[proofplan] For the forward direction, we build the projective completion explicitly from the affine plane by adding one point for each parallel class and one line at infinity. We then verify the projective incidence axioms by splitting pairs of points and pairs of lines into old-old, old-new, and new-new cases. For the converse direction, we remove a projective line $m$ and restrict all other lines to the remaining points; the projective joining and intersection axioms give the affine joining and parallel axioms, while the order counts come from deleting exactly one point from each projective line not equal to $m$. [/proofplan]

[step:Construct the completion from the affine parallel classes] Let $\mathcal A = (P, \mathcal L, I)$ be an [affine plane](/page/Affine%20Plane) of order $n$. Let $\Pi$ denote the set of parallel classes of lines in $\mathcal L$, where two affine lines are in the same class if they are equal or disjoint. For each $C \in \Pi$, choose a new point $p_C$ with $p_C \notin P$, and choose a new line $\ell_\infty$ with $\ell_\infty \notin \mathcal L$. Define the completed point set by \begin{align*} P^* &= P \cup \{p_C : C \in \Pi\}. \end{align*} Define the completed line set by \begin{align*} \mathcal L^* &= \mathcal L \cup \{\ell_\infty\}. \end{align*} Define the completed incidence relation $I^*$ as follows. If $x \in P$ and $\ell \in \mathcal L$, then $x I^* \ell$ exactly when $x I \ell$. If $C \in \Pi$ and $\ell \in \mathcal L$, then $p_C I^* \ell$ exactly when $\ell \in C$. Finally, $p_C I^* \ell_\infty$ for every $C \in \Pi$, and no point of $P$ is incident with $\ell_\infty$. Every affine line belongs to exactly one parallel class, so each old line $\ell \in \mathcal L$ receives exactly one new point, namely $p_C$ where $C$ is the parallel class containing $\ell$. Since $\ell$ has exactly $n$ old points, it has exactly $n+1$ points in the completed structure. The affine plane of order $n$ has exactly $n+1$ parallel classes, because through any fixed point there is exactly one line in each parallel class and exactly $n+1$ lines through the point. Hence $\ell_\infty$ has exactly $n+1$ points. [/step]

[step:Verify that any two completed points determine a unique completed line]Let $a,b \in P^*$ be distinct points. If $a,b \in P$, then the affine plane axiom gives a unique old line $\ell \in \mathcal L$ incident with both $a$ and $b$. The new line $\ell_\infty$ is incident with no old point, so $\ell$ is also the unique line in $\mathcal L^*$ incident with both. If $a \in P$ and $b = p_C$ for some $C \in \Pi$, then the affine parallel axiom gives a unique line $\ell \in C$ incident with $a$. By construction, $p_C I^* \ell$, so $\ell$ is incident with both $a$ and $p_C$. No other old line in $C$ passes through $a$, and no old line outside $C$ is incident with $p_C$. Also $\ell_\infty$ is not incident with $a$. Therefore $\ell$ is unique. The same argument covers the case $a = p_C$ and $b \in P$. If $a = p_C$ and $b = p_D$ with $C,D \in \Pi$ and $C \neq D$, then both points are incident with $\ell_\infty$. An old line belongs to only one parallel class, so no old line is incident with both $p_C$ and $p_D$. Thus $\ell_\infty$ is the unique line incident with both points.[/step]

[guided]We must check the projective joining axiom in every possible type of pair, because the completed point set has two kinds of points: old affine points and new points at infinity. First suppose $a,b \in P$ are distinct old points. Since $\mathcal A$ is an affine plane, there is a unique affine line $\ell \in \mathcal L$ with $a I \ell$ and $b I \ell$. The completed incidence relation agrees with the old incidence relation on $P \times \mathcal L$, so $a I^* \ell$ and $b I^* \ell$. The only new line is $\ell_\infty$, and by definition no old point is incident with $\ell_\infty$. Hence no new line can create another line through $a$ and $b$, and the old affine line remains unique. Next suppose $a \in P$ and $b = p_C$ for a parallel class $C \in \Pi$. The affine parallel axiom says that through the point $a$ there is a unique line parallel to any representative of $C$; equivalently, there is a unique line $\ell \in C$ incident with $a$. Since $\ell \in C$, the definition of $I^*$ makes $p_C$ incident with $\ell$. Thus $\ell$ joins $a$ and $p_C$. If another old line $\ell' \in \mathcal L$ also joined them, then $p_C I^* \ell'$ would force $\ell' \in C$, and $a I^* \ell'$ would say that $\ell'$ is another line of $C$ through $a$, contradicting uniqueness in the affine parallel axiom. The line $\ell_\infty$ cannot join them because it is not incident with $a$. Finally suppose $a = p_C$ and $b = p_D$ with $C \neq D$. Both points lie on $\ell_\infty$ by definition. If an old affine line $\ell \in \mathcal L$ contained both, then $\ell \in C$ and $\ell \in D$, which is impossible because parallel classes partition $\mathcal L$. Therefore no old line joins them, and $\ell_\infty$ is the unique completed line through the two new points.[/guided]

