Proof. We label a copy of $H_{0}$ in G as below and let $X = \mathopen{}\left\{a_{0}, a_{1}, \dotsc, a_{6}\right\}\mathclose{}$ .

Let $U_{4}$ be the set of vertices with exactly four neighbours in X. Remark 3.2 shows that no vertex has five neighbours in a copy of $H_{0}$ , so every non- $U_{4}$ vertex has at most three neighbours in X. Hence,

\begin{equation*}7/2 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} < 7 \delta (G) \leqslant e(X, G) \leqslant 4 \mathopen{}\left\vert U_{4} \right\vert\mathclose{} + 3(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert U_{4} \right\vert\mathclose{}) = 3 \mathopen{}\left\vert G \right\vert\mathclose{} + \mathopen{}\left\vert U_{4} \right\vert\mathclose{},\end{equation*}

and so

\begin{equation*}\mathopen{}\left\vert U_{4} \right\vert\mathclose{} > 1/2 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}.\end{equation*}

Now $\mathopen{}\left\vert U_{4} \right\vert\mathclose{} + d(a_{0}) > \mathopen{}\left\vert G \right\vert\mathclose{}$ and so some vertex v is adjacent to $a_{0}$ and has exactly four neighbours in X. Note that v cannot be adjacent to both $a_{1}$ , $a_{2}$ as otherwise $va_{0}a_{1}a_{2}$ is a $K_{4}$ , so by symmetry we may assume that v is not adjacent to $a_{1}$ . Similarly, we may assume that v is not adjacent to $a_{5}$ . But v has four neighbours in X so must be adjacent to at least one of $a_{2}$ , $a_{6}$ – by symmetry, we may assume v is adjacent to $a_{2}$ .

There are two possibilities: v is adjacent to $a_{0}, a_{2}, a_{3}, a_{4}$ , or v is adjacent to $a_{0}, a_{2}, a_{6}$ and one of $a_{3}$ , $a_{4}$ . In the latter case, we may assume by symmetry that v is adjacent to $a_{3}$ . Hence, there are two possibilities for $\Gamma(v) \cap X$ : $\mathopen{}\left\{a_{0}, a_{2}, a_{3}, a_{4}\right\}\mathclose{}$ and $\mathopen{}\left\{a_{0}, a_{2}, a_{3}, a_{6}\right\}\mathclose{}$ . In both cases, v cannot be any $a_{i}$ except for possibly $a_{1}$ .

If $\Gamma(v) \cap X = \mathopen{}\left\{a_{0}, a_{2}, a_{3}, a_{4}\right\}\mathclose{}$ , then $G[v, a_{3}, a_{4}, a_{5}, a_{6}, a_{0}, a_{2}]$ is a copy of $H_{1}$ . If $\Gamma(v) \cap X = \mathopen{}\left\{a_{0}, a_{2}, a_{3}, a_{6}\right\}\mathclose{}$ , then G contains the following graph where, in particular, v is not adjacent to $a_{4}$ .

Vertex $a_{6}$ is not adjacent to $a_{3}$ else $G_{a_{6}}$ contains the 5-cycle $a_{0}va_{3}a_{4}a_{5}$ . Hence, $va_{6}a_{4}a_{3}$ is an induced 4-cycle in G. By Lemma 3.7, at least one of the pairs $v, a_{4}$ and $a_{3}, a_{6}$ is dense. By symmetry, we may assume that $v, a_{4}$ is dense: let ${a^{\prime}_{2}}{a^{\prime}_{3}}$ be an edge in $G_{v, a_{4}}$ . Vertex v is not adjacent to $a_{5}$ , so $a_{5}$ is neither ${a^{\prime}_{2}}$ nor ${a^{\prime}_{3}}$ . Note that $a_{0}$ is not adjacent to $a_{4}$ (else $a_{0}a_{4}a_{5}a_{6}$ is a $K_{4}$ ) and so $a_{0}$ is neither ${a^{\prime}_{2}}$ nor ${a^{\prime}_{3}}$ . If $a_{6} = {a^{\prime}_{2}}$ , then $G_{a_{6}}$ contains the 5-cycle ${a^{\prime}_{3}}va_{0}a_{5}a_{4}$ , which is impossible. Similarly, $a_{6} \neq {a^{\prime}_{3}}$ . Hence, ${a^{\prime}_{2}}, {a^{\prime}_{3}}$ are distinct from $a_{4}, a_{5}, a_{6}, a_{0}, v$ and so $G[a_{6}, a_{0}, v, {a^{\prime}_{2}}, {a^{\prime}_{3}}, a_{4}, a_{5}]$ is a copy of $H_{1}$ .

Proof. Consider a copy of $H_{1}$ with vertices $X = \mathopen{}\left\{a_{0}, a_{1}, \dotsc, a_{6}\right\}\mathclose{}$ . We assign a weighting $\omega \colon X \rightarrow \mathbb{Z}_{\geqslant 0}$ as shown in the diagram below, so, for example $\omega(a_{0}) = 2$ and $\omega(a_{1}) = 1$ (recall this notation from Section 1.1). We will often use diagrams to give weightings in this way. For each vertex $v \in G$ , let f(v) be the total weight of the neighbours of v in X.

We will assume that G does not contain $H_{2}$ . We first show that any vertex v has $f(v) \leqslant 6$ and further that if $f(v) = 6$ , then $\Gamma(v) \cap X = \mathopen{}\left\{a_{1}, a_{2}, a_{5}, a_{6}\right\}\mathclose{}$ .

