝑥) = 𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} (𝑥 - 𝑎) .$

In particular, $𝑓 (𝑎) = 𝑐$ and $𝑓 (𝑏) = 𝑑$ . We check that $𝑓$ is well-defined, injective, and surjective.

1. Well-definedness.

Let $𝑥 \in [𝑎, 𝑏]$ . Since $𝑏 - 𝑎 > 0$ ,

$𝑎 \leq 𝑥 \leq 𝑏 \Rightarrow 0 \leq \frac{𝑥 - 𝑎}{𝑏 - 𝑎} \leq 1 .$

Multiplying by $𝑑 - 𝑐 > 0$ and adding $𝑐$ gives

$𝑐 \leq 𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} (𝑥 - 𝑎) \leq 𝑑 .$

Thus $𝑓 (𝑥) \in [𝑐, 𝑑]$ , so the formula really does define a function with the required domain and codomain.

2. Injectivity (one-to-one).

Let $𝑥_{1}, 𝑥_{2} \in [𝑎, 𝑏]$ and suppose $𝑓 (𝑥_{1}) = 𝑓 (𝑥_{2})$ . Then

$𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} (𝑥_{1} - 𝑎) = 𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} (𝑥_{2} - 𝑎) .$

Subtracting $𝑐$ and dividing by the nonzero number $(𝑑 - 𝑐) / (𝑏 - 𝑎)$ yields

$𝑥_{1} - 𝑎 = 𝑥_{2} - 𝑎, hence 𝑥_{1} = 𝑥_{2} .$

Therefore, $𝑓$ is injective: distinct points cannot share a partner.

3. Surjectivity (onto).

Let $𝑦 \in [𝑐, 𝑑]$ be arbitrary. Choose

$𝑥 = 𝑎 + \frac{𝑏 - 𝑎}{𝑑 - 𝑐} (𝑦 - 𝑐) .$

Since $𝑑 - 𝑐 > 0$ ,

$𝑐 \leq 𝑦 \leq 𝑑 \Rightarrow 0 \leq \frac{𝑦 - 𝑐}{𝑑 - 𝑐} \leq 1 .$

Multiplying by $𝑏 - 𝑎 > 0$ and adding $𝑎$ shows that

$𝑎 \leq 𝑎 + \frac{𝑏 - 𝑎}{𝑑 - 𝑐} (𝑦 - 𝑐) \leq 𝑏 .$

Hence $𝑥 \in [𝑎, 𝑏]$ . Substituting this choice into $𝑓$ gives

$\begin{matrix} 𝑓 (𝑥) & = 𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} (𝑥 - 𝑎) \\ = 𝑐 + \frac{𝑑 - 𝑐}{𝑏 - 𝑎} \frac{𝑏 - 𝑎}{𝑑 - 𝑐} (𝑦 - 𝑐) \\ = 𝑐 + (𝑦 - 𝑐) \\ = 𝑦 . \end{matrix}$

Thus every $𝑦 \in [𝑐, 𝑑]$ has a preimage in $[𝑎, 𝑏]$ , so $𝑓$ is surjective.

Since $𝑓$ is both injective and surjective, it is a bijection. Therefore,

$| [𝑎, 𝑏] | = | [𝑐, 𝑑] | .$

∎

The pairing can be reversed explicitly:

$𝑓^{- 1} (𝑦) = 𝑎 + \frac{𝑏 - 𝑎}{𝑑 - 𝑐} (𝑦 - 𝑐), 𝑦 \in [𝑐, 𝑑] .$

Conclusion

Any two nondegenerate closed intervals of real numbers therefore have the same cardinality, regardless of their lengths or positions on the real line. The nondegeneracy assumption matters: $[𝑎, 𝑎]$ has only one point and cannot have the same cardinality as an interval with distinct endpoints.