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Consider this fact that we all know from school mathematics: There is a one to one correspondence between real numbers and points of a line. But the problem is I have never seen a rigorous proof of this fact.

apostol

This is what Apostol writes. And similar is the case with other modern analysis books.

So my question is, where can I find the proof of this one to one correspondence. I don't care if geometry is in Hilbert's axioms or Tarski's or anyone else all I care about is correspondence.

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    $\begingroup$ What precise definitions are you using for "real numbers" and "line"? Personally I would take this as motivation for a choice of definition(s) rather than a technical result to be proved. That said, see the discussion here. $\endgroup$ Commented Aug 14, 2023 at 16:38
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    $\begingroup$ What definition of "line" are you using? If that definition refers to an ambient space (such as "the points in a plane that satisfy..."), what definition of that ambient space are you using? $\endgroup$ Commented Aug 14, 2023 at 16:38
  • $\begingroup$ Let real numbers be defined by fields +order +completeness axioms and line , point , betweeness and other geometric notions be axiomatically defined under any suitable axiom system (for example Hilbert's or Tarski's). Then prove this equivalence or correspondence $\endgroup$ Commented Aug 14, 2023 at 16:42
  • $\begingroup$ You can use $f(x)=\frac{1}{1+e^{-x}}$ and its inverse $g(y)=\log_e(y)-\log_e(1-y)$ to give a bijection between $\mathbb R$ and the interval $(0,1)$. A bijection $\mathbb R \leftrightarrow [0,1]$ is also possible but a little more complicated and not continuous. $\endgroup$
    – Henry
    Commented Aug 14, 2023 at 16:49
  • $\begingroup$ But this bijection does not prove the correspondence $\endgroup$ Commented Aug 14, 2023 at 16:50

3 Answers 3

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You may find the following useful:

  1. The nlab page on euclidean geometry.

    Here they mention that a model of Tarski's axioms is a product of a real closed field with itself. And vice versa, any product with itself of a real closed field is a model of Tarski's axioms.

  2. An answer of math stackexchange that has a link to a paper with the proof.

Importantly, real closed fields are not the same as the real numbers.

The real numbers are a model, but by Löwenheim-Skolem you get models of other cardinalities. Therefore the equivalence you were looking for actually only goes one way if the question is interpreted this way.

Here's the nlab page on real closed fields with some extra examples.

I understand that when one reads "the real numbers are often represented geometrically as points on a line", it is meant that the plane $\mathbb R\times \mathbb R$ is a model.

Once that's fixed, you are in set theory and know how to talk about a line as a subset of $\mathbb R\times \mathbb R$, and it is easy to see it is in correspondence with $\mathbb R$. So in this sense you do have a correspondence, but only once you fix a nice enough model, and I believe that's what is meant by "under an appropiate set of axioms".

But there's room for interpretation.

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  • $\begingroup$ I think you didn't understand the question. I want the equivalence (of points of line with real number) of two systems of axioms where is the room for philosophy here? $\endgroup$ Commented Aug 14, 2023 at 16:59
  • $\begingroup$ @PriyanshuKalal excuse me, indeed I saw that after answering and am editing. $\endgroup$
    – Julián
    Commented Aug 14, 2023 at 17:03
  • $\begingroup$ @PriyanshuKalal Edited. Hope that helps. Cheers! $\endgroup$
    – Julián
    Commented Aug 14, 2023 at 17:35
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You can build a model of $\Bbb R$ on a line $\ell$ in the plane. The steps follow this pattern: (Note that I am ignoring the issues already mentioned by others about the exact axioms in place for either the real numbers or Euclidean geometry. Those issues are real, but I am brushing them under the rug.)

