Trouble proving that a plane in (synthetic) projective space containing two points must contain the line between them

Question

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:

1. Two points determine a unique line.

2. Three non-collinear points determine a unique plane.

3. Any line intersects any plane in at least one point.

4. Any two planes intersect in at least one line.*

5. There exist four non-coplanar points, with no three among them collinear.

6. 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?

Answer 1

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).

Answer 2

I'm not convinced by user21820's answer to this question. I have found a different approach which does work, however, which I will explain in detail. I will denote points in lowercase, and lines/planes in uppercase. I will write $x\in S$ to denote that $x$ is a point lying on $S$. I write $A\subseteq B$ to mean $\forall x, x\in A \implies x\in B$. For $x\neq y$, I'll write $xy$ to denote the unique line containing $x,y$. Similarly $xyz$ is the unique plane containing each of $x,y,z$, provided they're not colinear. For clarity, a triplet of points is defined to be colinear precisely when there exists some line containing all three. A similar definition applies to "coplanarity". Our objective is to prove the implication $(x\neq y \land x\in P \land y\in P) \implies xy\subseteq P$ where $P$ is any plane.

---

Lemma: If $L$ is a line with $a,b\in L$ and $a\neq b$, then $L=ab$.
Lemma: If $P$ is a plane with $a,b,c\in P$ not colinear, then $P=abc$.
proof: First lemma follows from axiom 1, the second from axiom 2.
$\square$

Lemma: Given planes $P,Q\ni x,y$, with $x\neq y$, either $P=Q$ or $P\cap Q = xy$.
Lemma: Given planes $xyz$ and $xyw$, either $xyz=xyw$ or else $xyz\cap xyw = xy$.
proof: By axiom 4, there's a line $L$ with both $L\subseteq P$ and $L\subseteq Q$. By axiom 6, find $a,b\in L$ with $a\neq b$, then $L=ab$. If there were $p\notin L$ with $p\in P$ and $p\in Q$, then $P=abp=Q$. Otherwise there's no such $p$ and thus $P\cap Q = L$, and since $x,y\in P\cap Q$ then $x,y\in L$ so that $L=xy$, consequently $P\cap Q = xy$. This proves our first lemma. The second lemma is immediate using $P=xyz$ and $Q=xyw$.
$\square$

In light of the above, to prove the target claim, it suffices to show that for each plane $P$ and $x\neq y$ in $P$, there's a second plane $Q\neq P$ with both $x,y\in Q$. Then $xy=P\cap Q$ implies $xy\subseteq P$. This will be important in the last proof below.

Lemma: If $x,y,z,w$ are not coplanar, and no three are colinear, then no two among the planes $xyz,xyw,xzw,yzw$ are equal. Similarly the six lines $xy,xz,xw,yz,yw,zw$ are distinct.
proof: Since no three among $x,y,z,w$ are colinear, then each among $xyz,xyw,xzw,yzw$ is at least an extant plane. If any two were equal, say $xyz=xyw$, then all four $x,y,z,w\in xyz$ are coplanar, impossible. Similarly if any two of our lines were equal, we'd get a colinear triplet, impossible.
$\square$

Theorem: Given plane $P$ and $a,b\in P$ with $a\neq b$, necessarily $ab\subseteq P$
proof: In pursuit of contradiction, suppose not. Notice that for any plane $Q\neq P$, if we had $a,b\in Q$ then $P\cap Q = ab$, but this implies $ab\subseteq P$, a contradiction. Thus, $P$ is the unique plane containing the points $a,b$. Furthermore, given any $p\notin ab$, we find a plane $abp$ containing $a,b$, and thus $abp=P$ so that $p\in P$. Hence, every point $p$ either has $p\in ab$ or else $p\in P$.
Apply axiom 5 to find four points $x,y,z,w$ not coplanar, no three colinear. If any two among them were on $ab$, say $x,y\in ab$, then we find $ab=xy=xyz\cap xyw$, but then $xyz=P=xyw$, a contradiction. It follows that at least three among $x,y,z,w$ are not in $ab$, let's say $x,y,z\notin ab$. Then instead $x,y,z\in P$, hence $P=xyz$. Since $w\notin xyz = P$, then instead $w\in ab$.
Since $w\notin xyz=P$ while $a,b\in P$, then $w\neq a,b$. Since also $w\in ab$, then $ab=aw$. Since $x\notin ab=aw$, then $x,a,w$ are not colinear, so there's a plane $axw$. Similarly, there's a plane $ayw$. If these planes aren't equal, then $ab=aw=axw\cap ayw$, but then $axw=P=ayw$, a contradiction. It follows that $axw=ayw$ and thus $x,y,w\in ayw$. Consequently $xyw=ayw$ so that $a\in xyw$. By a symmetrical argument, we place $b\in xyw$ and thus infer $P=xyw$. But now $xyz=P=xyw$, a contradiction.
$\square$

