We've shown elsewhere that the rational numbers are countable; that is, we can put them into a one-to-one correspondence with the counting numbers. We call this number ℵ_{0} (read as "aleph sub naught"), the "smallest" infinity. We've also shown that the total number of points, both rational and irrational, in the number line is uncountable; that is, we cannot put them into such a correspondence. We call this number c = 2^{ℵ0}.

Thus, in a manner of speaking, there are more points in a finite line segment than there are integers. What is remarkable and somewhat counterintuitive is that there are as many points in the number line as there are in a plane.

Without loss of generalization, consider the unit line segment [0,1) and the unit square whose vertices are the points (1,0), (0,1), (1,1), and the origin.

To show: The number of points in the unit interval is equal to the number of points in the unit square.

We will show that for every point in the unit square, there's a unique point on the unit interval, and vice versa.

Let (x, y) be an arbitrary point in the unit square, and consider the decimal expansions of its coordinates:

x = 0.a_{1}a_{2}a_{3}... and y = 0.b_{1}b_{2}b_{3}...

Define a transformation T: ([0,1), [0,1)) → [0,1) on x and y as follows:

x = 0.a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}...

To show that T is one-to-one and onto.

All pairs (x, y) map to distinct values of x in [0,1), so T is one-to-one.

Furthermore, the decimal expansion of any value x in the unit interval can be decomposed to describe a unique point in the unit square. Its abscissa is determined by the odd decimal places, and its ordinate by the even places. So T maps onto the unit interval.

T is therefore a bijection.

QED

Note: By extending the above proof, we can establish that the number of points in any continuous space of any dimension is also c.

Sources: Amir D Aczel, *The Mystery of the Aleph*, 2000, and George Gamow, *One Two Three … Infinity*, 1961