Does the Axiom of Constructibility make Pasch's Axiom unnecessary?

Question

I conjecture that we can weaken or replace Pasch's axiom while still admitting the n-dimensional Cartesian spaces over Euclidean ordered fields. Am I correct?

Tarski's axioms include the Continuity Schema. However, it is established that the Continuity Schema can be replaced with a single axiom, the Segment-Circle Axiom, and still be sufficient for classical Euclidean geometry:

Tarski observed that a set of axioms for this geometry can be obtained... by replacing all instances of the Continuity Schema, As. 11, with a single sentence, the Circle Axiom. This sentence asserts that any segment which joins two points, one inside and one outside a given circle (with which the segment is coplanar), must intersect that circle. Let us denote the resulting set of axioms by CG(n). Its models are, up to isomorphisms, just the n-dimensional Cartesian spaces over Euclidean ordered fields — ordered fields in which every positive number has a square root.

-- Tarski's System of Geometry, Alfred Tarski and Steven Givant, Bulletin of Symbolic Logic / Volume 5 / Issue 02 / June 1999, pp 175 - 214 DOI: 10.2307/421089

The Segment-Circle Axiom can be expressed as follows (see also):

Given any points $C$ (i.e. the center of a circle) and $R$ (a point on the radius), $X$ between $C$ and $R$ (inside the circle), $Y$ such that $R$ is between $C$ and $Y$ (outside the circle), $P$ such that $CP \cong CX$ (inside the circle), $Q$ such that $CP \cong CY$ (outside the circle), there exists $Z$ between $P$ and $Q$ such that $CZ \cong CR$. (That is, for any segment with one end inside the circle and the other outside, there is a point of intersection between the segment and the circle.)

Given that simplification, can we replace Pasch's axiom with something simpler:

Axiom of Constructability: Only points whose existence is a result of these axioms exist.

That is, the only axiom which initially posits the existence of any points is the Lower n-Dimensional Axiom. Once we have it, we have at least $n - 1$ distinct points; from there, we can use the other axioms, such as Axiom of Segment Constructability, to show other points.

I conjecture that if we replace the Continuity Schema with the Segment-Circle Axiom, as Tarski writes is possible, we can replace Pasch's Axiom with the Axiom of Constructability, and still have a set of axioms for geometry whose models, up to isomorphism, are the n-dimensional Cartesian spaces over Euclidean ordered fields. Am I correct?

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Update

I understand that my Axiom of Constructability is not in first order logic, and is therefore substantially more complicated than Pasch's axiom. Nonetheless, I'm very interested in understanding if it suffices, in particular because of its logical and intuitive implications.

Answer 1

Apologies in advance if this is off-topic.

Let $r$ denote the length of $CR$. Applying the ruler postulate, we can assign the numbers $0$ and $1$ to $P$ and $Q$, respectively. According to the ruler postulate, for any point $W$ lying on the straight line passing through $P$ and $Q$, there exists exactly one real number $w$ satisfying the following properties:

(1) $PW=|w|, QW=|1-w|$.

(2) if $W$ lies between $P$ and $Q$, then $PW=w$ and $QW=1-w$.

(3) if $Q$ lies between $P$ and $W$, then $PW=w$ and $QW=w-1$.

(4) if $P$ lies between $Q$ and $W$, then $PW=-w$ and $QW=1-w$.

Let $f$ be a function such that if point $W$ corresponds to the real number $w$, then $f(w)$ exists and equals to the length of $CW$. Consequently, for any real number $w$, $f(w)$ is existent and always positive.

Because $CP=CX\lt CR \lt CY=CQ$, $f(0)\lt r \lt f(1)$.

Suppose that the real numbers $a$ and $b$ correspond to points $A$ and $B$, respectively. Applying the triangle inequality, we can show that $|f(a)-f(b)|=|CA-CB|\lt AB=|a-b|$, which implies that $f$ is a Lipschitz function with Lipschitz constant 1. Thus, $f$ is continuous.

Since $f(0)\lt r \lt f(1)$, according to the intermediate value theorem, there exists a real number $z$ such that $0 \lt z \lt 1$ and $f(z)=r$, which implies The Segment-Circle Axiom.

The intermediate value theorem is proved by the least-upper-bound property, which is hidden in the ruler postulate. As a result, we may use the ruler postulate and the triangle inequality to deduce The Segment-Circle Axiom.

Therefore, in my opinion, the question may boil down to whether the notion of completeness, when taken as axiomatic, makes Pasch's axiom unnecessary. Nonetheless, I cannot prove or disprove it.

EDIT

Apparently Szmielew proved that Pasch's axiom is a consequence of the circle axiom, but it remains a mystery how she proved it since I do not have a copy.