Can a line in a projective plane have only one point?

Question

Here is the definition of a projective plane from Stillwell's The Four Pillars of Geometry (2005):

Let $\cal P$ ("points") be a set, and let $\cal L$ ("lines") be a set of subsets of $\cal P.$ We say $({\cal P},{\cal L})$ a projective plane to mean

1. Any two distinct points lie on exactly one line;
2. Any two distinct lines intersect at exactly one point; and
3. There exist four points with the property that no three of these points are collinear.

Can a line have exactly one element?

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Below, we show that lines must have at least one element.

As notation, for any two distinct points $a$ and $b,$ let $ab$ be the line containing $a$ and $b.$ (Existence and uniqueness guaranteed by Axiom 1.)

Theorem. $6\le|\cal L|.$

Proof. Choose points $a,b,c,d$ such that no three of $a,b,c,d$ are collinear, and note that $a,b,c,d$ are distinct. Now, it suffices to show $ab,ac,ad,bc,bd,cd$ are distinct. To this end, let $S=\{a,b,c,d\},$ let $e,f\in S$ such that $e\ne f,$ and let $g,h\in S$ such that $g\ne h$ and $\{e,f\}\ne\{g,h\}.$

Note $\{e,f\}\cap\{g,h\}\subseteq\{e,f\}$ and $|\{e,f\}|=2=|\{g,h\}|.$ Thus, if $|\{e,f\}\cap\{g,h\}|=2$ then $\{e,f\}\cap\{g,h\}=\{e,f\},$ so $\{e,f\}\subseteq\{g,h\},$ so $\{e,f\}=\{g,h\},$ contradiction; thus $|\{e,f\}\cap\{g,h\}|<2.$ Thus, from the principle of inclusion-exclusion that $|\{e,f\}\cup\{g,h\}|=2+2-|\{e,f\}\cap\{g,h\}|>2,$ so $3\le|\{e,f\}\cup\{g,h\}|,$ so we can choose $T\subseteq\{e,f\}\cup\{g,h\}$ such that $|T|=3.$

Now, assume for the sake of contradiction that $ef=gh.$ Then the three points of $T$ are collinear, contradicting our choice of $a,b,c.$ Thus $ef\ne gh.$ The desired result now follows.$\square$

Theorem. Every line has cardinality at least 1.

Proof. Let $L\in\cal L,$ and choose a line $M\ne L.$ (Existence guaranteed by previous Theorem.) From Axiom 2, we have $1=|L\cap M|\le|L|.\square$

Answer 1

Here is the solution suggested by @Steven :

No. Indeed, assume there exists a line, call it $L,$ with cardinality 1, and let $p=\cup L$ (thus $L=\{p\}$). Then, choose points $a,b,c,d$ such that no three of $a,b,c,d$ are collinear.

Applying the existence portion of Axiom 2 to $L,ab$ and $L,ac$ yields $p\in ab,p\in ac.$ Then, applying the uniqueness portion of Axiom 2 to $ab,ac$ yields $p=a.$

Applying the existence portion of Axiom 2 to $L,bd$ and $L,cd$ yields $p\in bd,p\in cd.$ Then, applying the uniqueness portion of Axiom 2 to $bd,cd$ yields $p=d.$

Thus $a=d,$ so $a,b,d$ are collinear, contradiction.