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For fixed positive integers $r,t$, a graph $G$ on $n$ vertices with $e(G)=(1-1/r+o(1))n^2/2$ and no balanced complete multipartite blow-up $K_{r+1}(t)$ contains an $r$-partite subgraph $H$ with $n-o(n)$ vertices, balanced classes of size $n/r+o(n)$, and minimum degree of a graph $\delta (H)=(1-1/r+o(1))n$.

One proof first removes a vanishing fraction of vertices of low degree of a vertex, finds a slowly growing $K_r(L)$ by the Erdős-Stone theorem, and assigns almost every remaining vertex to a root class in which it has fewer than $t$ neighbours of a vertex. After discarding a further vanishing fraction, a $K_{t,t}$ inside an assigned class would extend to $K_{r+1}(t)$ using the root classes. Thus internal edges and missing cross-class edges both number $o(n^2)$. Removing the vertices with unusually many missing cross-class edges gives the stated minimum degree of a graph.

A topological graph is a CW complex with only zero-dimensional and one-dimensional cells: its vertices form the zero-skeleton, and each edge is a copy of an open interval attached by its two endpoints to vertices, possibly the same vertex. It has the weak topology of a CW complex. A subgraph is a CW subcomplex, obtained by choosing vertices and edges and retaining the endpoints of every chosen edge. Loops and multiple edges are allowed.

A spanning tree is a connected spanning subgraph containing no cycle in a graph. To construct one in any connected topological graph $\Gamma$, consider all acyclic spanning subgraphs, ordered by inclusion; the subgraph with every vertex and no edges is one. The union of a chain is still acyclic, since any cycle uses only finitely many edges and would already occur in a member of the chain. The Zorn lemma therefore gives a maximal such subgraph $\Delta$.

If $\Delta$ were disconnected, a finite edge path in $\Gamma$ between vertices of two of its components would contain an edge joining different components. Adding that edge creates no cycle, contradicting maximality. Thus $\Delta$ is a spanning tree. Fix a root vertex. Each point of $\Delta$ has a unique finite path to the root; give every edge length one and move the point a fraction $s$ of the way along that path at time $s\in [0,1]$. This defines a contraction of a topological space of $\Delta$. On each closed edge times $[0,1]$, the construction takes place in a finite tree and is continuous. Since the interval is a locally finite CW complex, the product with $\Delta$ has the corresponding weak topology of a CW complex, so these restrictions establish continuity globally. ConsequentlyThis argument includes infinite graphs and does not assume local finiteness of $\Gamma$.

If $p:\tilde{\Gamma }\to \Gamma$ is a covering map, take $p^{-1}({\Gamma }^0)$ as the vertices upstairs. Each component of the inverse image of an open edge is an open interval mapping homeomorphically to that edge, because an interval is simply connected. Its parametrization lifts along the closed edge to attach its endpoints to lifted vertices, by path lifting theorem. Small star neighborhoods at vertices are lifted homeomorphically by the covering map, and give exactly the neighborhoods of the resulting topological graph. Thus the cell structure has the topology of the original covering space. The total space may be disconnected, but every component is a topological graph.

The Nielsen–Schreier formula isFor a proof, realize $F_n$ as the fundamental group of the wedge sum of $n$ circles. Construct its connected covering graph with vertices the right cosets $Hg$, and an oriented edge labeled $x_i$ from $Hg$ to $Hg{x}_i$ for each free generator $x_i$. There is exactly one incoming and one outgoing edge of each label at every vertex, which establishes the local covering map property. Reading a word from $H$ ends at its coset, so the closed paths at $H$ give exactly the words in $H$. The induced map of fundamental groups is injective by the homotopy lifting property of a covering map, and its image is $H$.

For index $d$, this covering graph has $d$ vertices and $dn$ edges. A spanning tree in it has $d-1$ edges. The permitted rank formula for a finite connected graph now givesIn particular this proves both freeness and the index formula.

If $n\ge 2$ and $d>1$, then $1+d(n-1)>n$, so a proper finite-index subgroup cannot be isomorphic to $F_n$. For $n=1$, $F_1\cong \mathbb{Z}$, and $d\mathbb{Z}$ is a proper index-$d$ subgroup isomorphic to $\mathbb{Z}$ for every $d\ge 2$. Hence

Write $\alpha =1-1/r$. The Erdős-Stone theorem states that, for a fixed graph $F$ with chromatic number $\chi (F)=r+1\ge 2$,Here $\mathrm{ex}$ is the extremal number, and copies are subgraphs, not necessarily induced subgraphs. We write $K_a(b)$ for the balanced complete multipartite blow-up with $a$ classes of $b$ vertices each. All asymptotic errors below concern $n\to \infty$ with the forbidden graphs fixed.

