[proofplan] Start with a Steiner triple system $(X,\mathcal{B})$ of order $v$ and replace each point $x \in X$ by two labelled copies $(x,0)$ and $(x,1)$, then add one new point $\infty$. The new blocks are of two kinds: vertical blocks joining $\infty$ to the two copies of one old point, and lifted blocks obtained from each old block by choosing second coordinates in $\mathbb{F}_2$ whose sum is zero. We prove that every pair of new points lies in exactly one new block by separating the possibilities: pairs involving $\infty$, pairs with equal first coordinate, and pairs with distinct first coordinates. [/proofplan]

[step:Construct the doubled point set and the proposed block family]Let $(X,\mathcal{B})$ be a [Steiner triple system](/page/Steiner%20Triple%20System) of order $v$. Thus $|X|=v$, every block $B \in \mathcal{B}$ is a $3$-element subset of $X$, and every $2$-element subset of $X$ is contained in exactly one block of $\mathcal{B}$. Let $\mathbb{F}_2=\{0,1\}$ denote the field with addition taken modulo $2$. Define the new point set \begin{align*} Y := (X \times \mathbb{F}_2) \cup \{\infty\}, \end{align*} where $\infty$ is a point not contained in $X \times \mathbb{F}_2$. Since \begin{align*} |X \times \mathbb{F}_2|=2|X|=2v, \end{align*} we have $|Y|=2v+1$. For each point $x \in X$, define the vertical block \begin{align*} V_x := \{\infty,(x,0),(x,1)\}. \end{align*} For each old block $B=\{a,b,c\}\in\mathcal{B}$ and each triple $(i,j,k)\in\mathbb{F}_2^3$ satisfying $i+j+k=0$, define the lifted block \begin{align*} L_{B,i,j,k} := \{(a,i),(b,j),(c,k)\}. \end{align*} Finally define \begin{align*} \mathcal{C} := \{V_x : x \in X\} \cup \{L_{B,i,j,k} : B=\{a,b,c\}\in\mathcal{B},\ i,j,k\in\mathbb{F}_2,\ i+j+k=0\}. \end{align*} Every member of $\mathcal{C}$ has three distinct points: this is immediate for $V_x$, and for $L_{B,i,j,k}$ the first coordinates $a,b,c$ are distinct because $B$ is a block of the Steiner triple system $(X,\mathcal{B})$.[/step]

[guided]We begin with a [Steiner triple system](/page/Steiner%20Triple%20System) $(X,\mathcal{B})$ of order $v$. This means that $X$ is a set with $|X|=v$, each block $B \in \mathcal{B}$ is a $3$-element subset of $X$, and every $2$-element subset of $X$ occurs in exactly one block of $\mathcal{B}$. Let $\mathbb{F}_2=\{0,1\}$ be the field with addition modulo $2$. The construction replaces each old point $x \in X$ by two labelled points $(x,0)$ and $(x,1)$, and then adds one new point $\infty$ which is not in $X \times \mathbb{F}_2$. Thus we define \begin{align*} Y := (X \times \mathbb{F}_2) \cup \{\infty\}. \end{align*} Since \begin{align*} |X \times \mathbb{F}_2|=2|X|=2v, \end{align*} the new point set has cardinality $|Y|=2v+1$. There are two kinds of blocks. For each $x \in X$, define the vertical block \begin{align*} V_x := \{\infty,(x,0),(x,1)\}. \end{align*} This block ties together the two copies of $x$ with the new point. For each old block $B=\{a,b,c\}\in\mathcal{B}$ and each triple $(i,j,k)\in\mathbb{F}_2^3$ satisfying $i+j+k=0$, define the lifted block \begin{align*} L_{B,i,j,k} := \{(a,i),(b,j),(c,k)\}. \end{align*} The condition $i+j+k=0$ is the parity rule that makes the doubling construction work. Finally set \begin{align*} \mathcal{C} := \{V_x : x \in X\} \cup \{L_{B,i,j,k} : B=\{a,b,c\}\in\mathcal{B},\ i,j,k\in\mathbb{F}_2,\ i+j+k=0\}. \end{align*} Every member of $\mathcal{C}$ has three distinct points. For $V_x$, the points $\infty$, $(x,0)$, and $(x,1)$ are distinct because $\infty \notin X\times\mathbb{F}_2$ and $0\neq 1$ in $\mathbb{F}_2$. For $L_{B,i,j,k}$, the old block $B=\{a,b,c\}$ has three distinct elements, so the first coordinates of $(a,i)$, $(b,j)$, and $(c,k)$ are distinct; hence these three ordered pairs are distinct.[/guided]