[step:Verify that any two completed lines meet in a unique completed point]Let $\ell,k \in \mathcal L^*$ be distinct lines. If $\ell,k \in \mathcal L$ and they meet in the affine plane, then their affine intersection point is the unique completed point incident with both, because distinct intersecting affine lines are not parallel and therefore do not receive the same point at infinity. If $\ell,k \in \mathcal L$ and they are disjoint in the affine plane, then they lie in the same parallel class $C \in \Pi$. By construction, both are incident with $p_C$. They have no old common point because they are disjoint, and they cannot share any other new point because each old line belongs to exactly one parallel class. Thus their unique completed intersection point is $p_C$. If $\ell \in \mathcal L$ and $k = \ell_\infty$, let $C \in \Pi$ be the parallel class containing $\ell$. Then $\ell$ and $\ell_\infty$ both contain $p_C$. No old point lies on $\ell_\infty$, and no point $p_D$ with $D \neq C$ lies on $\ell$. Hence $p_C$ is the unique completed intersection point.[/step]

[guided]We verify the projective line-intersection axiom by separating the same three cases according to whether the completed lines are old affine lines or the new line at infinity. First let $\ell,k \in \mathcal L$ be distinct affine lines that meet in the affine plane. Let $x \in P$ be their unique affine intersection point. Since $I^*$ agrees with $I$ on $P \times \mathcal L$, the point $x$ is incident with both $\ell$ and $k$ in the completed structure. Because the two affine lines meet, they are not parallel and therefore belong to distinct parallel classes. Hence they receive distinct points at infinity, so no new point is incident with both. Thus $x$ is the unique completed point common to $\ell$ and $k$. Next let $\ell,k \in \mathcal L$ be distinct affine lines that are disjoint in the affine plane. By the definition of affine parallel classes, they lie in one parallel class $C \in \Pi$. The construction makes $p_C$ incident with every affine line in $C$, so $p_C$ is incident with both $\ell$ and $k$. There is no old common point because the affine lines are disjoint. There is no second new common point because each affine line belongs to exactly one parallel class. Therefore $p_C$ is their unique completed intersection point. Finally let $\ell \in \mathcal L$ and $k = \ell_\infty$. Let $C \in \Pi$ be the unique parallel class containing $\ell$. The point $p_C$ lies on $\ell$ by construction and lies on $\ell_\infty$ by the definition of the line at infinity. No old point lies on $\ell_\infty$, and if $D \neq C$, then $p_D$ is not incident with $\ell$ because $\ell \notin D$. Hence $p_C$ is the unique completed intersection point of $\ell$ and $\ell_\infty$.[/guided]

[step:Check the order and nondegeneracy of the completed plane] We have shown that any two distinct points determine a unique line and any two distinct lines meet in a unique point. Each completed line has exactly $n+1$ points: old lines have their $n$ affine points plus one point at infinity, and $\ell_\infty$ has one point for each of the $n+1$ parallel classes. The completed plane is nondegenerate. Choose an affine line $\ell \in \mathcal L$, two distinct old points $a,b \in P$ on $\ell$, and an old point $c \in P$ not on $\ell$, which exists because an affine plane of order $n \geq 2$ has more than one line and not all points lie on a single line. Then $a,b,c$ are not collinear in the completed structure. Hence the completed incidence structure is a projective plane of order $n$. [/step]