Let v be a vertex with $f(v) \geqslant 6$ . No vertex has five neighbours in a copy of $H_{1}$ (as noted in Remark 3.2), so v is adjacent to at most four of the $a_{i}$ . Thus, v is adjacent to at least two of the vertices of weight two, that is, to at least two of $a_{0}$ , $a_{2}$ , $a_{5}$ . If v is adjacent to all of $a_{0}, a_{2}, a_{5}$ , then $G_{a_{0}}$ contains the odd circuit $va_{5}a_{6}a_{1}a_{2}$ . Thus, v is adjacent to exactly two of $a_{0}$ , $a_{2}$ and $a_{5}$ .

Suppose v is adjacent to $a_{0}$ . By symmetry, we may assume that v is adjacent to $a_{2}$ but not to $a_{5}$ . Then,v is not adjacent to $a_{1}$ , else $va_{0}a_{1}a_{2}$ is a $K_{4}$ . Similarly, v cannot be adjacent to both $a_{3}$ and $a_{4}$ . Hence,v is adjacent to $a_{0}$ , $a_{2}$ , $a_{6}$ and one of $a_{3}$ , $a_{4}$ . By symmetry, we may assume v is adjacent to $a_{3}$ . Now v cannot be $a_{4}$ nor $a_{5}$ as it would then have five neighbours in X. Hence, $G[a_{5}, a_{6}, a_{0}, v, a_{2}, a_{3}, a_{4}]$ is a copy of $H_{2}$ .

Thus, v is not adjacent to $a_{0}$ and so is adjacent to both $a_{2}$ and $a_{5}$ . Note v cannot be adjacent to both $a_{3}$ , $a_{4}$ else $va_{2}a_{3}a_{4}$ is $K_{4}$ , so we may assume that v is adjacent to $a_{1}$ . If v were adjacent to one of $a_{3}$ , $a_{4}$ , then we may assume by symmetry that v is adjacent to $a_{4}$ and not $a_{3}$ . Now v cannot be $a_{0}$ nor $a_{6}$ as it would then have five neighbours in X. But then, $G[a_{5}, a_{6}, a_{0}, a_{1}, a_{2}, v, a_{4}]$ is a copy of $H_{2}$ . Therefore, v is adjacent to neither $a_{3}$ nor $a_{4}$ . But $f(v) \geqslant 6$ , so v is adjacent to $a_{1}$ , $a_{6}$ and thus $\Gamma(v) \cap X = \mathopen{}\left\{a_{1}, a_{2}, a_{5}, a_{6}\right\}\mathclose{}$ .

So every vertex v has $f(v) \leqslant 6,$ and furthermore, all $v \in \Gamma(a_{0}) \cup \Gamma(a_{3}) \cup \Gamma(a_{4})$ have $f(v) \leqslant 5$ . We first claim that there is $i \in \mathopen{}\left\{3, 4\right\}\mathclose{}$ such that $\Gamma(a_{0}, a_{2}, a_{i}) = \varnothing$ . If not, there is ${a^{\prime}_{1}} \in \Gamma(a_{0}, a_{2}, a_{3})$ and ${a^{\prime\prime}_{1}} \in \Gamma(a_{0}, a_{2}, a_{4})$ . But then, $G_{a_{2}}$ contains the odd circuit ${a^{\prime}_{1}} a_{3} a_{4} {a^{\prime\prime}_{1}} a_{0}$ . Similarly there is $j \in \mathopen{}\left\{3, 4\right\}\mathclose{}$ such that $\Gamma(a_{0}, a_{5}, a_{j}) = \varnothing$ .

Next we claim that there is $i \in \mathopen{}\left\{3, 4\right\}\mathclose{}$ with $\Gamma(a_{0}, a_{2}, a_{i}) = \Gamma(a_{0}, a_{5}, a_{i}) = \varnothing$ . If not, then without loss of generality there is ${a^{\prime}_{1}} \in \Gamma(a_{0}, a_{2}, a_{3})$ and ${a^{\prime}_{6}} \in \Gamma(a_{0}, a_{5}, a_{4})$ . Certainly, $\Gamma({a^{\prime}_{1}}) \cap X \neq \mathopen{}\left\{a_{1}, a_{2}, a_{5}, a_{6}\right\}\mathclose{}$ , so $f({a^{\prime}_{1}}) \leqslant 5$ . But $\omega(a_{0}) + \omega(a_{2}) + \omega(a_{3}) = 5$ , so $\Gamma({a^{\prime}_{1}}) \cap X = \mathopen{}\left\{a_{0}, a_{2}, a_{3}\right\}\mathclose{}$ . Similarly $\Gamma({a^{\prime}_{6}}) \cap X = \mathopen{}\left\{a_{0}, a_{4}, a_{5}\right\}\mathclose{}$ . In particular, all of $a_{0}, \dotsc, a_{6}, {a^{\prime}_{1}}, {a^{\prime}_{6}}$ are distinct. But then, $G[a_{0}, {a^{\prime}_{1}}, a_{2}, a_{3}, a_{4}, a_{5}, {a^{\prime}_{6}}]$ is a copy of $H_{2}$ . Thus, without loss of generality, $\Gamma(a_{0}, a_{2}, a_{3}) = \Gamma(a_{0}, a_{5}, a_{3}) = \varnothing$ . Then, $\Gamma(a_{3})$ , $\Gamma(a_{0}, a_{2})$ , $\Gamma(a_{0}, a_{5})$ are pairwise disjoint (we already showed that $\Gamma(a_{0}, a_{2}, a_{5}) = \varnothing$ ). Hence, the set $Y = \Gamma(a_{3}) \cup \Gamma(a_{0}, a_{2}) \cup \Gamma(a_{0}, a_{5})$ has size

\begin{equation*}\mathopen{}\left\vert Y \right\vert\mathclose{} = d(a_{3}) + d(a_{0}, a_{2}) + d(a_{0}, a_{5}) \geqslant \delta(G) + 2(2\delta(G) - \mathopen{}\left\vert G \right\vert\mathclose{}) = 5 \delta(G) - 2 \mathopen{}\left\vert G \right\vert\mathclose{}.\end{equation*}

