In one of my Elementary Geometry previous exams, one of the questions was the following:

Study if the following statements are equivalent to the paralellism axiom:

$(i)$ Any three straight lines have a common transversal

$(ii)$ Any four points are in the interior of a triangle

I am absolutely clueless about this. Definitely, assuming Playfair's axiom (or some equivalent form, like the right angle axiom) we should be able to prove $(ii)$ (after some work). Is the converse true? It seems pretty odd and unrelated to parallel lines. How should we proceed to prove or disprove this claim?

I would really appreciate a detailed answer, for I am learning just from scratch (I've had absolutely no previous experience with this kind of topics).

EDIT: I am supposed to work in a Hilbert plane without assuming or denying the Archimedes' axiom, nor the circle-circle intersection property, nor any other additional assumptions, like the existence of limiting parallel lines for any line and any point outside of it.

  • $\begingroup$ To show that something is not equivalent to the parallel axiom, you might exhibit a model of a geometry without parallelism and with that something holding $\endgroup$ – Hagen von Eitzen May 5 at 20:15
  • $\begingroup$ The negation of (ii) is not "for any configuration of four points, there is no triangle that contains them". It's enough to find one configuration of four points not contained in a triangle. (But also, note that showing that (i) or (ii) fails in the Poincaré model does not show that it's equivalent to the parallel postulate - not in the absence of Archimedes's axiom.) $\endgroup$ – Misha Lavrov May 5 at 20:36
  • 1
    $\begingroup$ In planar elliptic geometry any two lines intersect so, given 3 lines, any other line will intersect all 3 of them $\endgroup$ – DanielWainfleet May 5 at 20:47
  • 2
    $\begingroup$ @DanielWainfleet That's not a Hilbert plane, though: it violates the betweenness axioms. We can show that in any Hilbert plane, parallel lines do exist. $\endgroup$ – Misha Lavrov May 5 at 20:58
  • $\begingroup$ @Akerbeltz What is your definition of a transversal? I think I know the solution to the first one but I'd rather know your definition for a detailed answer. $\endgroup$ – Kulisty May 6 at 22:55

Both (i) and (ii) are strictly weaker than the parallel postulate. Probably the easiest way to see this is using coordinates. Given any Pythagorean ordered field $F$ and a convex subring $F_0\subseteq F$, the full subplane $F_0^2$ of the coordinate plane $F^2$ is a Hilbert plane which will always satisfy (i) and (ii), and will not satisfy the parallel postulate unless $F_0=F$. (If $F$ is non-Archimedean, you can always find such an $F_0$ which is not all of $F$, for instance by taking $F_0$ to be the set of all "finite" elements of $F$, i.e. elements bounded by integers.) Conversely, every Hilbert plane satisfying the parallel postulate is isomorphic to the coordinate plane over some Pythagorean ordered field, and so will satisfy (i) and (ii).

Here is a proof that $F_0^2$ satisfies (i) and (ii). For (i), suppose we have a non-vertical line with equation $y=ax+b$. If the slope $a$ is in $F_0$, then $ax\in F_0$ for all $x\in F_0$, and thus we must have $b\in F_0$ for the line to have any points in $F_0^2$ at all, in which case the line intersects every vertical line. On the other hand, if the slope $a$ is not in $F_0$ (or if our line is vertical), then $a$ must be infinitely large (since $F_0$ is a convex subring of $F$, it contains all elements which are bounded by an integer), so $a^{-1}$ is infinitesimal and is in $F$. Rewriting the equation as $x=a^{-1}(y-b)$, the argument from the first case now shows that our line intersects every horizontal line.

Thus, every line in $F_0^2$ either intersects every vertical line or intersects every horizontal line. Given three lines, we may thus assume without loss of generality that two of them intersect every vertical line. Now take any point on the third line, and the vertical line on that point will be a transversal to all three lines. (Unless the third line is itself vertical, in which case you can take its intersection with the first line and draw the line from it to any other point on the second line, choosing the latter point so that the line is not vertical.)

For (ii), suppose we have four (or indeed any finite collection of) points in $F_0^2$. Choose $a<b$ in $F_0$ such that both coordinates of all four points are strictly between $a$ and $b$. Then the right triangle with vertices $(a,a)$, $(a,2b-a)$, and $(2b-a,a)$ contains all four points. (Note that this argument actually requires only that $F_0$ is a subgroup of $F$, not a subring as was needed for (i).)


Following Akerbeltz's definition: I define that a line $K$ is a transversal to line $L$ whenever $K$ cuts through $L$ (in other words $K$ and $L$ have a point in common). It is not required that these lines are distinct. It is also not required that the intersection points are distinct in case that a line is a transversal to two or more other lines.

Recall some axioms:

Axiom 1. For any line there exist (at least) two distinct points on this line.

