Translating Tarski's Axiomatization/Logic of $\mathbb R$ to the Theory of Magnitudes

Question

Update: This has become a project, but I need help. All answers will now be definitions, propositions, theorems, etc. that build on the theory. I will marks some of my own answers as community wiki so that they can be improved/expanded/clarified/fixed.

I changed the tags. See below for why we added the 'operator-theory' tag.

Current Problem: Show that the endomorphins (= automorphisms) on $M$ commute. Once this is done we will be able to define multiplication (given a selected unit of measure).

Now of course you can always throw up your hands, going back and constructing the real numbers with multiplication, but that is cheating. Perhaps something can be found in Tarski's work; I haven't looked at it but if someone has access his logic might work here.

It might be necessary to develop extensive techniques from real analysis, or maybe even create the theory of topological spaces. We are searching for the honed blend of algebra and analysis that can make for an elegant exposition.

My work - I went right for defining multiplication, and that might be the best route. But my proof is sketchy.

Here is the question: Prove the following

Theorem: Any two automorphisms of $M$ commute.

This can be deduced from logic using only the properties of $M$, but how much ancillary mathematical machinery do you have to build up to prove it?

In my work I also started looking at employing Dini's theorem, but gave up. And of course the composition of automorphisms, with a selected unit of measure, corresponds to finding the area of a rectangle, and the area doesn't change when we rotate it.

I also observed with a chosen $1 \in M$, if $\phi$ and $\psi$ are two automorphisms, the same is true of $p\phi + q\psi$, for positive integers $p$ and $q$. Now if $\Delta$ is any automorphism not equal to the identity, it is either an dilation or contraction. It, along with its inverse $\Delta^{-1}$, generates a commutative 'module' $\mathcal U$ that can be identified with a 'dense' commutative subalgebra $U$ of $(M,1,+)$, so intuitively, we can 'approximate' any two automorphisms with two commuting automorphisms.

Due to the above paragraph, I added in the operator theory tag. If any experts in this area think that is not appropriate, they can remove it.

The punch line of course is that the automorphism group of $M$ is isomorphic to $\mathbb R$.

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Definition: Let $M$ be a set with a binary operation $+$ satisfying the following properties:

P-0: The operation $+: M \times M \to M$ is both associative and commutative.

P-1: $\text{For every } x,y,z \in M \text{, if } z + x = z + y \, \text{ then } \, x = y$.

P-2: $\text{For every } x,y,z \in M \text{, if } z = x + y \, \text{ then } \, z \ne x$.

P-3: $\text{For every } x,y \in M \text{, if } x \ne y \, \text{ then } \, [\exists u \; | \, x = y +u] \text{ or } [\exists u \; | \, y = x +u]$.

P-4: $\text{For every } x \in M \; \exists \, y,z \in M \, \text{ such that } \; x = y + z$.

P-5:
$\text{For all } X, Y \subsetneq M$
$\quad \text{such that } (\forall x \in X) \; (\forall y \in Y) \; (\exists u \in M) \; y = x + u$
$\exists \, z \in M \text{ such that }$
$\quad \forall x \in X \; \; [\,x = z \text{ or } (\exists u \in M \text{ such that } x + u = z)\,]$
$\quad \text{and}$
$\quad \forall y \in Y \; \; [\,y = z \text{ or } (\exists u \in M \text{ such that } z + u = y)\,]$

Then $(M,+)$ is said to be a system of magnitudes and must also be non-empty.

Theorem: Let $(M,+)$ and $(N,+)$ be two systems of magnitudes and pick any element $m \in M$ and any $n \in N$. Then there exist a unique morphism $\gamma: M \to N$ such that $m \mapsto n$.
Moreover, this mapping $\gamma$ must also be an isomorphism.

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Before sketching out my ideas, realize that the above is a translation of the work in logic

Tarski's axiomatization of the reals

to the semigroup of magnitudes. Interestingly, the last words in that wikipedia article are

$\quad$... has its origins in Eudoxus' definition of magnitude.

Work Sketch

By taking $X,Y \subset M$ to both be empty sets, $\text{P-5}$ implies that $M$ is non-empty. Intuitively, selecting any point in $M$ then becomes an 'act' of selecting the unit of measure on an abstract line of points.

I also proved the following result ($s \lt t$ means $s + u = t$):

Proposition: If $x,y \in M$ there exists a $n \in \mathbb N$ with $n \gt 0$ such that $nx \gt y$.
Proof
Let $A = \{nx \, |\, n \gt 0 \}$. Assume that $y$ is an upper bound for $A$. Using $\text{P-5}$ the least upper bound $\alpha$ must exist for $A$. Since $x \lt \alpha$, we can write $x + u = \alpha$ and so $u \lt \alpha$. Since $u$ can't be an upper bound, for some $m$, $u \lt mx$. Adding $x$ to both sides of the inequality and using the The Law of Monotonicity, we get $x + u \lt (m+1)x$. But $x + u$ is $\alpha$ and we get a contradiction. $\quad \blacksquare$

This proof is an adaption of Theorem 1.20-(a) found in Walter Rudin's Principles of Mathematical Analysis, $\,3^{rd}$ Edition.

