How to Prove Two Nondegenerate Closed Intervals Have Equal Cardinality

If $[a,b]$ and $[c,d]$ are closed intervals with $a<b$ and $c<d$, they have equal cardinality. For bounded intervals, the proof is especially direct: a linear function maps one interval bijectively onto the other. In fact, each such interval has the cardinality of the real numbers.

The condition that the intervals have positive length matters. A degenerate interval such as $[a,a]=\{a\}$ is a singleton, so it does not have the same cardinality as an interval containing more than one point.

Bounded intervals: construct a bijection

Let $a<b$ and $c<d$. Define

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

This affine function sends the endpoints to the corresponding endpoints: $f(a)=c$ and $f(b)=d$. Its slope $(d-c)/(b-a)$ is positive, so it is strictly increasing. It follows that values of $x$ between $a$ and $b$ map to values between $c$ and $d$.

To check the map is onto, take any $y\in[c,d]$ and solve $y=f(x)$:

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

Because $y$ lies between $c$ and $d$, this $x$ lies between $a$ and $b$. Thus every point in $[c,d]$ has a preimage in $[a,b]$. The positive slope also makes $f$ one-to-one. So $f$ is a bijection, which proves the intervals have equal cardinality.

Worked example

Consider $[2,5]$ and $[-1,7]$. Here $a=2$, $b=5$, $c=-1$, and $d=7$. The formula gives

$$f(x)=-1+\frac{7-(-1)}{5-2}(x-2) =-1+\frac{8}{3}(x-2).$$

At the endpoints, $f(2)=-1$ and $f(5)=7$. Since the slope $8/3$ is nonzero, the map is one-to-one; and as $x$ ranges over $[2,5]$, $f(x)$ ranges over all of $[-1,7]$. This is a specific bijection, not merely a comparison of the lengths of the intervals. Equal cardinality does not mean the intervals have equal length.

What if an interval is unbounded?

The same affine formula applies when both intervals are rays with corresponding finite endpoints. For example, $[a,\infty)$ maps onto $[c,\infty)$ by $f(x)=x-a+c$. But it does not directly map intervals of different forms, such as a bounded interval and the whole real line.

For the more general claim, use the cardinality of $\mathbb{R}$. Every interval in $\mathbb{R}$ is a subset of $\mathbb{R}$, so its cardinality is at most that of $\mathbb{R}$. If an interval contains two distinct points $u<v$, it contains the whole closed interval $[u,v]$. There is a bijection from $\mathbb{R}$ onto the open interval $(u,v)$, for example

$$g(t)=\frac{u+v}{2}+\frac{v-u}{\pi}\arctan(t).$$

Since $(u,v)$ is inside the original interval, this gives an injection from $\mathbb{R}$ into it. The interval therefore has cardinality at least that of $\mathbb{R}$. Together, the two comparisons show that its cardinality is exactly the continuum. This applies to bounded intervals, rays, and $\mathbb{R}$ itself, provided the interval has more than one point.

Common mistakes

• Forgetting singleton intervals: $[a,a]$ contains exactly one element. It cannot be bijective with $[c,d]$ when $c<d$.
• Assuming equal length is required: Cardinality concerns the existence of a bijection, not geometric length. The intervals in the worked example have different lengths but equal cardinality.
• Using the bounded-interval formula without checking its domain: The endpoint formula applies to two bounded intervals with positive length. Use the continuum argument for intervals of different forms, such as a bounded interval and a ray.
• Claiming a bijection without checking onto: A function being one-to-one is not enough. A bijection must also reach every element of the target interval.