Theorem. Any Two Nondegenerate Closed Intervals Have the Same Cardinality

Glomzzz
2026-10-26
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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:

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