So $M$ satisfies the Archimedean property. Contrast this with known theory on linearly ordered groups,

Otto Hölder showed that every Archimedean group (a bi-ordered group satisfying an Archimedean property) is isomorphic to a subgroup of the additive group of real numbers.

The rest of my work is a matter of showing that once a 'unit of measure is chosen', we get an imbedding of $\{\frac{m}{2^n}\}$ into $M$ and, using $\text{P-5}$, everything 'comes for the ride', as far as proving the $\gamma$ isomorphism.

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While working on this project I asked five related (at least to me) questions:

Algebraically Constructing the Natural Numbers Using a Binary Operation Satisfying Some Properties

Automorphisms on $(\mathbb R, +)$ and the Axiom of Choice

Is it Useful Knowing that Automorphisms on (R>0,+) Are Always Continuous?

Examples of Commutative Semigroups Where the Cardinality of the Carrier Set is Greater Than c.

In a fifth question I asked for counterexamples showing that the properties for $M$ don't always lead to $\mathbb R^{\gt 0}$. It served us well allowing us to 'stress test' the theory of magnitudes, but I deleted it since the 'action' is now here.

Also, I would like to thank @JohnHughes who helped me remove syntax error and 'brush up' the formulation of the properties. Also, @M.Nestor's offline work showed that, indeed, we can only get $\mathbb R^{\gt 0}$. He also asked a question viewing this theory from another angle:

Is R the only complete ordered Abelian group?

Answer 1

A little Euclid [1] helps; so if, for the purposes of this project, you "threw away all your math books", at least retrieve that one! :)

The proposition can be proved from P-0 to P-4 and the Archimedean property; completeness (P-5) is only used to prove the latter, and denseness (P-4) is not needed at all. (Of course, P-4 and P-5 are both needed for characterising $\mathbb{R}_{>0}$, or complete systems of magnitudes in general.)

It is clear from the definition of the order relation in $M$, in conjunction with the axioms of associativity and commutativity, that addition respects order: that is, if $x < y$, then $w + x < w + y$, and so on.

For present purposes, I'll take $\mathbb{N}$ to be the set of positive integers, i.e. zero is excluded. I'll take for granted the usual properties of $\mathbb{N}$, as well as the binary operation $\mathbb{N} \times M \to M$, $(n, x) \mapsto nx$, defined recursively in the usual way for semigroups. In particular, $1x = x$, $m(nx) = (mn)x$, $(m + n)x = mx + nx$, and because of the commutativity of addition in $M$, $n(x + y) = nx + ny$, for all $x, y \in M$ and $m, n \in \mathbb{N}$. (Thus, the map $M \to M$, $x \mapsto nx$ is an endomorphism of $M$.)

If $x < y$, then by definition there exists $u$ with $x + u = y$, so $nx + nu = ny$, so $nx < ny$. If $m < n$, then there exists $p \in \mathbb{N}$ with $m + p = n$, so $mx + px = nx$, so $mx < nx$. Similarly $mx > nx$ if $m > n$; so we have $m < n$ or $m = n$ or $m > n$ according as $mx < nx$ or $mx = nx$ or $mx > nx$.

By induction on $n$, if $x < y$, then $nx < ny$, and if $x > y$, then $nx > ny$; so we have $x < y$ or $x = y$ or $x > y$ according as $nx < ny$ or $nx = ny$ or $nx > ny$. Thus we can "divide by $n$" when handling inequalities or equations in $M$.

For $x, y \in M$, define the ratio of $x$ to $y$ to be the binary relation on $\mathbb{N}$, $$ x \mathbin{:} y = \{ (n, m) : nx > my \}. $$

^{Under a more general definition of a system of magnitudes, $\mathbb{N}$ itself is a system of magnitudes, Archimedean, but not complete. One can define the rational number $\tfrac{m}{n}$ as the ratio $m \mathbin{:} n$ in that system. Then one can prove that the set of all ratios is totally ordered by inclusion. Under a still more general definition, in which the order relation is given as a primitive concept instead of being defined in terms of addition, the set of integers $> 1$ is a system of magnitudes, also Archimedean, and also not complete, having multiplication as its "addition" operation. One can can define $\log_nm$ as the ratio $m \mathbin{:} n$ in that system. Incidentally, one need not even have an operation of addition of magnitudes. See section 3.10.1, "Extensive Multiples", in Krantz et al. [2] I won't pursue any of these thoughts further here, because they lead far afield! I develop only enough theory to prove that any two endomorphisms of $M$ commute. But if you have an appetite for more, and can get hold of Scott's unpublished notes [3]: he develops a theory along quite similar lines. His approach is not the only one possible. For instance, one can characterise those binary relations on $\mathbb{N}$ that are ratios, without initially referring to systems of magnitudes at all, and then prove that ratios form a complete system of magnitudes. But I found that that approach gets a bit messy, or at least in my hands it does! Please excuse this digression.}