[step:Delete a projective line and define affine parallelism] Conversely, let $\mathcal P = (P,\mathcal L,I)$ be a [projective plane](/page/Projective%20Plane) of order $n$, and fix a line $m \in \mathcal L$. Define the retained point set by \begin{align*} P_0 &= \{x \in P : x \text{ is not incident with } m\}. \end{align*} Define the retained line set by \begin{align*} \mathcal L_0 &= \mathcal L \setminus \{m\}. \end{align*} The restricted incidence relation $I_0$ is defined by $x I_0 \ell$ exactly when $x \in P_0$, $\ell \in \mathcal L_0$, and $x I \ell$. For distinct lines $\ell,k \in \mathcal L_0$, define $\ell \parallel k$ if their unique projective intersection point lies on $m$; also declare $\ell \parallel \ell$ for every $\ell \in \mathcal L_0$. This relation is the affine parallelism induced by the deleted line. Each line $\ell \in \mathcal L_0$ meets $m$ in a unique point, because distinct projective lines meet in a unique point. Since every projective line has $n+1$ points, the restricted affine line $\ell \cap P_0$ has exactly $n$ points. Thus every affine line in the deleted structure has order $n$. [/step]

[step:Verify the affine incidence and parallel axioms after deletion]Let $a,b \in P_0$ be distinct points. In the projective plane there is a unique line $\ell \in \mathcal L$ incident with both $a$ and $b$. This line cannot be $m$, since neither $a$ nor $b$ lies on $m$. Hence $\ell \in \mathcal L_0$, and it is the unique affine line through $a$ and $b$. Let $\ell \in \mathcal L_0$ and let $a \in P_0$ be a point not incident with $\ell$. Let $q$ be the unique projective intersection point of $\ell$ and $m$. The projective joining axiom gives a unique projective line $k$ incident with $a$ and $q$. Since $a \notin m$, this line is not $m$, so $k \in \mathcal L_0$. Since $q \in m$, the projective intersection point of $k$ and $\ell$ lies on $m$, so $k \parallel \ell$ in the deleted structure. If $k' \in \mathcal L_0$ is another affine line through $a$ parallel to $\ell$, then the projective intersection point of $k'$ and $\ell$ lies on $m$. Because $\ell$ meets $m$ only at $q$, that intersection point is $q$. Thus $k'$ is a projective line through both $a$ and $q$. By uniqueness of the projective line through two distinct points, $k' = k$. Therefore through every affine point outside a line there is a unique parallel to that line. Finally, if distinct retained lines $\ell,k \in \mathcal L_0$ are not parallel, then their unique projective intersection point is not incident with $m$ by the definition of $\parallel$. Hence that intersection point lies in $P_0$, so $\ell$ and $k$ meet in a retained affine point. This verifies the affine intersection alternative for nonparallel lines.[/step]

[guided]The deleted line $m$ plays exactly the role of the line at infinity. To prove the affine joining axiom, take two retained points $a,b \in P_0$. The projective plane gives a unique projective line $\ell$ through them. This line cannot be the deleted line $m$, because points of $P_0$ were defined precisely as points not incident with $m$. Therefore $\ell$ survives as an element of $\mathcal L_0$, and it remains unique after deletion. Now fix an affine line $\ell \in \mathcal L_0$ and an affine point $a \in P_0$ not on $\ell$. We need to construct the unique affine line through $a$ parallel to $\ell$. Let $q$ be the unique point where $\ell$ meets the deleted line $m$. The point $q$ is the direction of $\ell$. The projective joining axiom gives a unique line $k$ through $a$ and $q$. Since $a$ is not on $m$, the line $k$ cannot equal $m$, so $k$ remains in $\mathcal L_0$. Its projective intersection with $\ell$ is $q$, which lies on $m$, so by the definition of the induced parallel relation, $k \parallel \ell$. For uniqueness, suppose $k' \in \mathcal L_0$ is another line through $a$ with $k' \parallel \ell$. Parallelism means that the projective intersection point of $k'$ and $\ell$ lies on $m$. But $\ell$ meets $m$ in the single point $q$, so $k'$ must pass through $q$. Thus both $k$ and $k'$ are projective lines through the two distinct points $a$ and $q$. The projective uniqueness axiom forces $k'=k$. This proves the affine parallel axiom. We also record the complementary intersection statement for retained lines. If distinct lines $\ell,k \in \mathcal L_0$ are not parallel, then by definition their unique projective intersection point does not lie on $m$. Therefore that point belongs to $P_0$, and the two retained lines meet in the deleted structure. Thus the induced relation has exactly the affine dichotomy: distinct lines either meet in a retained point or are parallel with their projective intersection on the deleted line.[/guided]