The stability subgraph. We prove the high-minimum-degree multipartite stability subgraph assertion directly from the Erdős-Stone theorem. For $r=1$, the edgeless spanning subgraph suffices. Suppose $r\ge 2$.

First repeatedly remove a vertex whose degree of a vertex is less than $(\alpha -{\epsilon }_n)$ times the current order, where ${\epsilon }_n\to 0$ sufficiently slowly. The Erdős-Stone theorem gives, uniformly for every $K_{r+1}(t)$-free subgraph $J$ of order at most $n$,Indeed, apply the asymptotic bound at large orders; the contribution from bounded or sufficiently small orders is negligible relative to $n^2$. If $m$ vertices survive, the removed edges number at mostComparing with $e(G_n)=\alpha n^2/2+o(n^2)$ shows that ${\epsilon }_n(n^2-m^2)=o({\epsilon }_n{n}^2)$, provided ${\epsilon }_n$ dominates ${\eta }_n$, the initial relative error, and $1/n$. Consequently $m=n-o(n)$. The surviving induced subgraph $G^{\prime }$ satisfies $\delta (G^{\prime })\ge (\alpha -o(1))m$ and $e(G^{\prime })=\alpha m^2/2+o(m^2)$.

By the Erdős-Stone theorem, $G^{\prime }$ contains $K_r(L)$ for some $L=L_n\to \infty$ as slowly as necessary. This follows because $\alpha$ exceeds the extremal number density for each fixed $K_r(L)$ by the positive gap $1/(r(r-1))$. Choose $L$ so slowly that $rL=o(m)$ and ${\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$, and denote its root classes by $A_1,\dots ,A_r$.

An outside vertex having at least $t$ neighbours of a vertex in every $A_i$ determines one choice of a $t$-subset in each $A_i$. At most $t-1$ outside vertices can determine any fixed choice, since otherwise these vertices and the chosen root sets form $K_{r+1}(t)$. Thus at most $(t-1){\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$ outside vertices have this property. Exclude those vertices and the root classes; put each remaining vertex in a class $C_i$ for which it has fewer than $t$ neighbours of a vertex in $A_i$.

Write $\delta (G^{\prime })\ge (\alpha -{\gamma }_n)m$, with ${\gamma }_n\to 0$. Counting missing incidences with $A_i$ givesSince ${\sum }_i|C_i|=m-o(m)$, it follows that $|C_i|=m/r+o(m)$ for every $i$. The total missing incidences from all root vertices to assigned vertices are at most $(1+r{\gamma }_n)Lm$. At least $(L-t+1)(m-o(m))$ of them are incidences with a vertex's own assigned root class. Hence only $o(Lm)$ missing incidences go to other root classes.

Choose ${\rho }_n\to 0$ slowly and discard the $o(m)$ assigned vertices with more than ${\rho }_{n}L$ missing incidences to other root classes. Each remaining vertex of $C_i$ is adjacent to all but at most ${\rho }_{n}L$ vertices of every $A_j$, $j\ne i$. If its class contained a complete bipartite graph $K_{t,t}$, the $2t$ vertices of this copy would have at least $L-2t{\rho }_{n}L\ge t$ common neighbours of a vertex in each other root class. These sets extend the copy to $K_{r+1}(t)$, a contradiction. By the bipartite case of the Erdős-Stone theorem, the number of internal edges in each surviving class is $o(m^2)$.

The surviving graph still has $\alpha m^2/2+o(m^2)$ edges, whereas the total possible cross-class edges, with these balanced classes, is $\alpha m^2/2+o(m^2)$. Thus only $o(m^2)$ cross-class edges are missing. Discard the $o(m)$ vertices missing more than ${\sigma }_{n}m$ cross-class edges, with ${\sigma }_n\to 0$ slowly, and delete all internal edges. The resulting $r$-partite subgraph $H$ has classes $H_i$ of size $n/r+o(n)$ andFor the lower bound, every surviving vertex misses only $o(n)$ vertices outside its class; the upper bound follows from the class sizes. This also records the uniform cross-class error that we need next.

Minimum degree in an extremal graph. Fix $F$ with $\chi (F)=r+1$. Choose $t$ large enough that $F$ embeds in $K_{r+1}(t)$. An extremal $F$-free graph $G$ is therefore $K_{r+1}(t)$-free, and its edge count has the required asymptotics by the Erdős-Stone theorem. For $r\ge 2$ take $H$ as above.

For any vertex $v$, remove $v$ and introduce a new vertex $v^{\ast }$ whose vertex neighbourhood consists precisely of the vertices in $H_2\cup \cdots \cup H_r$ other than $v$. The new degree of a vertex is $\alpha n-o(n)$. This new graph remains $F$-free. Indeed, a new copy of $F$ must use $v^{\ast }$, and all its neighbours of a vertex in that copy lie in the other classes of $H$. There are only boundedly many of them. Each misses only $o(n)$ vertices in $H_1$, so an unused vertex $w\in H_1$ is adjacent to all of them. Replacing $v^{\ast }$ by $w$ would give $F$ in $G-v$, which is impossible.

