Prove that the union of all lines containing a point A is the plane.

Question

I need to prove this with the knowledge of incidence and order axioms.

Let $X$ be a set with all the points forming all the lines containing the point $A$, and let $Y$ the set of all points in the plane. I want to prove that they have the same elements, so I will show that $X \subset Y$ and $Y \subset X$.

Given any line $r \subset X$, presumably, $r \subset Y$ because I want to show that the union of all of them is exactly equal to the plane, so each line is IN the plane itself, then $X \subset Y$.

Now choose a point $Y_0 \in Y$ and let $s \subset X$ be a line. By one of the axioms, there's a point $X_0$ out of $s$, make $X_0 = Y_0$ and by the first incidence axiom, given two points, in this case, $X_0 = Y_0$ and $A$, there's a line containing both of these points, thus $Y_0$ is a point in a line contaning $A$. Therefore, $Y \subset X$.

Finally, it's proved that $X = Y$.

Just to clafiry on the geometry I'm dealing with:

Axioms of incidence:

I) Given two distinct points, there's a single line containing both.
II) In any given a single line, there are at least two distinct points in it.
III) There are three points such that there's not a line containing all of them simultaneously.

Axioms of Order:

$A * B * C$ means that point B is between $A$ and $C.$

I) If $A * B * C$, then $A, B$ and $C$ are three distinct points in the same line and $C * B * A$.
II) Given three points in a line, one, and only one of them is in between the other two.
III) Given two points $B$ and $D$, there exists points $A, C$ and $E$ such that $A * B * D$, $B * C * D$, $B * D * E$.
IV) For any line $m$ and for any given three points $A, B$ and $C$ out of $m$:

• i) If $A$ and $B$ are in the same side of $m$ and $B$ and $C$ are in the same side of $m$, then $A$ and $C$ are in the same side of $m$.
  
• ii) If $A$ and $B$ are in opposite sides of $m$ and $B$ and $C$ are in opposite sides of $m$, then $A$ and $C$ are in the same side of $m$.

PS: Two points $A$ and $B$ being in the same side of a line $m$ means that the segment $AB$ has no intersection with the line $m$. If they're in opposite sides, then $AB$ intersects $m$.

Answer 1

Given any line $r \subset X$, presumably, $r \subset Y$ because I want to show that the union of all of them is exactly equal to the plane, so each line is IN the plane itself, then $X \subset Y$.

This argument sounds a bit cyclical: “because I want to show something, the thing that I want to show must be true.”

But showing the first direction using the cited axioms is indeed problematic. None of your axioms talk about a concept called “plane”, so you can't directly apply the axioms to prove anything about $Y$.

If you say that your axioms are meant to characterized the plane, then you'd probably say all the points there are belonging to the plane. $Y$ would be the set of all the points, and any set of points would be a subset of that. So this first direction might be obvious in that sense, but it depends on a definition that should have been explicit but wasn't stated as such.

Now choose a point $Y_0 \in Y$ and let $s \subset X$ be a line. By one of the axioms, there's a point $X_0$ out of $s$, make $X_0 = Y_0$ and by the first incidence axiom, […]

I find this argument hard to follow. The order appears wrong, because you pick $Y_0$ and $s$ at the same time and apparently independent from one another, yet you then expect $X_0$ to be both equal to $Y_0$ and incident with $s$. To make that work, you need to pick the point first, and the line second and taking the point into account.

The second half of that paragraph actually appears to do just that, so best to drop the first half.

given two points, in this case, $X_0 = Y_0$ and $A$, there's a line containing both of these points, thus $Y_0$ is a point in a line contaning $A$. Therefore, $Y \subset X$.

This is a sound argument, and the core of the whole proof.