[step:Confirm noncollinearity and conclude the affine plane has order $n$]It remains to check that the deleted structure has non-collinear points. Since the projective plane has order $n \geq 2$, the line $m$ has $n+1 \geq 3$ points. Choose two distinct points $q_1,q_2 \in P$ incident with $m$. Choose a projective line $\ell_1 \in \mathcal L$ through $q_1$ with $\ell_1 \neq m$, and choose a projective line $\ell_2 \in \mathcal L$ through $q_2$ with $\ell_2 \neq m$. Such lines exist because every point of a projective plane of order $n$ is incident with $n+1 \geq 3$ lines. The lines $\ell_1$ and $\ell_2$ are distinct, since $\ell_1 \cap m = \{q_1\}$ and $\ell_2 \cap m = \{q_2\}$. Let $c \in P$ be the unique projective intersection point of $\ell_1$ and $\ell_2$. The point $c$ is not incident with $m$, because otherwise $c \in \ell_1 \cap m = \{q_1\}$ and $c \in \ell_2 \cap m = \{q_2\}$, contradicting $q_1 \neq q_2$. Hence $c \in P_0$. Choose $a \in P$ incident with $\ell_1$ such that $a$ is incident with neither $m$ nor equal to $c$, and choose $b \in P$ incident with $\ell_2$ such that $b$ is incident with neither $m$ nor equal to $c$. These choices are possible because each of $\ell_1$ and $\ell_2$ has $n+1 \geq 3$ points, and on each line we exclude exactly the point where it meets $m$ and the point $c$. Then $a,b,c \in P_0$, with $a \neq c$ and $b \neq c$; moreover $a \neq b$, since a common point of $\ell_1$ and $\ell_2$ must equal $c$. The points $a,b,c$ are not collinear in the deleted structure. If one retained line contained all three, then as a projective line it would contain $a$ and $c$, hence would equal $\ell_1$ by uniqueness of the projective line through two distinct points. The same retained line would contain $b$ and $c$, hence would equal $\ell_2$. This would force $\ell_1=\ell_2$, contradicting $q_1 \neq q_2$ and $\ell_i \cap m = \{q_i\}$ for $i=1,2$. Thus the deleted incidence structure satisfies the affine plane axioms. Since every remaining line has exactly $n$ points, it is an affine plane of order $n$. This completes both directions of the theorem.[/step]

[guided]We need three retained points that are not collinear. The only subtlety is that the intersection point of two chosen projective lines must not accidentally coincide with one of the other two retained points. Since the projective plane has order $n \geq 2$, the deleted line $m$ contains $n+1 \geq 3$ points. Choose two distinct points $q_1,q_2 \in P$ on $m$. Through each point of a projective plane of order $n$ there are $n+1 \geq 3$ lines, so we may choose a line $\ell_1 \neq m$ through $q_1$ and a line $\ell_2 \neq m$ through $q_2$. These two lines are distinct: if they were equal, then their common line would meet $m$ in both $q_1$ and $q_2$, contradicting the uniqueness of the intersection of two projective lines. Let $c$ be the unique point of intersection of $\ell_1$ and $\ell_2$. This point is retained. Indeed, if $c$ lay on $m$, then $c$ would belong to both $\ell_1 \cap m$ and $\ell_2 \cap m$. But $\ell_1 \cap m = \{q_1\}$ and $\ell_2 \cap m = \{q_2\}$, so this would force $q_1=q_2$, contrary to our choice. Hence $c \in P_0$. Now choose the other two retained points away from $c$. The line $\ell_1$ has $n+1 \geq 3$ points. On $\ell_1$ we exclude the point $q_1$ on $m$ and the point $c$, leaving at least one point; choose such a point and call it $a$. Similarly, on $\ell_2$ we exclude $q_2$ and $c$, and choose a remaining point $b$. Then $a,b,c$ are all in $P_0$, and they are pairwise distinct. The equality $a=b$ would put that common point on both $\ell_1$ and $\ell_2$, so it would have to be their unique intersection point $c$, contrary to the choice of $a$ and $b$. Finally, suppose a retained line contained $a,b,c$. Viewed as a projective line, it contains the distinct points $a$ and $c$, so by projective uniqueness it must be $\ell_1$. It also contains the distinct points $b$ and $c$, so it must be $\ell_2$. This would imply $\ell_1=\ell_2$, impossible because the two lines meet $m$ at the distinct points $q_1$ and $q_2$. Therefore $a,b,c$ are noncollinear retained points.[/guided]