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Let $\mathcal{A} = (\mathcal{P},\mathcal{L})$ where $\mathcal{P}$ is a set of points and $\mathcal{L}$ consists of subsets of $\mathcal{P}$ be a affine plane if

  1. each two points of $\mathcal{P}$ lie in a unique line
  2. Let $L\in \mathcal{L}$ be a line and $x\not \in L$ a point; then there is a unique line $M\in \mathcal{L}$ such that $x\in M$ and $L\cap M = \varnothing$
  3. There are three points which do not belong to a line.

I need to answer:

Show that 1. cannot be simplified as

1'. each two points of $\mathcal{P}$ lie in a (not not necessarily unique) line

I guess the strategy would be to show that any structure where this is valid somehow results in a contradiction.

I'm given the following as solution, but I don't see why this is or contributes to a solution.

Proposed solution / hint (?)

Let $\mathcal{P} = \mathbb{R}^3, \mathcal{L} = \{p : p $ is a plane in $ \mathbb{R}^3\}$.

Notice how $(\mathcal{P},\mathcal{L})$ fullfills 1', 2 but does not fulfill 3.

Now let $\mathcal{L}' = \{p : p $ is a plane in $ \mathbb{R}^3\} \setminus \{p:p $ is a plane in $ \mathbb{R}^3$ parallell to the $x,y$-plane $\}$

Here 1', 2 and 3 are valid.

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The point isn't that $1'$ is contradictory, but rather that it's strictly weaker than $1$: the exercise is asking you to show that the axioms $1', 2, 3$ don't describe the same class of structures as the axioms $1, 2,3$.

So how would you show this? Well, you'd have to find a structure satisfying one set of axioms but not the other. Since $1$ implies $1'$, this means we can't find something satisfying $1, 2, 3$ but not $1', 2, 3$; so we have to find something satisfying $1', 2, 3$ but not $1, 2, 3$. This is what the hint is aiming you towards.

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