Now all $v \in Y$ have $f(v) \leqslant 5$ . We bound $\omega(X, G)$ from both directions (recalling this notation from Section 1.1) to get

\begin{equation*}10 \delta(G) \leqslant \sum_{x \in X} \omega(x) d(x) = \sum_{v \in G} f(v) \leqslant 5 \mathopen{}\left\vert Y \right\vert\mathclose{} + 6(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert Y \right\vert\mathclose{}) = 6 \mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert Y \right\vert\mathclose{} \leqslant 8 \mathopen{}\left\vert G \right\vert\mathclose{} - 5 \delta(G),\end{equation*}

which contradicts $\delta(G) > 8/15 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Proof. Label the configuration as follows and write $v_{0}$ for v (we consider indices modulo $2k + 1$ ). Note $C = \mathopen{}\left\{v, v_{1}, \dotsc, v_{2k}\right\}\mathclose{}$ .

Consider a vertex x which is adjacent to u. Suppose that x is adjacent to two vertices in C which are not two apart: x is adjacent to $v_{i}$ and $v_{i + r}$ where $r \in \mathopen{}\left\{1, 3, 4, \dotsc, k\right\}\mathclose{}$ . Firstly, if $r = 1$ , then either G contains the $K_{4}$ $uxv_{i}v_{i + 1}$ or $G_{u, v}$ contains an edge (if one of $v_{i}$ or $v_{i + 1}$ is v) which contradicts the sparsity of u, v. Secondly, if $r > 1$ is odd, then $C' = x v_{i} v_{i + 1} \dotsb v_{i + r}$ is an odd circuit which is shorter than C. But then, C $^{\prime}$ contains an odd cycle $C^{\prime\prime}$ which is shorter than C. Either $C^{\prime\prime}$ is in $G_{u}$ (if $v \not \in C''$ ) contradicting the local bipartiteness of G or we have found a shorter odd cycle than C which satisfies the properties of C (if $v \in C''$ ). Finally, if $r > 2$ is even, then $C' = xv_{i + r}v_{i + r + 1} \dotsb v_{i - 1} v_{i}$ is an odd circuit which is shorter than C. Again C $^{\prime}$ must contain an odd cycle $C^{\prime\prime}$ which is shorter than C. We either obtain an odd cycle in $G_{u}$ or contradict the minimality of C. Hence, every neighbour of u has at most two neighbours in C, and if two, then they are $v_{i}$ , $v_{i + 2}$ for some i.

All of $v_{1}, \dotsc, v_{2k}$ are neighbours of u so have two neighbours in C. Hence, C is induced.

Proof. Consider a sparse pair u, v that is the missing spoke of a $(2k + 1)$ -wheel. Choose the odd wheel so that k is minimal. Without loss of generality, u is the central vertex and v is in the outer $(2k + 1)$ -cycle which we call C. Lemma 3.11 shows that $k > 2$ . We are done if we can show that $k = 3$ .

By Lemma 3.12, every neighbour of u has at most two neighbours in C. All vertices have at most 2k neighbours in C as otherwise G contains a $(2k + 1)$ -wheel and so is not locally bipartite. Hence,

\begin{align*}(2k + 1) \delta(G) & \leqslant e(G, C) \leqslant 2 d(u) + 2k(\mathopen{}\left\vert G \right\vert\mathclose{} - d(u)) \\[3pt]& = 2k \mathopen{}\left\vert G \right\vert\mathclose{} - (2k - 2) d(u) \leqslant 2k \mathopen{}\left\vert G \right\vert\mathclose{} - (2k - 2) \delta(G),\end{align*}

so

\begin{equation*}\frac{8}{15} < \frac{\delta(G)}{\mathopen{}\left\vert G \right\vert\mathclose{}} \leqslant \frac{2k}{4k - 1},\end{equation*}

which implies that $k < 4$ .

Proof. Let G be a locally bipartite graph which does not contain $H_{0}$ and either satisfies $\delta(G) > 7/13 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ , or $\delta(G) > 8/15 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ and G does not contain $T_{0}$ . In a slight abuse of notation, we will say that a vertex u is sparse to a cycle C if u is adjacent to $\mathopen{}\left\vert C \right\vert\mathclose{} - 1$ vertices of C and is in a sparse pair with the final vertex. We are required to show that there is no vertex u and no 7-cycle C with u sparse to C.

Proof of Claim. First consider the induced 4-cycle $uv_{6}v_{0}v_{1}$ . The pair $u, v_{0}$ is sparse, so $\Gamma(u, v_{6}, v_{0}) = \Gamma(v_{0}, v_{1}, u) = \varnothing$ . Also $\Gamma(v_{1}, u, v_{6}) = \varnothing$ as otherwise $G_{u}$ contains an odd circuit. Hence, all $z \not \in \Gamma(v_{6}, v_{0}, v_{1})$ have at most two neighbours in $\mathopen{}\left\{u, v_{6}, v_{0}, v_{1}\right\}\mathclose{}$ .

Thus,

\begin{equation*}4 \delta(G) \leqslant e(\mathopen{}\left\{v_{0}, v_{1}, u, v_{6}\right\}\mathclose{}, G) \leqslant 3 \mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{} + 2(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{}),\end{equation*}

so $\mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{} \geqslant 4 \delta(G) - 2 \mathopen{}\left\vert G \right\vert\mathclose{}$ .

