I'm working through Hartshorne's Foundations of Projective Geometry, and have hit a sticking point with problem 7(a). For clarity, I'll reproduce the axiom system presented for projective 3-space:
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Two points determine a unique line.
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Three non-collinear points determine a unique plane.
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Any line intersects any plane in at least one point.
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Any two planes intersect in at least one line.*
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There exist four non-coplanar points, with no three among them collinear.
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Every line has at least three points.
*Note that if we have two planes, their intersection contains some line $l$ by this axiom, and by axiom 6 that line has some points on it, say $S$ and $T$. If the intersection also contained $U \notin l$, then both planes would contain $S,T,U$ non-collinear, and would thus have to be equal by axiom 2; therefore, in these axioms, we still have that two distinct planes intersect in exactly one line.
Problem 7(a) then asks the reader to prove:
"If two distinct points $P$, $Q$ lie in a plane $\Sigma$ then the line joining them is contained in $\Sigma$."
Here's an attempted proof, and what goes wrong with it.
Proof. Let $P, Q \in \Sigma$. There must exist a point $R$ which is neither in $\Sigma$ nor in the line $PQ$. But then $PQR \cap \Sigma$ is exactly a line by above, and that line must be $PQ$ since both planes contain the points $P$ and $Q$. Therefore $\Sigma$ contains $PQ$.
The statement in bold doesn't feel justifiable—there certainly has to be a point outside of $\Sigma$ by axiom 5, but what precludes the possibility that all points outside of $\Sigma$ happen to lie on $PQ$? Or, is there a better approach that I'm just not seeing?
Best Answer
Take any distinct points $P,Q$ on any plane $S$. Let $A,B,C,D$ be non-coplanar points such that no three of them are collinear. Then not all of $A,B,C,D$ are on $S$. By symmetry we can assume that $A$ is not on $S$. Let $T$ be a plane containing $P,Q,A$. Then $S ≠ T$. Let $L = S∩T$. Let $X,Y$ be distinct points on $L$. Then $P,X,Y$ are contained in both $S,T$. If $P$ is not on $L$, then $P,X,Y$ are contained within a unique plane, contradicting $S ≠ T$. Thus $P$ is on $L$. Symmetrically, $Q$ is on $L$. Thus $P,Q$ are contained in both $PQ,L$. Thus $PQ = L$.
[Edit: It was pointed out that it's not so simple if $P,Q,A$ are collinear, so here is how we can deal with that special case. Let $E$ be a point not on $PA$. Let $U$ be a plane containing $P,A,E$. Then by the same reasoning as above we have $PA = S∩U$. Similarly let $V$ be a plane containing $Q,A,E$, and then $QA = S∩V$. Thus $PA$ is contained in both $U,V$. If $U ≠ V$, then $PA = U∩V = AE$ (with second equality by the same reasoning again), yielding contradiction. Therefore $U = V$ and that plane contains $P,Q,A$.]
By the way, using bare "let" is not a good idea when it conflates ∃elim with ∀subcontext-reasoning (i.e. reasoning about an arbitrary given object of a stipulated type).