Cantor-Dedekind Hypothesis

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.

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.

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.

Then let the image of $x$ be a point to the right of the origin. Define the magnitude of its displacement be the absolute value of $x$.

Suppose $x$ is negative.

Then let the image of $x$ be a point to the left of the origin. Define the magnitude of its displacement be the absolute value of $x$ multiplied by $-1$.

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

By the method of construction, there exists a point 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.

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.

Also see
Real Number Plane