---

Axioms 3 was not necessary for this proof. Axiom 6 can be weakened to only require two points on a line. Axiom 4 can also be weakened to require two shared points on our two planes, before inferring they also share some line.

Answer 3

Here's a detailed step-by-step version of user21820's proof, as best I could render it. My numbering scheme is probably inconsistent in places, but that makes no real difference. I believe that this proof is incomplete, as noted in the "reasons" column below.

Let me start by stating Hartshorne's axioms.

Note that we have a set $X$ of points, a collection of subsets called lines, and another collection of subsets called planes. A statement like ``$Q$ lies on $l$" means that $Q$ is in the subset $l$.

S1. Two distinct points $P$ and $Q$ lie on one and only one line $l$.

S2. Three non-collinear points $P,Q,R$ lie on a unique plane.

S3. A line $l$ meets a plane $S$ in at least one point.

S4. Two planes $S$ and $T$ have at least a line in common

S5. There exist four non-coplanar points, no three of which are collinear.

S6. Every line $l$ has at least three points.

I've added names in S3 and S6, so that I can write ``S6[of $m$]" as shorthand for the statement "the line $m$ has at least three points," for instance.

Following convention, I'm denoting the unique line through distinct points $P$ and $Q$ by $PQ$, and assuming the usual definitions of collinearity, coplanarity, etc.

We also have ``Fargle's Lemma":

Lemma F: If $S$ and $T$ are distinct planes, then there is a line $k$ such that $k = S \cap T$.

I've marked 3.2.3, 3.2.6 as not-yet-justified, and 3.2.8.1.2 as something I couldn't easily piece together, although I suspect it's fine (modulo 3.2.3 and 3.2.6).

To prove: If (i) $S$ is a plane, (i) $P, Q \in X$, (iii) $P \ne Q$, (iv) $P, Q \in S$ then the line $PQ \subset S$.

