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For an undirected finite graph, its forest polytope is the matroid independence polytope of its graphic matroid. It has the exact inequality descriptionsHere $\kappa (V,A)$ counts all connected components of a graph, including isolated vertices, and $E(S)$ denotes the edges internal to $S$. A maximal forest has one fewer edge than vertices in each component, giving the rank $|V|-\kappa (V,A)$. The matroid independence polytope theorem therefore proves the first description. It implies the second since the rank of $E(S)$ is at most $|S|-1$. Conversely, split any $A$ into its vertex components $S_j$; nonnegativity gives $x(A)\le {\sum }_{j}x(E(S_j))\le {\sum }_j(|S_j|-1)=|V|-\kappa (V,A)$. No spanning requirement is imposed, so an optimum with negative edge weights can leave vertices isolated.

For $A\subseteq E$, write $x(A)={\sum }_{e\in A}{x}_e$, and let $\kappa (A)$ count the connected components of $(V,A)$, including isolated vertices. A maximal forest in each component has one fewer edge than vertices, so $r(A)=|V|-\kappa (A)$. The required forest polytope isWe prove both inclusions, using the matroid independence polytope argument explicitly.

If $I$ is a forest, $I\cap A$ is an independent subset of $A$, so $1_I(A)=|I\cap A|\le r(A)$. Its indicator is nonnegative. All convex combinations of these indicators therefore lie in $P$.

For the converse, take any real objective vector $c$. List just its positive-weight edges as $e_1,\dots ,e_k$ with $c_1\ge \cdots \ge c_k>0$, put $c_{k+1}=0$, and set $S_i=\{e_1,\dots ,e_i\}$. Run the forest greedy algorithm on these edges and call the output $I$. At every prefix, the accepted edges form a maximal forest of $(V,S_i)$, soIndeed a previously rejected edge still has its endpoints joined later; therefore no rejected prefix edge can augment the chosen forest. A maximal forest spans every prefix component and has the stated size.

For every $x\in P$, dropping nonpositive objective terms and telescoping now givesThe last equality uses the same telescoping identity for the greedy indicator. If there are no positive coefficients, the empty forest attains the bound zero. Thus every linear objective over $P$ is maximized by a forest indicator.

The singleton inequalities give $0\le x_e\le r(\{e\})\le 1$, so $P$ is compact. The convex hull $Q$ of the finitely many forest indicators is also compact. If some $x\in P$ were outside $Q$, the finite-dimensional separating hyperplane theorem would give $c$ with $c^{T}x>{\max }_{y\in Q}{c}^{T}y$, contradicting the bound just proved. Therefore $P\subseteq Q$, completing the convex-hull proof.

An equivalent system indexed by vertex sets iswhere $E(S)$ contains all edges with both ends in $S$. The rank system implies this since $r(E(S))\le |S|-1$. Conversely, split any edge set $A$ into its vertex components $S_1,\dots ,S_t$. Nonnegativity and the vertex inequalities give $x(A)\le {\sum }_{j}x(E(S_j))\le {\sum }_j(|S_j|-1)=|V|-\kappa (A)$. Hence this second system describes the same polytope.