If the claim is false, then all vertices in $\Gamma(v_{6}, v_{0}, v_{1})$ have at most five neighbours in C. Also note that $\Gamma(u)$ and $\Gamma(v_{6}, v_{0}, v_{1})$ are disjoint and that all vertices have at most six neighbours in C (otherwise G contains a 7-wheel) so,

\begin{align*}7 \delta(G) & \leqslant e(C, G) \leqslant 2 d(u) + 5\mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{} + 6(\mathopen{}\left\vert G \right\vert\mathclose{} - d(u) - \mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{}) \\[3pt]& = 6 \mathopen{}\left\vert G \right\vert\mathclose{} - 4d(u) - \mathopen{}\left\vert \Gamma(v_{6}, v_{0}, v_{1}) \right\vert\mathclose{} \leqslant 6 \mathopen{}\left\vert G \right\vert\mathclose{} - 4 \delta(G) - 4 \delta(G) + 2 \mathopen{}\left\vert G \right\vert\mathclose{},\end{align*}

which contradicts $\delta(G) > 8/15 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Proof of Claim. Let vertex u be sparse to the 7-cycle C and let x be a vertex with at least six neighbours in C. Since G is locally bipartite, x has exactly six neighbours in C. Let y be the vertex of C to which x is not adjacent. There is a vertex u $^{\prime}$ which has six neighbours in C and is adjacent to y. Indeed, if u is adjacent to y then take $u' = u$ and if u is not adjacent to y, then Claim 3.15 gives the desired u $^{\prime}$ .

Now $\Gamma(u',x)$ contains two consecutive vertices of C, so the pair u $^{\prime}$ , x is dense. But u $^{\prime}$ is adjacent to y so, by Lemma 3.8, the pair y, x cannot be dense. In particular, x, y is sparse and so x is sparse to C.

Proof of Claim. Suppose for contradiction that $v_{1}$ and $v_{6}$ are both lonely: let $u_{1}$ and $u_{6}$ be sparse to C with both the pairs $u_{1}, v_{1}$ and $u_{6}, v_{6}$ sparse. Lemma 3.12 shows that any neighbour of $u_{1}$ (or $u_{6}$ ) has at most two neighbours in C. Let $X = \mathopen{}\left\{u_{1}, u_{6}, v_{0}, \dotsc, v_{6}\right\}\mathclose{}$ and give a weighting $\omega$ to the vertices in X as shown below.

For each vertex $v \in G$ , let f(v) be the total weight of the neighbours of v in X. We shall show that if $v \not \in \Gamma(u_{6}, u_{1}, v_{0})$ , then $f(v) \leqslant 10$ and if $v \in \Gamma(u_{6}, u_{1}, v_{0})$ , then $f(v) = 13$ . Let v be a vertex with $f(v) \geqslant 11$ . It suffices to show that v is adjacent to $u_{6}$ , $u_{1}$ , $v_{0}$ and none of $v_{1}$ , …, $v_{6}$ .

If v is adjacent to neither $u_{1}$ nor $u_{6}$ , then $f(v) \leqslant 11$ with equality only if v is adjacent to all of C which would give a 7-wheel so, in fact, $f(v) \leqslant 10$ . Thus, we may assume v is adjacent to $u_{1}$ . But then, v must have at most two neighbours in C. If neither of these is $v_{0}$ , then $f(v) \leqslant 4 + 4 + 1 + 1 = 10$ . Hence, we may assume v is adjacent to both $v_{0}$ and $u_{1}$ .

Since $f(v) \geqslant 11$ and v has at most two neighbours in C, v must be adjacent to $u_{6}$ as well. Now, v is adjacent to both $u_{1}$ and $v_{0}$ so, by Lemma 3.12, the only other possible neighbour of v in C is one of $v_{2}$ and $v_{5}$ . However, if v is adjacent to $v_{2}$ , then $G_{u_{1}}$ contains the odd circuit $v_{0}vv_{2}v_{3}\dotsc v_{6}$ while if v is adjacent to $v_{5}$ , then $G_{u_{6}}$ contains the odd circuit $v_{0}v_{1}\dotsc v_{5}v$ . In conclusion, v is adjacent to $u_{1}$ , $u_{6}$ , $v_{0}$ , and no other vertices of C.

Thus,

\begin{equation*}19 \delta(G) \leqslant \omega(X, G) = \sum_{v \in G} f(v) \leqslant 13 \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{} + 10 (\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{}),\end{equation*}

so

(1) \begin{equation}3 \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{} \geqslant 19 \delta(G) - 10 \mathopen{}\left\vert G \right\vert\mathclose{}.\end{equation}

Any $v \in \Gamma(v_{0}, u_{1}, u_{6})$ satisfies $f(v) = 13$ so has only one neighbour in C. Also any neighbour of $u_{1}$ has at most two neighbours in C. Hence,

\begin{align*}7 \delta(G) & \leqslant e(C, G) \leqslant \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{} + 2(d(u_{1}) - \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{}) + 6(\mathopen{}\left\vert G \right\vert\mathclose{} - d(u_{1})) \\[3pt]& = 6 \mathopen{}\left\vert G \right\vert\mathclose{} - 4d(u_{1}) - \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{},\end{align*}

so

(2) \begin{equation}\mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{} \leqslant 6 \mathopen{}\left\vert G \right\vert\mathclose{} - 11 \delta(G).\end{equation}

Combining inequalities (1) and (2) gives

\begin{equation*}19 \delta(G) - 10 \mathopen{}\left\vert G \right\vert\mathclose{} \leqslant 3 \mathopen{}\left\vert \Gamma(v_{0}, u_{1}, u_{6}) \right\vert\mathclose{} \leqslant 18 \mathopen{}\left\vert G \right\vert\mathclose{} - 33 \delta(G),\end{equation*}

so $\delta(G) \leqslant 7/13 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ and so G is $T_{0}$ -free.