Lemma 1 For all $x, y \in M$, and for all $r \in \mathbb{N}$, $$ x \mathbin{:} y = (rx) \mathbin{:} (ry). $$

Proof \begin{align*} x \mathbin{:} y & = \{ (n, m) : nx > my \} && \text{by definition} \\ & = \{ (n, m) : r(nx) > r(my) \} && \text{by ``division by $r$'' (see above)} \\ & = \{ (n, m) : (rn)x > (rm)y \} \\ & = \{ (n, m) : (nr)x > (mr)y \} \\ & = \{ (n, m) : n(rx) > m(ry) \} \\ & = (rx) \mathbin{:} (ry) && \text{by definition.} \end{align*} $\square$

Proposition 2 For systems of magnitudes $M, N$, $x, y \in M$, $u, v \in N$, and $p, q \in \mathbb{N}$, $$ \text{if } x \mathbin{:} y = u \mathbin{:} v, \text{ then } (px) \mathbin{:} (qy) = (pu) \mathbin{:} (qv). $$

Proof \begin{align*} (px) \mathbin{:} (qy) & = \{ (n, m) : n(px) > m(qy) \} && \text{by definition} \\ & = \{ (n, m) : (np)x > (mq)y \} \\ & = \{ (n, m) : (np)u > (mq)v \} && \text{because } x \mathbin{:} y = u \mathbin{:} v \\ & = \{ (n, m) : n(pu) > m(qv) \} \\ & = (pu) \mathbin{:} (qv) && \text{by definition.} \end{align*} $\square$

^{This is another of Euclid's results. It can be used to define multiplication of ratios in general by rational numbers in particular. We don't actually need it, but I thought I'd throw it in anyway. So sue me. :)}

Lemma 3 For all $x, y, u \in M$, if $x < y$, then there exist $n, m \in \mathbb{N}$ such that $nx < mu < ny$.

Proof By hypothesis, there exists $t$ such that $y = x + t$. By the Archimedean property (as proved in the question, or postulated without also postulating completeness), there exists $n \in \mathbb{N}$ such that $nt > u$. Hence: $$ ny = nx + nt > nx + u. $$ Again by the Archimedean property, there exists $m \in \mathbb{N}$ such that $mu > nx$. Let $m$ be the smallest integer satisfying this condition. If $m = 1$, then $$ nx < u < ny. $$ On the other hand, if $m > 1$, then by the definition of $m$, $(m - 1)u \leqslant nx$. Therefore, $$ mu = [(m - 1) + 1]u = (m - 1)u + u \leqslant nx + u < ny, $$ as required. $\square$

Corollary 4 For all $x, y, u \in M$, if $x \mathbin{:} u = y \mathbin{:} u$, then $x = y$.

Proof If $x \ne y$, then $x < y$ or $x > y$. Supposing that $x < y$, the lemma implies that $(n, m) \in y \mathbin{:} u$ but $(n, m) \notin x \mathbin{:} u$, therefore $x \mathbin{:} u \ne y \mathbin{:} u$. Similarly if $x > y$. $\square$

Corollary 5 For all $x, y, u \in M$, if $x \mathbin{:} u = x \mathbin{:} v$, then $u = v$.

Proof If $u \ne v$, then $u < v$ or $u > v$. If $u < v$, then by applying the lemma to $u, v, x$, instead of $x, y, u$, we find that there are $n, m \in \mathbb{N}$ such that $$ mu < nx < mv, $$ so $(n, m) \in x \mathbin{:} u$ but $(n, m) \notin x \mathbin{:} v$, so $x \mathbin{:} u \ne x \mathbin{:} v$. Similarly if $u > v$. $\square$

Corollary 6 For all $x, y, u, v \in M$, if $x \mathbin{:} u = y \mathbin{:} v$, then $x < y$ or $x = y$ or $x > y$ according as $u < v$ or $u = v$ or $u > v$.

Proof The previous corollary has dealt with the case $x = y$. If $x < y$, take $n, m$ as in the lemma. Because $x \mathbin{:} u = y \mathbin{:} v$ and $nx < mu$, we have $(n, m) \notin y \mathbin{:} v$, i.e. $ny \leqslant mv$, whence $mu < mv$, whence on "dividing by $m$", $u < v$. Interchanging the roles of $x$ and $y$, and of $u$ and $v$, in this argument, we find that if $x > y$ then $u > v$. $\square$

Theorem 7 For all $x, y, u, v \in M$, $x \mathbin{:} y = u \mathbin{:} v$ if and only if $x \mathbin{:} u = y \mathbin{:} v$.