[step:Cover every pair involving the new point or two copies of the same old point]Let $P \subset Y$ be a $2$-element subset. First suppose $\infty \in P$. Then there is a unique point $(x,i)\in X\times\mathbb{F}_2$ such that \begin{align*} P=\{\infty,(x,i)\}. \end{align*} The block $V_x=\{\infty,(x,0),(x,1)\}$ contains $P$. If $V_y$ is another vertical block containing $P$, then $(x,i)\in V_y$, so $y=x$. No lifted block contains $\infty$ by construction. Hence $P$ is contained in exactly one block of $\mathcal{C}$. Next suppose $\infty \notin P$ and the two points of $P$ have the same first coordinate. Since $P$ has two distinct elements, there is a unique $x\in X$ such that \begin{align*} P=\{(x,0),(x,1)\}. \end{align*} Again $V_x$ contains $P$. No lifted block contains two points with the same first coordinate, because each lifted block is built from an old block $\{a,b,c\}$ whose elements are distinct. Also $V_y$ contains both $(x,0)$ and $(x,1)$ only when $y=x$. Therefore $P$ is contained in exactly one block of $\mathcal{C}$.[/step]

[guided]Let $P \subset Y$ be a $2$-element subset. We first handle all pairs whose uniqueness is forced by the vertical blocks. Suppose $\infty \in P$. Since $P$ has two elements and the only points of $Y$ other than $\infty$ are ordered pairs in $X\times\mathbb{F}_2$, there are unique $x\in X$ and $i\in\mathbb{F}_2$ such that \begin{align*} P=\{\infty,(x,i)\}. \end{align*} The vertical block \begin{align*} V_x=\{\infty,(x,0),(x,1)\} \end{align*} contains this pair. If another vertical block $V_y$ contained $P$, then it would contain $(x,i)$, which forces $y=x$ from the definition of $V_y$. A lifted block contains only points of $X\times\mathbb{F}_2$ and never contains $\infty$, so no lifted block contains $P$. Hence $P$ is contained in exactly one block of $\mathcal{C}$. Now suppose $\infty \notin P$ and the two points of $P$ have the same first coordinate. Because the two elements of $P$ are distinct and the only two elements of $\mathbb{F}_2$ are $0$ and $1$, there is a unique $x\in X$ such that \begin{align*} P=\{(x,0),(x,1)\}. \end{align*} Again $V_x$ contains $P$. No lifted block can contain two points with the same first coordinate, because a lifted block comes from an old block $\{a,b,c\}$ whose entries are distinct. If a vertical block $V_y$ contains both $(x,0)$ and $(x,1)$, then its first coordinate must be $x$, so $y=x$. Therefore this pair is also contained in exactly one block of $\mathcal{C}$.[/guided]

[step:Use the old Steiner triple system to cover pairs with distinct first coordinates]Now suppose \begin{align*} P=\{(x,i),(y,j)\} \end{align*} where $x,y\in X$, $x\neq y$, and $i,j\in\mathbb{F}_2$. Since $(X,\mathcal{B})$ is a Steiner triple system, the pair $\{x,y\}$ is contained in a unique old block $B\in\mathcal{B}$. Write \begin{align*} B=\{x,y,z\}, \end{align*} where $z\in X$ is the unique third point of that block. Define \begin{align*} k := i+j \in \mathbb{F}_2. \end{align*} Because addition in $\mathbb{F}_2$ has characteristic $2$, this is exactly the unique element satisfying \begin{align*} i+j+k=0. \end{align*} Thus the lifted block \begin{align*} \{(x,i),(y,j),(z,k)\} \end{align*} belongs to $\mathcal{C}$ and contains $P$. To prove uniqueness, let $C\in\mathcal{C}$ be any block containing $P$. The block $C$ cannot be vertical, because a vertical block contains only one first coordinate from $X$, while $P$ has distinct first coordinates $x$ and $y$. Therefore $C$ is a lifted block coming from some old block $B'\in\mathcal{B}$ containing both $x$ and $y$. By uniqueness in the old Steiner triple system, \begin{align*} B'=B=\{x,y,z\}. \end{align*} Once $B$ is fixed, the second coordinate of the third point is forced by the equation $i+j+k=0$ in $\mathbb{F}_2$, so it must be $k=i+j$. Hence $C=\{(x,i),(y,j),(z,k)\}$, and $P$ is contained in exactly one block of $\mathcal{C}$.[/step]