Inequality (1) and $\delta(G) > 8/15 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ show that $\Gamma(v_{0}, u_{1}, u_{6})$ is non-empty. Let v be a common neighbour of $v_{0}$ , $u_{1}$ , $u_{6}$ . As G is $T_{0}$ -free, v must be one of the $v_{i}$ . But $C = v_{0}v_{1} \dotsc v_{6}$ is induced, so v must be one of $v_{1}$ , $v_{6}$ . This means one of the edges $u_{1}v_{1}$ , $u_{6}v_{6}$ is present. However, these are both sparse pairs giving the required contradiction.

Proof. By Lemma 3.8 and the definition of dense, $D_{u} \cup \mathopen{}\left\{u\right\}\mathclose{}$ is an independent set. However, it contains the maximal independent set I, so must equal it.

Proof. Fix a vertex $u \in I$ and let v be any other vertex which is not adjacent to u. It suffices to show that $v \in I$ . If the pair u, v is (quasi)dense, then $v \in I$ , so we may assume that u, v is sparse (and not quasidense). Thus, u, v appears in one of the configurations given in Figure 5 (with labels u and v possibly swapped). However, in each of Figure 5(a), (b), (e) and (g) to (j) the pair u, v is quasidense. Hence, we may assume that u, v appear in one of Figure 5(c), (d) and (f).

We consider Figure 5(c) and (d) together. For ease, we label some more of the vertices as follows.

In both cases, the pair u $^{\prime}$ , w is dense, and so, by Lemma 3.8, the pair u $^{\prime}$ , v $^{\prime}$ is not dense. However, u $^{\prime}$ v $^{\prime}$ is not an edge, as the pair u, v is sparse, and so u $^{\prime}$ , v $^{\prime}$ is a sparse pair. But then, uu $^{\prime}$ vv $^{\prime}$ is an induced 4-cycle in which both non-edges are sparse which contradicts Lemma 3.7.

Finally, we consider Figure 5(f) which we label as follows. Let $X = \mathopen{}\left\{u, v, v', v_{1}, v_{2}, v_{3}, v_{4}\right\}\mathclose{}$ and give a weighting $\omega$ to the vertices of X as shown.

The pair v, v $^{\prime}$ is dense, so, if u, v $^{\prime}$ is quasidense, then u, v is quasidense, a contradiction. Also, as G is $H_{0}$ -free, uv $^{\prime}$ is not an edge. Hence, the pair u, v $^{\prime}$ is sparse and not quasidense. Now v, v $^{\prime}$ is dense and so, by Lemma 3.8, the pair $vv_{4}$ is not. But $vv_{4}$ is not an edge (else u, v is dense), so $v, v_{4}$ is a sparse pair. Similarly, $v', v_{1}$ is a sparse pair. To summarise, the pairs u, v and u, v $^{\prime}$ are sparse and not quasidense and the pairs $v, v_{4}$ and $v', v_{1}$ are sparse. It follows that G[X] contains no more edges than shown.

If a vertex x has five neighbours in X, then it is adjacent to three consecutive vertices round the 7-cycle, so x is either adjacent to a triangle or to all of $u, v_{1}, v$ or to all of $v', v_{4}, u$ . The first gives a $K_{4}$ while the latter two contradict the pairs u, v and u, v $^{\prime}$ being sparse. Hence, all vertices have at most four neighbours in X.

For each vertex x in G, let f(x) be the total weight of the neighbours of x in X. Now

\begin{equation*}\sum_{x \in G} f(x) = \omega(X, G) \geqslant 15 \delta(G) > 8 \mathopen{}\left\vert G \right\vert\mathclose{},\end{equation*}

so some vertex x has $f(x) \geqslant 9$ . All vertices of X have f value at most eight, so $x \not \in X$ . As x has at most four neighbours in X, either x is adjacent to both $v_{1}$ and $v_{4}$ or x is adjacent to exactly one of $v_{1}$ , $v_{4}$ and both of $v_{2}$ , $v_{3}$ .

First suppose that x is adjacent to both $v_{1}$ and $v_{4}$ . As $v, v_{4}$ is sparse, x is not adjacent to v. Similarly x is not adjacent to v $^{\prime}$ . If x is adjacent to $v_{2}$ , then x, v is dense (the edge $v_{1}v_{2}$ ). But also u, x is dense (the edge $v_{1}v_{4}$ ), so u, v is quasidense, a contradiction. Hence, x is not adjacent to $v_{2}$ . Similarly x is not adjacent to $v_{3}$ , and so $f(x) = 8$ , a contradiction.

In the second case, we may assume, by symmetry, that x is adjacent to $v_{1}$ , $v_{2}$ , $v_{3}$ but not to $v_{4}$ . Then, x, v and x, v $^{\prime}$ are both dense pairs (the edge $v_{2}v_{3}$ ) so x is adjacent to neither v nor v $^{\prime}$ . Finally, if x is adjacent to u, then $x, v_{4}$ is dense (the edge $uv_{1}$ ). But then, the edge $v_{4}v'$ is in $D_{x}$ , contradicting Lemma 3.8. Hence, $f(x) = 8$ , a contradiction.

Proof of Theorem 3.3. Let $u \in I$ . Proposition 3.21 gives $G[V(G) \setminus I] = G_{u}$ , so $G[V(G) \setminus I]$ is 2-colourable. Using a third colour for the independent set I gives a 3-colouring of G.