Axiom 2. For any two points $a,b$ there exists a line $L$ such that $a,b\in L$.

Axiom of parallelism. For any line $L$ and a point $p\notin L$ there exists at most one line disjoint with $L$ passing through $p$.

Proof of (i) in Euclidean geometry:

Let $L_1,L_2,L_3$ be straight lines.

First case: One pair of these three lines has (at least one) point in common. Say $L_1$ and $L_2$ have a point in common (say $a$). By axiom 1 there exist a point $b\in L_3$. By axiom 2 there exists a line $K$ such that $a,b\in K$. $K$ is the desired transversal.

Second case: $L_1,L_2,L_3$ are pairwise disjoint. By axiom 1 there exists a point $a\in L_1$ and a point $b\in L_2$. By axiom 2 there exists a line $K$ such that $a,b\in K$. Observe that $a\notin L_2$ because $a\in L_1$ and $L_1,L_2$ are disjoint. By the same argument $b\notin L_3$. Next observe that $K\neq L_2$ because $a\in K$ and $a\notin L_2$. So $b\notin L_3$ and $K,L_2$ are distinct lines passing through $b$ and $L_2$ is disjoint with $L_3$. By axiom of parallelism $K$ cuts $L_3$ and $K$ is the desired transversal.

Note. I did not use any other axioms that 3 mentioned before the proof.

For proving that (i) implies the axiom of parallelism I will need way more axioms and properties. That's why I'll present something more like a sketch.

Recall that an angle is an unordered pair $\{A,B\}$ or rays emanating from the same origin which are distinct and not complementary. The interior of the angle $\{A,B\}$ is the intersection of two halfplanes $M,N$ where the boundary of $M$ is $L(A)$ (i.e. the line containing $A$) and $B\subset M$ and the boundary of $N$ is $L(B)$ and $A\subset N$.

Proof of the negation of (i) in hyperbolic geometry:

There exist two distinct lines $K,L$ which have exactly one point $o$ in common. Let $A,B$ be rays emanating from $o$ contained in $K,L$ respectively (choice is arbitrary). By $A^*,B^*$ I'll denote the rays complementary to $A,B$. Then $\{A,B\},\{B,A^*\},\{A^*,B^*\}$ are angles. By $M$ I'll denote the halfplane with boundary $K=L(A)=L(A^*)$ such that $B\subset M$. By $N$ I'll denote the halfplane with boundary $L=L(B)=L(B^*)$ such that $A^*\subset N$. Then the interior of $\{A,B\}$ is $M\cap N^*$, the interior of $\{B,A^*\}$ is $M\cap N$ and the interior of $\{A^*,B^*\}$ is $M^*\cap N$.

Now I will use the following lemma which is the only place where the negation of parallelism axiom is needed i.e. it holds only in hyperbolic geometry. I won't give a proof for this and I'm not sure whether some continuity axioms (such as Archimedes' axiom) aren't necessary.

Lemma. For any angle $\{A,B\}$ there exists a line $L$ which is contained is the interior of this angle.

Applying lemma to angles $\{A,B\},\{B,A^*\},\{A^*,B^*\}$ we get lines $L_1,L_2,L_3$ respectively. They will be the desired lines. We have $L_1\subset M\cap N^*, L_2\subset M\cap N,L_3\subset N\cap M^*$. So $L_1,L_3$ lie on opposite sides of line $L$ and $L_1,L_2$ lie on opposite sides of line $L$. Therefore $L_3,L,L_2$ lie on the same side of $L_1$. Similarly $L_1,K,L_2$ lie on the same side of $L_3$. Next $L_1,L_2$ lie on opposite sides of line $L$. Therefore $L_1$ and point $o$ lie on the same side of $L_2$. Similarly $L_3,L_2$ lie on opposite sides of line $K$ and $L_3$ and point $o$ lie on the same side of $L_2$. Hence $L_1,L_3$ on the same side of $L_2$.

To summarize this part of the proof:

  • $L_1,L_2$ lie on the same side of $L_3$.
  • $L_1,L_3$ lie on the same side of $L_2$.
  • $L_2,L_3$ lie on the same side of $L_1$.

These three properties will make it impossible for $L_1,L_2,L_3$ to have a common tranversal. Suppose that there exists a line tranversal to $L_1,L_2,L_3$ (RAA hypothesis). It cuts lines $L_1,L_2,L_3$ in points $a,b,c$ respectively. Points $a,b,c$ are collinear and distinct. By one of the axioms $$B(abc)\vee B(bac) \vee B(acb)$$ where $B$ is the betweenness relation (one of the primitive notions). Assumptions are symmetric so assume without loss of generality that $B(abc)$. It means (from definition of halfplane) that $a$ and $c$ lie on opposite sides of line $L_2$ and we have $a\in L_1, c\in L_3$. It's a contradiction because $L_1,L_3$ lie on the same side of $L_2$.

