Real Number Line

Theorem

The points on an infinite straight line are in one-to-one correspondence with the set $\R$ of real numbers.

Hence the set of all points on an infinite straight line and $\R$ are equinumerous.

Proof

Step 1

We will show that there exists a mapping from the infinite straight line $L$ to the set of real numbers $\R$.

Let us establish a relation $h$ between points on $L$ and elements of $\R$.

We allow the Axiom of Choice to set up a choice function to allow the points of $L$ to be selected systematically.

Pick any point on $L$, and label it $z$. This point can be referred to informally as the origin.

Map $z$ to zero, that is, by allowing $\left({z, 0}\right) \in h$.

Using the choice function pick any other point on $L$.

From Euclid's first postulate, we can draw a line segment between the two points.

If the second point is to the right of the origin, let its length be positive.

If the second point is to the left of the origin, let its length be negative.

The existence of the choice function allows that this process can be done for any point on $L$

Hence:

$\forall p \in L: \exists x \in \R: \left({p, x}\right) \in h$

That is, $h$ is left-total.

The method of construction of $h$ is such that every point on $L$ is assigned to exactly one element of $\R$.

Thus it is seen that $h$ is functional.

Hence, by definition, $h$ is a mapping.

Since every point mapped to is associated to exactly one element of $L$, $h$ is injective.

$\Box$

Step 2

Now we need to demonstrate that there exists a mapping from $\R$ to $L$.

Let us establish a relation $h$ between elements of $\R$ and points on $L$.

We allow the Axiom of Choice to set up a choice function to allow the elements of $\R$ to be selected systematically.

Pick any point on $L$, and label it $z$.

Map zero to $z$ that is, by allowing $\left({0, z}\right) \in g$.

Using the choice function pick any other element $x$ of $\R$.

Suppose $x$ is positive.

From Real Numbers form Vector Space, we can create a vector on $L$ with magnitude $x$ pointing to the right of $z$.

Suppose $x$ is negative.

From Real Numbers form Vector Space, we can create a vector on $L$ with magnitude $x$ pointing to the left of $z$.

The union of all such vectors, with the zero vector from $z$ to $z$, is the infinite straight line.

By the method of construction, there exists a vector on $L$ for all elements of $\R$, the relation $g$ is left-total.

The method of construction of $g$ is also such that every element of $\R$ is assigned to exactly one point on $L$.

Thus it is seen that $g$ is functional.

Hence, by definition, $g$ is a mapping.

Since every element of $L$ mapped to is associated to exactly one point on $\R$, $g$ is injective.

$\Box$

Step 3

Since:

$h: L \hookrightarrow \R$ is an injection

$g: \R \hookrightarrow L$ is an injection

by the Cantor-Bernstein-Schroeder Theorem, there exists a one-to-one mapping between them.

Hence by definition, $L$ and $\R$ are equinumerous.

$\blacksquare$

Also see

Real Number Plane

Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Sources

• Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (1936) $\S 5.30-33$