Proof. We will prove this for $b = 3$ and then use the lifting lemma for larger b.

For $b = 3$ , G is a locally tripartite graph with $\delta(G) > 15/22 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ and u, v is a 3-sparse pair in G. Suppose the conclusion does not hold. Each neighbourhood of G is 3-colourable so does not contain an odd wheel. In particular, G does not contain $K_{2} + C_{\text{odd}}$ , and so G contains one of the following two configurations (corresponding to whether only one of u, v is in the clique or they both are). In the left figure, x is adjacent to all vertices inside the ring, and in the right figure, u and v are adjacent to all vertices inside the ring. In the future, we will use rings in this way.

We deal with the left-hand configuration first. Odd wheels, $H_{0}$ and $T_{0}$ are not 3-colourable and so $G_{x}$ is locally bipartite, $H_{0}$ -free and $T_{0}$ -free. Applying Lemma 4.2 with $X = \mathopen{}\left\{x\right\}\mathclose{}$ and $\gamma = 1/7$ gives $\delta(G_{x}) > (1 - 1/(2 + 1/7)) \cdot \mathopen{}\left\vert G_{x} \right\vert\mathclose{} = 8/15 \cdot \mathopen{}\left\vert G_{x} \right\vert\mathclose{}$ so, by Corollary 3.18, $G_{x}$ does not contain a sparse pair which is the missing spoke of an odd wheel. In particular, the pair u, v is dense in $G_{x}$ so must be 3-dense in G, a contradiction.

Now consider the right-hand configuration. We may assume that k is minimal. As the pair u, v is 3-sparse, $k > 1$ . Let $X = \mathopen{}\left\{u, v, v_{0}, v_{1}, \dotsc, v_{2k}\right\}\mathclose{}$ . We consider indices modulo $2k + 1$ .

First consider a vertex x adjacent to both u and v. We claim that x is adjacent to at most two of the $v_{i}$ so has at most four neighbours in X. For $r \not \in \mathopen{}\left\{0, \pm 2\right\}\mathclose{}$ , x cannot be adjacent to both $v_{i}$ and $v_{i + r}$ . Indeed if $r = \pm 1$ , then $xv_{i}v_{i + r}$ is a triangle in $G_{u,v}$ while other r give a shorter odd cycle in $G_{u,v}$ .

Now consider any other vertex x: this is not adjacent to at least one of u or v. If this is the only non-neighbour of x in X, then $G_{x}$ contains an odd-wheel, which is not 3-colourable. Thus, all vertices have at most $2k + 1$ neighbours in X. Hence,

\begin{align*}(2k + 3) \delta(G) & \leqslant e(G, X) \leqslant 4 \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{} + (2k + 1)(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{}) \\[3pt]& \leqslant (2k + 1) \mathopen{}\left\vert G \right\vert\mathclose{} - (2k - 3)(2 \delta(G) - \mathopen{}\left\vert G \right\vert\mathclose{}) = (4k -2) \mathopen{}\left\vert G \right\vert\mathclose{} - (4k - 6) \delta(G),\end{align*}

which contradicts $\delta(G) > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Suppose now that $b \geqslant 4$ and that $G + uv$ does contain a $K_{b - 1} + C_{\text{odd}}$ where at least one of u, v is in the $K_{b - 1}$ . The $(b - 1)$ -clique contains a $(b - 3)$ -clique, L, with $u, v \not \in L$ . Thus L is a $(b - 3)$ -clique in G.

Now $G_{L}$ is a 4-colourable (so locally tripartite) graph in which u, v is 3-sparse (by Remark 4.3). Applying Lemma 4.2 with $X = L$ , $\gamma = 1/7$ gives $\delta(G_{L}) > (1 - 1/(b - (b - 3) + 1/7)) \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{} = 15/22 \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{}$ . Finally, $G_{L} + uv$ contains a $K_{2} + C_{\text{odd}}$ where at least one of u, v is in the 2-clique. This contradicts the result for $b = 3$ .

Proof. We split into two cases depending upon whether only one of u, v is in the $(b - 2)$ -clique or both are. In each case, we will prove the result for small b and then use the lifting lemma for larger b.

Suppose only one of u, v is in the $(b - 2)$ -clique. We first prove the result for $b = 3$ : G is locally tripartite, the pair u, v is 3-sparse, and $\delta(G) > 15/22 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ . Suppose the conclusion does not hold. Since G does not contain $K_{1} + H_{0}$ , G contains the configuration shown in Figure 6(a) where u is adjacent to all of the copy of $H_{0}$ except for one vertex (v) to which is it 3-sparse.

Figure 6.

Configurations in Proposition 4.6.

Let X be the vertex set of the $H_{0}$ . First, consider a vertex x adjacent to u: x cannot be adjacent to a triangle or 5-cycle in G[X] otherwise $G + uv$ contains a 2-clique ux joined to an odd cycle which contradicts Proposition 4.5. In particular, any neighbour of u has at most four neighbours in X.

Consider any vertex x: $\chi(G[X]) = 4$ so x has at most six neighbours in X. Hence,

\begin{equation*}7 \delta(G) \leqslant e(G,X) \leqslant 4 d(u) + 6(\mathopen{}\left\vert G \right\vert\mathclose{} - d(u)) \leqslant 6 \mathopen{}\left\vert G \right\vert\mathclose{} - 2\delta(G),\end{equation*}

which contradicts $\delta(G) > 15/22 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Now let $b \geqslant 4$ and suppose $G + uv$ does contain a $K_{b - 2} + H_{0}$ where exactly one of u, v (say u) is in the $(b - 2)$ -clique. Graph G is locally b-partite so does not contain $K_{b - 2} + H_{0}$ , so v is in the copy of $H_{0}$ . Let L be the $(b - 2)$ -clique without u: L is a $(b - 3)$ -clique in G and so $G_{L}$ is 4-colourable and so locally tripartite. Also, by Lemma 4.2, $\delta(G_{L}) > 15/22 \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{}$ and, by Remark 4.3, u, v is a 3-sparse pair in $G_{L}$ . Finally, $G_{L} + uv$ contains $u + H_{0}$ , which contradicts the result we just proved for $b = 3$ .