  1. Choose two points on $\ell$ to serve as $0$ and $1$. Call the ray from $0$ through $1$ the "positive side of $\ell$" and its opposite ray in $\ell$, the "negative side of $\ell$". For any point $a$ on $\ell$, "the distance $a$" means the distance from $0$ to $a$. Other common terminology for real numbers are defined on $\ell$ in similar fashion.
  2. For two points $a,b$ on $\ell$, define $a+b$ to be the point on $\ell$ that is the distance $a$ from $b$, and if $a \ne 0$, is the same direction from $b$ as $a$ is from $0$ (if $a = 0$, there is only one point that is the distance $0$ from $b$, namely $b$ itself).
  3. For a point $a \ne 0$ on $\ell$, define $-a$ to be the other intersection point between $\ell$ and the circle centered at $0$ and passing through $a$. Define $-0 = 0$.
  4. Prove that for all $a,b,c$ on $\ell,a+b=b+a,(a+b)+c=a+(b+c),a+(-a)=0$.
  5. For points $a,b$ on the line other than $0$, define $a\cdot b$ by constructing the perpendiculars to $\ell$ at $1$ and at $a$. On the perpendicular at $1$, construct a point $c$ at distance $b$ from $1$. Draw the line through $0$ and $c$, and find its intersection $d$ with the perpendicular at $a$. The circle about $0$ whose radius is the distance from $a$ to $d$ intersects $\ell$ in two points, one positive, and one negative. If $a$ and $b$ are on the same side of $0, a\cdot b$ is the positive intersection point. If $a$ and $b$ are on opposite sides of $0, a\cdot b$ is the negative intersection point. Finally, define $0\cdot a = a\cdot 0 = 0$.
  6. Show that $a\cdot b$ is the same no matter which point on the perpendicular to $1$ at a distance of $b$ from $1$ that you pick.
  7. For $a \ne 0$ on $\ell$, define $a^{-1}$ by constructing the perpendiculars to $\ell$ at $1$ and $a$. Construct a point $b$ on the perpendicular at $a$ at a distance of $1$ from $a$. Draw the line through $0$ and $b$, and let $c$ be its intersection with the perpendicular to $1$. Then $a^{-1}$ is the point on $\ell$ whose distance from $0$ is the distance from $c$ to $1$, and lies on the same side of $0$ as $a$. Show that $a^{-1}$ is the same point regardless of which point on the perpendicular at $a$ of distance $1$ from $a$ that you chose for $b$.
  8. Show that for all $a,b,c$ on $\ell$, that $a\cdot b = b\cdot a, a\cdot(b\cdot c) = (a\cdot b)\cdot c, a\cdot (b + c) = (a\cdot b) + (a\cdot c)$ and if $a \ne 0$, then $a\cdot a^{-1} = 1$.
  9. For $a, b$ on $\ell$, define $a < b$ if $b + (-a)$ is positive (i.e., on the positive side of $\ell$).
  10. Show for $a,b,c$ on $\ell$ that $a < a$ is false; that exactly one of $a < b$, $b < a$, $a = b$ is true; and if $a < b$ and $b < c$, then $a < c$.
  11. Show that if $a < b$, then $a + c < b + c$. And if $c > 0$, then $a\cdot c < b\cdot c$.

Those are all basic construction proofs. However, the final step needs some of the axioms found in modern treatments of Euclidean geometry which Euclid himself never recognized. The exact axioms needed will differ depending on whose axiom system you've chosen. But any complete system for Euclidean geometry will have some means of accomplishing this.

  1. Show that if $S$ is a collection of points on $\ell$ and if $a \in \ell$ is a point such that every $s \in S$ satisfies $s < a$, then there is a unique point $m\in \ell$ such that for every $s \in S$, either $s = m $ or $s < m$; and for every point $b < m$, there is some $s \in S$ with $b < s$.

These definitions and demonstrations show that $\ell$ is a complete ordered field. All complete ordered fields are isomorphic to $\Bbb R$.

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Starting from geometric axioms, it is indeed quite demanding to construct a correspondence from a line to the real numbers. I will assume Hilbert's axioms and sketch how it is usually done. In fact I will use one-dimensional axioms only.

Let $L$ be a geometric line together with the betweenness relation $B$ and congruence relation $\equiv$ satisfying all the axioms.

The key point in the reasoning is proving the existence of segment measure. A function $\mu\colon L\times L\rightarrow [0,\infty)$ is called a measure, whenever the following hold:

  1. $\mu(ab)=0 \iff a=b$.
  2. $ab\equiv a'b' \implies \mu(ab)=\mu(a'b')$.
  3. $B(abc) \implies \mu(ab)+\mu(bc)=\mu(ac)$.

To construct the measure we take the following steps:

  1. Consider the family of all free segments, i.e. the equivalence classes of the congruence relation.
  2. Define multiplication of a free segment by a positive real number, which we do successively from natural and rational numbers.
  3. Prove that for any two free segments, one is the product of the other and a unique real positive number.
  4. Fix one segment that will serve as a unit and express any other segment as a multiple of the unit one.
  5. Assign this value to any free segment and carry it to the points on the line.

Once the measure is constructed, we can pick a point $o\in L$ together with a ray originating at $o$ and define $$\xi(p):=\begin{cases}\mu(op) & , p\in A\\ -\mu(op) & , p\in A^*.\end{cases}$$ Then $\xi\colon L\rightarrow \mathbb R$ is the desired correspondence function.

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