  • $\begingroup$ Will have to read this carefully when I go back home. Thanks. $\endgroup$ – Akerbeltz May 7 at 16:40
  • $\begingroup$ Ok, this proof is pretty good. Just a comment on the lemma you gave; it can be proved with the Hilbert plane axioms and the additional axiom $(L)$ that ensures the existence of limiting parallels (cf. Hartshorne, page 374, 378 prop 40.6). However, I have found no other reference that ensures the validity of this statement just in a Hilbert plane where it is granted that the parallel postulate fails. The axiom $(L)$ is, according to Hartstong, a pretty strong assumption, so it could be possible to prove the lemma you gave under weaker hypothesis. $\endgroup$ – Akerbeltz May 7 at 21:58
  • $\begingroup$ However, I don't think this statement follows naturally from the denial of the parallel postulate. Good question for another post, I guess. $\endgroup$ – Akerbeltz May 7 at 22:09
  • $\begingroup$ Isn't Poincare disc a model for all axioms of hyperbolic geometry? I'm pretty sure the lemma holds in this model (as well as axiom $(L)$ for example). $\endgroup$ – Kulisty May 7 at 22:39
  • $\begingroup$ Btw, what do you exactly mean by Hilbert plane? Model for all Hilbert axioms except parallel and continuity axioms? $\endgroup$ – Kulisty May 7 at 22:42

We can show that parallelism implies (ii) in our context:

First note that any three non-collinear points lie in the interior of a triangle. Indeed, if $A,B,C$ are the three points, find $A',B', C'$ such that $A$ between $C$ and $B'$, $B$ between $A$ and $C'$, $C$ between $B$ and $A'$. Apply Pasch's axiom to triangle $A'BC$ and line $AC$ (that cannot pass through $B$ or $C'$ because $C\notin AB=AC'$, and cannot pass through $A'$ because $A\notin BC=BA'$). As $AC$ cannot intersect the line segment $BC'$, we obtain a point $P\in AC$ such that $P$ is between $A'$ and $C'$. Then $A$ is between $B'$ and $P$, i.e., in the interior of $A'B'C$. By symmetry, the same holds for $B$ and $C$.

Now consider four points $A,B,C,D$ with no three of them collinear.

Case 1: $AB$ intersects $CD$ in a point $P$ such that $B$ is between $A,P$, and $D$ is between $B,C$. Then $A,,C,D$ are boundary points of triangle $ABP$ and hence interior of any triangle having $A,B,P$ as interior points

Case 2: $AB$ intersects $CD$ in a point $P$ such that $P$ is between $A,B$ and $D$ is between $C,P$. Then $D$ is an interior point of $ABC$, hence any triangle containing $A,B,C$ as interior points has $A,B,C,D$ as interior points.

Case 3: $AB\|CD$ and $AC\|BD$. Pick $A'$ with $A$ between $A',B$. Then a triangle cointaining $A',B,C,D$ also contains $A,B,C,D$. From $A'D\ne AD$ and parallelism, $A'D\not\|BC$. Let $P=A'D\cap BC$. As $BC$ does not intersect the line segments $AD$ (assumed parallel) and $A'A$ (intersection $B$ is not between $A',A$), we conclude from Pasch's axiom applied to $\Delta A'AD$ and line $BC$ that $BC$ does not intersect line segment $A'D$, i.e., $P$ is not between $A',D$. Then after relabelling, $A',B,C,D$ falls under either case 1 or case 2 above.

Case 4: We have interior intersection between opposing sides, i.e, wlog. $P$ between $A,B$ and between $C,D$. Then $AD$ does not intersect line segments $PB$ and $PC$, hence by Pasch's axiom does not intersect line segment $BC$. If $AD$ intersects $BC$ not between $b$ and $C$, we are (after relabelling) in case 1 or in case 2. Hence we may assume that $AD\|BC$. By the same argument, we may assume $AC\|BD$ and are in case 3.

We can also see that without parallelism, $(ii)$ may fail: Just consider the Poincaré disk and four points $r,ri,-r,-ri$, where $r=1-\epsilon$. Then any hyperbolic line outside these four points is an euclidean circular arc with end points at most slightly more than $90^\circ$ apart, hence three such arcs will not suffice to cover the disk boundary in their interiors, i.e., our points are not contained in a triangle.

  • $\begingroup$ How does case 1 make any sense? If $B\ast D\ast C$, then $B$ is in the line $DC$ (B1). But $B$ is in the line $AB$, so $B=P$ (I1). But by hypothesis, $A\ast B\ast P$, which contradicts (B3). $\endgroup$ – Akerbeltz May 13 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.