Proof By the symmetry of the result, we need only prove that if $x \mathbin{:} y = u \mathbin{:} v$ then $x \mathbin{:} u = y \mathbin{:} v$. If $x \mathbin{:} y = u \mathbin{:} v$, then for all $n, m \in \mathbb{N}$, by two applications of Lemma 1, we have $(nx) \mathbin{:} (ny) = (mu) \mathbin{:} (mv)$. By Corollary 6, therefore: $nx < mu$ or $nx = mu$ or $nx > mu$ according as $ny < mv$ or $ny = mv$ or $ny > mv$; and in particular $x \mathbin{:} u = y \mathbin{:} v$. $\square$

^{This proof shines as brightly now as when Euclid gave it two and half thousand years ago. (Unless I've managed to tarnish it, that is! I haven't been closely following the sources listed, or even my own old notes, having been more in the mood to work things out as I went along, even at the risk of messing up.)}

It is clear that if $M, N$ are systems of magnitudes, and $\phi: M \to N$ is a morphism of semigroups, then $\phi$ respects the order structures of $M, N$, and is injective. It follows immediately that: \begin{equation} \tag{1}\label{eq:1} \phi(x) \mathbin{:} \phi(y) = x \mathbin{:} y \text{ for all } x, y \in M. \end{equation} If $N = M$, Theorem 7 gives the corollary: \begin{equation} \tag{2}\label{eq:2} \phi(x) \mathbin{:} x = \phi(y) \mathbin{:} y \text{ for all } x, y \in M. \end{equation} If $\psi: M \to M$ is also a morphism, taking $y = \psi(x)$ in \eqref{eq:2} and using \eqref{eq:1} gives: $$ \phi(\psi(x)) \mathbin{:} \psi(x) = \phi(x) \mathbin{:} x = \psi(\phi(x)) \mathbin{:} \psi(x), $$ and Corollary 4 gives $\phi(\psi(x)) = \psi(\phi(x))$. Because $x$ was arbitrary, it follows that $\phi \circ \psi = \psi \circ \phi$. $\square$.

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References

[1] Euclid's Elements, Book V

[2] David H. Krantz et al., Foundations of Measurement, I: Additive and Polynomial Representations (Academic Press 1971, repr. Dover 2007)

[3] Dana Scott, A General Theory of Magnitudes (unpublished, but referred to in this answer) (1963)

Answer 2

We give a short proof of the theorem.

Let $M$ be a system of magnitudes and select any element in the carrier set and call it $1$, so that the set $M$ is a pointed set and the object of study becomes $(M,1,+)$. We also have an injective morphism

$\tag 1 \iota: \mathbb N^> = \mathbb N \setminus \{0\} \to M \text{ such that } 1 \mapsto 1$

so we can view the image of the imbedding as an inclusion, $\mathbb N^> \subset M$.

It is not difficult to show that for any $x \in M$ there exist a unique element $H(x)$ such that $H(x)+H(x)=x$. So we define

$\tag 2 U = \{mH^n(1) \; | \; m \in \mathbb N^> \text{ and } n \in \mathbb N\}$

where '$m \; \text{times}$' is shorthand for repeated addition.

Again we have an injective morphism and we can regard $U \subset M$, where $mH^n(1)$ is the new name for an element in $M$.

Theorem: Let $(M,1,+)$ and $(N,1,+)$ be two systems of magnitudes with selected units of measure. Then there exist one and only one morphism

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\phi: M \to N$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 1 \mapsto 1$

Moreover, this mapping is an isomorphism.

Proof
Note that any morphism of $M$ into $N$ must be an injection.

For any $s \in M$ there exist an $N_s \ge 0$ such that for all $n \ge N_s$ the equations $m H^n(1) + u = s$ have solutions. So we can take the maximum $m_{(s,n)}$ and define the set $X_s =\{m_{(s,n)}H^n(1)\} $ and set $Y_s = \{ m \in M \; | \; (\forall x \in X_s) (\exists u \in M) \,[x + u = m]\}$.

Invoking $\text{P-5}$ we can get a $z_s \in M$ that separates $X_s$ and $Y_s$; this element is clearly in $Y_s$ and is therefore unique. It is equal to $s$.

The subset $X_s$ is naturally identified with a subset of $N$ and defines a $Y^{'}$ in $N$ the same way, and, again, we get a unique element $t \in N$ separating these two sets. All that remains is to show that the mapping $s \mapsto t$ defines an isomorphism, which is not difficult to argue. $ \blacksquare$