Now consider the second case, where both u and v are in the $(b - 2)$ -clique: this means that $b \geqslant 4$ . We first prove the result for $b = 4$ : G is locally 4-partite, the pair u, v is 4-sparse, and $\delta(G) > 22/29 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 3/4 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ . If the result is false, then G contains the configuration shown in Figure 6(b).

Let $X = V(H_{0}) \cup \mathopen{}\left\{u\right\}\mathclose{}$ – a set of eight vertices. First, consider a vertex x adjacent to both u and v: x cannot be adjacent to a triangle or 5-cycle in $V(H_{0})$ as this would contradict Proposition 4.5. Hence, x has at most four neighbours in $V(H_{0})$ and so at most five in X.

Since G[X] is not 4-colourable, all vertices have at most seven neighbours in X. Hence,

\begin{align*}8 \delta(G) & \leqslant 5 \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{} + 7(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{}) \\[3pt]& \leqslant 7 \mathopen{}\left\vert G \right\vert\mathclose{} - 2(2 \delta(G) - \mathopen{}\left\vert G \right\vert\mathclose{}) = 9 \mathopen{}\left\vert G \right\vert\mathclose{} - 4 \delta(G),\end{align*}

which contradicts $\delta(G) > 22/29 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 3/4 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Finally, suppose $b \geqslant 5$ and $G + uv$ does contain a $K_{b - 2} + H_{0}$ where both of u, v are in the $(b - 2)$ -clique. Let L be the $(b - 2)$ -clique without u, v: L is a $(b - 4)$ -clique in G so, by Remark 4.3, $G_{L}$ is 5-colourable and so locally 4-partite. Also, by Lemma 4.2, $\delta(G_{L}) > 22/29 \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{}$ and u, v is a 4-sparse pair in $G_{L}$ . Finally, $G_{L} + uv$ contains $uv + H_{0}$ , which contradicts the result we obtained for $b = 4$ .

Proof. Again we split into two cases depending upon whether only one of u, v is in the $(b - 2)$ -clique or both are, and in each case, we will prove the result for small b and then use the lifting lemma for larger b.

Suppose only one of u, v is in the $(b - 2)$ -clique. We first prove the result for $b = 3$ : G is locally tripartite, the pair u, v is 3-sparse, and $\delta(G) > 15/22 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ . Suppose the conclusion does not hold. Since G does not contain $K_{1} + T_{0}$ , G contains the configuration shown in Figure 7(a) (labels have been added for convenience) where u is adjacent to all of the copy of $T_{0}$ except for one vertex (v) to which it is 3-sparse.

Figure 7.

Configurations in Proposition 4.7.

We first show that at least one of the pairs $u_{1}, v_{1}$ and $u_{6}, v_{6}$ is 3-sparse. Neither of these is an edge as otherwise $G + uv$ contains u joined to a 7-wheel which contradicts Proposition 4.5. If $v \not \in \mathopen{}\left\{u_{1}, u_{6}, v_{3}, v_{4}\right\}\mathclose{}$ , then $u_{1}, u_{6}$ is 3-dense (triangle $uv_{3}v_{4}$ ) and so $u_{1}, v_{1}$ is 3-sparse, by Lemma 4.4. On the other hand, suppose $v \in \mathopen{}\left\{u_{1}, u_{6}, v_{3}, v_{4}\right\}\mathclose{}$ – by symmetry we may assume $v \neq u_{6}$ . Then, $v_{1}, t$ is 3-dense (triangle $uv_{0}u_{6}$ ) and so $u_{1}, v_{1}$ is 3-sparse, by Lemma 4.4. From now on, we will assume that the pair $u_{1}, v_{1}$ is 3-sparse in G.

Let $G' = G + uv$ – we work in G $^{\prime}$ . From Proposition 4.5, ${G^{\prime}_{u}}$ contains no odd wheel, i.e. is locally bipartite, and from Proposition 4.6, ${G^{\prime}_{u}}$ is $H_{0}$ -free. Now $u_{1}v_{1}$ is not an edge (else there is a 7-wheel in ${G^{\prime}_{u}}$ ) and the pair $v_{1}, t$ is 2-dense in ${G^{\prime}_{u}}$ so $u_{1}, v_{1}$ is 2-sparse in ${G^{\prime}_{u}}$ , by Lemma 3.8. Note that $\delta(G') > 2/3 \cdot \mathopen{}\left\vert G' \right\vert\mathclose{}$ and so applying Lemma 4.2 with $b = 3$ , $\gamma = 0$ , $s = 1$ and $X = \mathopen{}\left\{u\right\}\mathclose{}$ gives

\begin{equation*}\delta({G^{\prime}_{u}}) > \big(1 - \tfrac{1}{3 - 1}) \cdot \lvert {G^{\prime}_{u}} \rvert = 1/2 \cdot \lvert {G^{\prime}_{u}} \rvert.\end{equation*}

Hence, we may apply Lemma 3.11 to show that $u_{1}v_{1}$ is not a missing spoke of a 5-wheel in ${G^{\prime}_{u}}$ . Hence, by Lemma 3.12, any neighbour of $u_{1}$ in ${G^{\prime}_{u}}$ is adjacent to at most two of the $v_{i}$ . In particular, any common neighbour of u and $u_{1}$ in G $^{\prime}$ is adjacent to at most two of the $v_{i}$ .