Extremality now forces $d_G(v)\ge \alpha n-o(n)$, uniformly in $v$. The upper bound for the minimum degree of a graph follows from $\delta (G)\le 2e(G)/n$. For $r=1$, $e(G)=o(n^2)$ already gives $0\le \delta (G)=o(n)$. ThusThis is the minimum degree of an extremal forbidden-subgraph graph principle.

The exact extremal graph for disjoint cliques. DefineThe plus sign denotes the join of graphs. The Turan graph contains no $K_{r+1}$, so every such clique in $J_s(n)$ uses a vertex of $K_{s-1}$. Therefore $J_s(n)$ contains no $s$ vertex-disjoint such cliques, and the extremal number is at least $b_s(n)$.

Induct on $s$, with $s=1$ supplied by the Turan theorem, including its equality characterization. Let $s\ge 2$ and let $G$ be extremal for $s{K}_{r+1}$. The preceding results give $\delta (G)\ge \alpha n-o(n)$ and the $r$-partite subgraph $H$. Keep its classes $H_i$ fixed and assign each of the remaining $o(n)$ vertices to a class in which it has fewest neighbours of a vertex in $H$. Write the resulting classes as $C_i$; all have size $n/r+o(n)$. Choose a positive ${\tau }_n\to 0$ so slowly that ${\tau }_{n}n$ dominates the outside count, every cross-class error in $H$, and every error in the minimum degree of a graph bound.

Suppose some vertex $x\in C_i$ has at least ${\tau }_{n}n$ internal neighbours of a vertex. If $x\notin H$, the assignment rule shows that $x$ has at least ${\tau }_{n}n-o(n)$ neighbours of a vertex in every $H_j$. If $x\in H_i$, it has this many in $H_i$ and $n/r-o(n)$ in the other $H_j$. If $G-x$ contained $s-1$ disjoint $K_{r+1}$, we could choose one vertex in each $N(x)\cap H_j$, successively avoiding those copies and the $o(n)$ missing cross-neighbours of the already chosen vertices. This produces a further $K_{r+1}$ through $x$. For $r=1$, this just chooses an unused neighbour of $x$. Hence $G-x$ is $(s-1)K_{r+1}$-free, and induction givesSince the candidate attains $b_s(n)$, equality holds throughout: $x$ is a universal vertex and $G-x$ is the inductively unique extremal graph. Thus $G$ is isomorphic to $J_s(n)$.

It remains to show that this case must occur. Otherwise all internal vertex degrees are less than ${\tau }_{n}n=o(n)$. Together with the lower bound on $\delta (G)$ and $|C_i|=n/r+o(n)$, this implies that every vertex misses only $o(n)$ vertices outside its own class. If some $C_i$ contains a matching in a graph of $s$ edges, extend its edges one by one to $K_{r+1}$, choosing one vertex from each other class. At every step only $o(n)$ cross-neighbours and boundedly many used vertices are excluded. This gives $s$ disjoint forbidden cliques, a contradiction.

A maximal matching in each class therefore has at most $s-1$ edges, and its at most $2(s-1)$ endpoints form a vertex cover of the internal edges. Their internal vertex degrees are all $o(n)$, so the total internal edge count is $o(n)$. The cross-class edge count is at most $e(T_r(n))$, whenceBut the balanced part sizes of the Turan graph givecontradicting extremality for $s\ge 2$. This argument also works for $r=1$. Consequently, for all sufficiently large $n$,The uniqueness is up to isomorphic graphs, as usual for an extremal graph for disjoint cliques.

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.

The Erdős-Stone theorem states precisely that, for every fixed graph $H$ with chromatic number $\chi (H)\ge 2$,If the bipartite graph $G$ has an edge, then $\chi (G)=2$. Substitution into the theorem givesIf $G$ has no edges, then every graph of sufficiently large order contains $G$ as a subgraph, so the conclusion is immediate.

For each edge $e$, let $I_e$ be the indicator random variable of the event that precisely one endpoint lies in $U$. Independent inclusion of the two endpoints with probability $1/2$ givesBecause $X={\sum }_{e\in E(G)}{I}_e$, linearity of expectation givesSome realization therefore has $X\ge m/2$. Retaining its crossing edges gives a subgraph that is bipartite, with vertex classes $U$ and $V(G)\setminus U$, proving the random cut lower bound.

Now suppose $n$ is an even number and instead choose $U$ uniformly among all subsets of size $n/2$. For a fixed edge $uv$,and the reverse orientation has the same probability. ThusAnother application of linearity of expectation givesAt least one such balanced choice attains this expectation, and its crossing edges give the required subgraph that is bipartite. This is the balanced random cut lower bound.