\begin{array}{l|l|l} 1 & \text{There are four noncoplanar points } K,L,M,N & \text{ S4 }\\ 2 & \text{There is a point} A \in \{K, L, M, N\} \text{ with } A \notin S& \text{By contradiction} \\ 2.1 & \text{Suppose for contradiction that} \{K, L, M, N\} \subset S & \text{Contradiction hypothesis} \\ 2.2 & \text{Then } K, L, M, N \text{ are coplanar}& \text{Definition of coplanar}\\ 2.3 & \text{This contradicts S1.} & \text{End of contradiction} \\ 3 & \text{Cases: }A \notin PQ\text{ or } A \in PQ& \text{Excluded middle.} \\ 3.1 & \text{case 1: we assume }A \notin PQ & \text{Statement 3} \\ 3.1.1& \text{Let }T\text{ be a plane containing }P,Q \text{ and }A & \text{3.1, S2[of $A$ $P$ $Q$]} \\ 3.1.2 & \text{$ S \ne T$ } & \text{Because $A \in T$ (3.1.1) and $A \notin S$ (2) }\\ 3.1.3 & \text{$k = S \cap T$ is a line }& \text{Lemma F[of S T], 3.1.2 }\\ 3.1.4 & \text{$P, Q \in k$ }& \text{Assumption (iv), 3.1.1 }\\ 3.1.5 & \text{$P, Q \in PQ$. }& \text{Axiom S1, defn of PQ.}\\ 3.1.6 & \text{$PQ = k$ }& \text{Axiom S1, uniqueness part.}\\ 3.1.7 & \text{qed }& \text{concludes proof for case 1. (3.1)}\\ 3.2 & \text{case 2: we assume $A \in PQ$ }& \text{statement 3 }\\ 3.2.1 & \text{Let $E$ be a point with $E \notin PA$ }& \text{by contradiction }\\ 3.2.1.1 & \text{Suppose, for contradiction, that EVERY point} &\\ & \text{is on $PA$, i.e., $X \subset PA$. }& \text{contradiction hypothesis}\\ 3.2.1.2 & \text{There are four noncoplanar points } & \\ & \text{$K,L,M,N$, no three collinear. }& \text{Axiom S4 }\\ 3.2.1.3 & \text{$K,L,M$ are noncollinear }& \text{3.2.1.2 }\\ 3.2.1.4 & \text{$K,L, M$ are collinear }& \text{because all lie in $PA$, by 3.2.2 }\\ 3.2.1.5 & \text{Contradiction }& \text{3.2.4, 3.2.5, finishes 3.2.1.1 }\\ 3.2.2 & \text{There is a plane $U$ containing $P, A, $ and $E$.} & \text{noncollinear by 3.2.1; apply S2[of $P$ $A$ $E$].} \\ 3.2.3 & \text{$PA = S \cap U$ }& \mathbf{Unjustified. } \text{ Apply LemmaF[of $S$ $U$]. Still need to show $S \ne U$.} \\ 3.2.4 & \text{$PA = QA$ }& \text{3.2, Axiom S1 }\\ 3.2.5 & \text{There is a plane $V$ containing $Q, A, $ and $E$.}& \text{3.2.4, 3.2.1, which says they're noncollinear; apply S2[of $Q$ $A$ $E$]. }\\ 3.2.6 & \text{$PA = S \cap V$ }& \mathbf{Unjustified. } \text{ Apply Lemma F[of S V]. Need to show $S \ne V$.}\\ 3.2.7 & \text{$PA \subset U \cap V$ }& \text{3.2.3, 3.2.6} \\ 3.2.8 & \text{cases: $U \ne V$ or $U = V$ }& \text{Excluded middle. } \\ 3.2.8.1 & \text{case i: $U \ne V$ }& 3.2.8 \\ 3.2.8.1.1 & \text{$PA = U\cap V$ }& \text{Lemma F, 3.2.8.1, 3.2.7 }\\ 3.2.8.1.2 & \text{$AE = U\cap V$ }& \mathbf{???} \text{ ``By the same reasoning as above" is too vague for me to work out here.} \\ 3.2.8.1.3 & \text{$PA = AE$, hence $P,A,E$ are collinear }& \text{3.2.8.1.1, 3.2.8.1.2, defn of collinear }\\ 3.2.8.1.4 & \text{qed}& \text{3.2.8.1.3, closes case i of 3.2.8 }\\ 3.2.8.2 & \text{case ii: $U = V$ }& \text{3.2.8} \\ 3.2.8.2.1 & \text{The plane $U$ contains $P,Q,A$ }& \text{3.2.3, 3.2.6} \\ 3.2.2 & \text{$U \ne S$ }& \text{Because $A \in U$ (3.2.8.2.1) but $A \notin S$ (2)} \\ 3.2.3 & \text{$U \cap S$ is a single line $k$ }& \text{3.2.2, Lemma F} \\ 3.2.4 & \text{$P,Q \in k$ }& \text{3.2.2,3.2.5, 3.2.8.1, assumption iv.}\\ 3.2.5 & \text{$PQ = k$ }& \text{3.2.4, assumption (iii), Axiom S1}\\ 3.2.6 & \text{qed} & \text{closes case i of 3.2.8, concludes proof for case 2 (3.1)}\\ \end{array}

By the way, just in case there's some doubt about whether I've got the logic right or am assuming something, here's an Isabelle/HOL representation of the axioms of a projective 3-space and the purported proof above; I've had to use the justification "sorry" for step 3.2.3 unfortunately. I recognize that most people don't know Isabelle, but it seemed like a good exercise just to be sure that all the steps were correct. I'm somewhat embarrassed by the proof of Fargle's lemma --- it feels as if it should be only a couple of lines --- but so be it.