[guided]We now handle the only non-vertical kind of pair: two copied points whose first coordinates are different. Let \begin{align*} P=\{(x,i),(y,j)\}, \end{align*} where $x,y\in X$, $x\neq y$, and $i,j\in\mathbb{F}_2$. The original system $(X,\mathcal{B})$ is a Steiner triple system, so the old pair $\{x,y\}$ lies in exactly one old block. Denote that block by \begin{align*} B=\{x,y,z\}, \end{align*} where $z\in X$ is the third point of the block. The construction tells us that lifted blocks over $B$ are obtained by assigning second coordinates whose sum in $\mathbb{F}_2$ is zero. Since the first two second coordinates are already fixed as $i$ and $j$, the third one must be the unique element $k\in\mathbb{F}_2$ satisfying \begin{align*} i+j+k=0. \end{align*} In $\mathbb{F}_2$, every element is its own additive inverse, so this element is \begin{align*} k=i+j. \end{align*} Therefore \begin{align*} \{(x,i),(y,j),(z,k)\} \end{align*} is one of the lifted blocks in $\mathcal{C}$, and it contains the pair $P$. It remains to prove that no other block can contain $P$. A vertical block has the form \begin{align*} V_t=\{\infty,(t,0),(t,1)\} \end{align*} for some $t\in X$, so all non-$\infty$ points in a vertical block have the same first coordinate $t$. Since $P$ contains points with distinct first coordinates $x$ and $y$, no vertical block contains $P$. Thus any block containing $P$ must be lifted from an old block $B'\in\mathcal{B}$. For such a lifted block to contain both $(x,i)$ and $(y,j)$, the old block $B'$ must contain both old points $x$ and $y$. The Steiner property of $(X,\mathcal{B})$ says that the old pair $\{x,y\}$ lies in exactly one old block, so \begin{align*} B'=B=\{x,y,z\}. \end{align*} With the old block fixed, the third second coordinate is also fixed, because the equation \begin{align*} i+j+k=0 \end{align*} has the unique solution $k=i+j$ in $\mathbb{F}_2$. Hence the only possible block containing $P$ is \begin{align*} \{(x,i),(y,j),(z,i+j)\}. \end{align*} So $P$ is contained in exactly one block of $\mathcal{C}$.[/guided]

[step:Conclude that the constructed incidence structure is a Steiner triple system]The preceding cases cover every $2$-element subset $P\subset Y$: either $P$ contains $\infty$, or it consists of two points of $X\times\mathbb{F}_2$ with equal first coordinate, or it consists of two points of $X\times\mathbb{F}_2$ with distinct first coordinates. In each case, $P$ is contained in exactly one block of $\mathcal{C}$. Since every block in $\mathcal{C}$ has cardinality $3$, the incidence structure $(Y,\mathcal{C})$ is a Steiner triple system. Since $|Y|=2v+1$, this is an $\operatorname{STS}(2v+1)$. Therefore the existence of an $\operatorname{STS}(v)$ implies the existence of an $\operatorname{STS}(2v+1)$.[/step]

[guided]The cases above exhaust all possible $2$-element subsets $P\subset Y$. Indeed, either $P$ contains the new point $\infty$, or both points lie in $X\times\mathbb{F}_2$. In the latter situation, the two first coordinates are either equal or distinct. The previous steps proved, in each of these mutually exclusive cases, that $P$ is contained in exactly one block of $\mathcal{C}$. We have also proved that every block in $\mathcal{C}$ has exactly three distinct points. Therefore $(Y,\mathcal{C})$ satisfies the defining property of a [Steiner triple system](/page/Steiner%20Triple%20System): every $2$-element subset of the point set occurs in exactly one $3$-element block. Since $|Y|=2v+1$, the constructed system is an $\operatorname{STS}(2v+1)$. Hence the existence of an $\operatorname{STS}(v)$ implies the existence of an $\operatorname{STS}(2v+1)$.[/guided]