Proving this axiom system is consistent.

Question

I want to make sure I understand the correct notation and expressions for proving that an axiom system is consistent. I have the axioms

1. Every line is a collection of points.
2. There are at least two points.
3. For any two points there is exactly one line containing them.
4. For any line there is a point not on the line.
5. For any line and point not on the line, there is a line parallel to the first line which contains the point.

To prove this consistent I want to exhibit a model. I this case, I think I want two points $P,Q$ that lay on one line, and an infinity of other (distinct) points $A_{1}, A_{2}, ...$ each of which lies on lines that are pairwise parallel. Each of these lines will, themselves, be a collection of points. While I could have each line be an infinity of points, a minimal model would have all lines contain just one point except for the first line I mentioned, which has two.

Do I need to say more than this? Is there some notation or form of expression that is more appropriate for the assignment?

Answer 1

$ \newcommand{L}{\mathbb L} \newcommand{P}{\mathbb P} $

Every line is a collection of points.

$$\forall l \in \L ~~ a \in l \rightarrow a \in \P$$

There are at least two points.

$$\exists a,b \in \P ~~ a \ne b$$

For any two points there is exactly one line containing them.

$$\forall a,b \in \P ~~\exists_1 l \in \L ~~a \in l \land b \in l $$

For any line there is a point not on the line.

$$\forall l \in \L~~ \exists p \in \P ~~p \not \in l$$

For any line and point not on the line, there is a line parallel to the first line which contains the point.

$$\forall l \in \L , p \in \P ~~ p \not \in \L \rightarrow \exists m \in \L ~~ p \in m \land Z(m,l)$$

These axioms require 2 sets, $\P$ and $\L$. They require 1 function $Z(,)$ (for parallel) which has to be defined over $\L$.

To show consistency in terms of model theory, you show a structure (a collection of objects,sets,functions) for which the axioms are true. This structure doesn't have to have anything to do with geometry as you intuitively think of it.

For example , a structure which satisfies the axioms is:

• $\P$ is the set of natural numbers $\{0,~ 1,~ 2,~ \dots\}$
• $\L$ is the set of pairs of natural numbers $\{(0,1),~ (0, 2),~ (1, 2),~ (0, 3),~ \dots \}$
• $Z( (a,b) , (x,y) )$ is true if and only if $a + 1 = x$ and $b + 1 = y$

each of which lies on lines that are pairwise parallel.

This seems to defy the axiom that all pairs of points can make a line. You have partitioned the points among lines, but the axioms that any 2 points make a line imply that there must be a "chain" among the line containment predicate : in other words, for any 2 points $p_1, p_2$, there are a collections of points $q$ and line $L$ such that $p_1 L_0 q_0 \land q_0 L_1 q_1 \land q_1 L_2 q_2 \land q_2 L_3 q_3 \dots q_n L_m p_2$. Where $pLq$ means points $p$ and $q$ are on $L$.

Each of these lines will, themselves, be a collection of points.

Why? What axiom is this satisfying?

a minimal model would have all lines contain just one point except for the first line I mentioned, which has two.

I believe a minimal model would be

• $\P = \{a, b, c\}$
• $\L = \{(a,b) , (b,c), (a,c)\}$
• $Z( x , y ) = \begin{cases} x = (a,b) \land y = (b,c) \\ \text{ or } \\ x = (b,c) \land y = (a,c) \\ \text{ or } \\ x = (a,c) \land y = (a,b) \end{cases}$

Since "For any line there is a point not on the line." demands that there is at least 3 points.