At the end, I've stopped bothering with the proof as written, because it's an easy jump from step 3.2.3 to the conclusion, if only we could find a proof for 3.2.3. (Isabelle's sledgehammer tool gives up, although it has no trouble with any of the earlier steps.)

theory MSE4550757
  imports Main
begin

locale P3 =
fixes Points ::"'p set" and Lines :: "'p set set" and Planes:: "'p set set" and join::"'p ⇒ 'p ⇒ ('p set)"
assumes
S0: "Points = (UNIV::'p set)" and
S1a: "⟦P ∈ Points; Q ∈ Points⟧ ⟹ (join P Q) ∈ Lines ∧ P ∈ (join P Q) ∧ Q ∈ (join P Q)" and
S1b: "⟦P ∈ Points; Q ∈ Points; P ≠ Q; k ∈ Lines; P ∈ k; Q ∈ k⟧ ⟹ k = (join P Q) " and
S2: "⟦P ∈ Points; Q ∈ Points; R ∈ Points; distinct[P,Q,R]; R ∉ join P Q⟧ ⟹
(∃! H ∈ Planes . (P ∈ H ∧ Q ∈ H ∧ R ∈ H))" and
S3: "⟦l ∈ Lines; S ∈ Planes⟧ ⟹ (∃ P ∈ Points . P ∈ (l ∩ S))" and
S4: "⟦T ∈ Planes; S ∈ Planes⟧ ⟹ (∃ k ∈ Lines .  k ⊆ (T ∩ S))" and
S5: "∃ A B C D . (A ∈ Points ∧ B ∈ Points ∧ C ∈ Points ∧ D ∈ Points ∧
(¬ (∃ H ∈ Planes . A ∈ H ∧ B ∈ H ∧ C ∈ H ∧ D ∈ H)) ∧
(¬ (∃ k ∈ Lines . A ∈ k ∧ B ∈ k ∧ C ∈ k)) ∧
(¬ (∃ k ∈ Lines . A ∈ k ∧ B ∈ k ∧ D ∈ k)) ∧
(¬ (∃ k ∈ Lines . A ∈ k ∧ C ∈ k ∧ D ∈ k)) ∧
(¬ (∃ k ∈ Lines . B ∈ k ∧ C ∈ k ∧ D ∈ k)))" and
S6: "k ∈ Lines ⟹ ∃ P Q R . P ∈ Points ∧ Q ∈ Points ∧ R ∈ Points ∧ distinct[P,Q,R] ∧ P ∈ k ∧ Q ∈ k ∧ R ∈ k"
begin
definition  collinear :: "'p ⇒ 'p ⇒ 'p ⇒ bool"
where "collinear A B C = (∃ l. l ∈ Lines ∧ A ∈ l ∧ B ∈ l ∧ C ∈ l)"
definition  coplanar :: "'p ⇒ 'p ⇒ 'p ⇒ 'p ⇒ bool"
where "coplanar A B C D = (∃ H. H ∈ Planes ∧ A ∈ H ∧ B ∈ H ∧ C ∈ H ∧ D ∈ H)"
lemma S5alt:
shows "∃ A B C D . (A ∈ Points ∧ B ∈ Points ∧ C ∈ Points ∧ D ∈  Points ∧

(¬ coplanar A B C D) ∧
(¬ collinear A B C) ∧
(¬ collinear A B D) ∧
(¬ collinear A C D) ∧
(¬ collinear B C D))" using S5 coplanar_def collinear_def by metis
lemma collinear_join [iff] :
assumes "P ≠ Q"
shows "collinear P Q X ⟷ X ∈ join P Q"
using S0 S1a S1b assms collinear_def by auto
lemma distinct3_alt:
"distinct[A,B,C] ⟷ (A ≠ B ∧ A ≠ C ∧ B ≠ C)"
by auto
lemma fargle:
fixes S T
assumes "S ≠ T"
assumes "S ∈ Planes ∧ T ∈ Planes"
shows "∃ k . (k ∈ Lines ∧ k = (S ∩ T))"
proof -
obtain  k where kfacts: "(k ∈ Lines ∧ k ⊆ (S ∩ T))" using assms S4 by blast
obtain  P Q where pqfacts: "P∈ Points ∧ Q ∈ Points ∧ P ∈ k ∧ Q ∈ k ∧ P ≠ Q" using kfacts S6[of k] by auto
have u: "P ∈ S ∧ Q ∈ S ∧ P ∈T ∧ Q∈ T" using kfacts pqfacts by auto
fix X
have  a0: "X ∈ S∩T ⟹ collinear P Q X" for X
proof (rule ccontr) (* classical contradiction *)
assume ch: "¬collinear P Q X"
assume minor: "X ∈ S ∩ T"
have d: "distinct[P,Q,X]" using pqfacts ch collinear_def kfacts by force
have b: "∃! H . H ∈ Planes ∧ P ∈ H ∧ Q ∈ H ∧ X ∈ H" using S2[of P Q X] d ch distinct3_alt[of P Q X]
by (simp add: S0 S1a S1b collinear_def)
then have "S = T" using assms u ch b minor by auto
then show False using assms by auto
qed
have a1: "X ∈ S∩T ⟹ X ∈ join P Q" using a0 pqfacts by blast
then have a2: "S ∩ T = join P Q" using S1b a0 kfacts pqfacts by force
then show ?thesis by (simp add: S1a pqfacts)
qed
theorem main:
fixes S P Q
assumes "S ∈ Planes ∧ P ∈ Points ∧ Q ∈ Points"
assumes "P ≠ Q"
assumes "P ∈ S ∧ Q ∈ S"
shows "join P Q ⊆ S"
proof -
obtain A where a2: "A ∈ Points ∧ A ∉ S" using assms S4 S5 by auto (* 2 )
consider (outside) "A ∉ join P Q" | (inside)  "A ∈ join P Q" by blast ( 3 )
then show ?thesis
proof (cases)
case outside ( 3.1 )
obtain T where tfacts: "T ∈ Planes ∧ A ∈ T ∧ P ∈ T ∧ Q ∈ T"
by (metis S2 a2 assms(1,2,3) distinct3_alt outside) ( 3.1.1 )
have "S ≠ T" using tfacts a2 by blast ( 3.1.2 )
then obtain k where  a313: "k ∈ Lines ∧ k = S ∩ T" using fargle[of S T]
using assms(1) tfacts by blast ( 3.1.3 )
have "P ∈ k ∧ Q ∈ k" by (simp add: a313 assms(3) tfacts)
then have a316: "join P Q = k" using S0 S1b a313 assms(2) by blast ( 3.1.6 )
then show ?thesis by (simp add: a313) ( 3.1.7 )
next
case inside ( 3.2 )
have "A ∈ join P Q" using inside by blast
obtain E where efacts: "E ∈ Points ∧ E ∉ join P A" by (smt (verit) S1a S5 a2 assms(1)) ( 3.2.1 -- 3.2.1.5 )
then obtain U where ufacts: "U ∈ Planes ∧ P ∈ U ∧ E ∈ U ∧ A ∈ U"
by (metis S1a S2 a2 assms(1,3) distinct3_alt) ( 3.2.2 )
then have pa1: "join P A = S ∩ U" sorry ( 3.2.3 )
show ?thesis using pa1 S0 S1a a2 by blast ( cut to the chase and conclude join P Q ⊆ S *)
qed
qed

Answer 4

N.B.: This answer is now moot, because we have Jade Vanadium's correct proof.

I believe I have a resolution. I took the simple expedient of writing to Robin Hartshorne, who looked at this and said that he thought that an additional axiom was needed, for example (3A) "a line not contained in a plane meets it in exactly one point," or (3B) "any plane in the projective three space together with the lines contained in it forms a projective plane."

In the meantime, two of my students are on a quest to find a structure that satisfies the axioms given in the question above, but not axiom 3A, because they want to show that their frustration with doing the exercise above was justified. :)

Certainly if you use 3A, then a whole lot of challenging-but-equivalent results fall out immediately.