Let $X = \mathopen{}\left\{u,u_{1},v_{0},v_{1}, \dotsc, v_{6}\right\}\mathclose{}$ . What we have just shown is that any common neighbour of u and $u_{1}$ in G $^{\prime}$ has at most four neighbours in X. Consider a vertex x which is not adjacent to both u and $u_{1}$ : x cannot be adjacent to all of $X \setminus \mathopen{}\left\{u_{1}\right\}\mathclose{}$ otherwise ${G^{\prime}_{u, x}}$ contains a 7-cycle which contradicts Proposition 4.5. Also, x cannot be adjacent to all of $X \setminus \mathopen{}\left\{u\right\}\mathclose{}$ as otherwise ${G^{\prime}_{x}} = G_{x}$ contains a 7-wheel missing the spoke $u_{1}v_{1}$ which is 3-sparse in G. This again contradicts Proposition 4.5. Hence, any vertex has at most seven neighbours in X. Thus,

\begin{align*}9 \delta(G) & \leqslant 9 \delta(G') \leqslant 4 \lvert {G^{\prime}_{u, u_{1}}} \rvert + 7(\mathopen{}\left\vert G' \right\vert\mathclose{} - \lvert {G^{\prime}_{u, u_{1}}} \rvert) = 7 \mathopen{}\left\vert G \right\vert\mathclose{} - 3 \lvert {G^{\prime}_{u, u_{1}}} \rvert \\& \leqslant 7 \mathopen{}\left\vert G \right\vert\mathclose{} - 3 \mathopen{}\left\vert G_{u, u_{1}} \right\vert\mathclose{} \leqslant 7 \mathopen{}\left\vert G \right\vert\mathclose{} - 3(2\delta(G) - \mathopen{}\left\vert G \right\vert\mathclose{}) = 10 \mathopen{}\left\vert G \right\vert\mathclose{} - 6 \delta(G),\end{align*}

which contradicts $\delta(G) > 2/3 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Now let $b \geqslant 4$ and suppose $G + uv$ does contain a $K_{b - 2} + T_{0}$ where exactly one of u, v (say u) is in the $(b - 2)$ -clique. Graph G is locally b-partite so does not contain $K_{b - 2} + T_{0}$ so v is in the copy of $T_{0}$ . Let L be the $(b - 2)$ -clique without u. We make use of Remark 4.3: L is a $(b - 3)$ -clique in G and so $G_{L}$ is 4-colourable and so locally tripartite. Also, by Lemma 4.2, $\delta(G_{L}) > 15/22 \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{}$ and u, v is a 3-sparse pair in $G_{L}$ . Finally, $G_{L} + uv$ contains $u + T_{0}$ , which contradicts the result we just proved for $b = 3$ .

Now consider the second case, where both u and v are in the $(b - 2)$ -clique: this means that $b \geqslant 4$ . We first prove the result for $b = 4$ : G is locally 4-partite, the pair u, v is 4-sparse, and $\delta(G) > 22/29 \cdot \mathopen{}\left\vert G \right\vert\mathclose{} > 3/4 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ . If the result is false, then G contains the configuration shown in Figure 7(b) (labels have been added for convenience).

By Propositions 4.5 and 4.6, $G_{u, v}$ is locally bipartite and $H_{0}$ -free. By Lemma 3.2, $\delta(G_{u, v}) > 1/2 \cdot \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{}$ . Since $G_{u, v}$ does not contain an odd wheel, $u_{1}v_{1}$ is not an edge. Also, $v_{1}, t$ is a 2-dense pair, $tu_{1}$ is an edge and $G_{u, v}$ is $H_{0}$ -free, so, by Lemma 3.8, $u_{1}, v_{1}$ is a 2-sparse pair in $G_{u, v}$ . Similarly, $u_{6}, v_{6}$ is 2-sparse in $G_{u, v}$ .

Now let $X = \mathopen{}\left\{u, u_{1}, u_{6}, v_{0}, \dotsc, v_{6}\right\}\mathclose{}$ (note that this does not contain t or v). Within $G_{u, v}$ , $u_{1}$ together with the $v_{i}$ form a 7-wheel missing the spoke $u_{1}v_{1}$ which is a sparse pair. Since $G_{u,v}$ is locally bipartite with $\delta(G_{u, v}) > 1/2 \cdot \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{}$ , Lemma 3.11 implies that $G_{u, v}$ does not contain any 5-wheels missing a sparse spoke. Hence, by Lemma 3.12, any neighbour of $u_{1}$ in $G_{u, v}$ is adjacent to at most two of the $v_{i}$ . Thus, any neighbour of u, v, $u_{1}$ has at most five neighbours in X (two amongst $v_{i}$ together with possibly $u_{1}, u_{6}, u$ ). Similarly, any neighbour of u, v, $u_{6}$ has at most five neighbours in X. Next, consider a vertex x adjacent to both u, v but to neither $u_{1}$ nor $u_{6}$ . As $G_{u,v}$ is locally bipartite, x is adjacent to at most six of the $v_{i}$ so x has at most seven neighbours in X. Finally, $\chi(G[X]) = 5$ so all vertices have at most nine neighbours in X. Hence,

\begin{align*}10 \delta(G) & \leqslant 5\mathopen{}\left\vert \Gamma(u, v, u_{1}) \cup \Gamma(u, v, u_{6}) \right\vert\mathclose{} + 7(\mathopen{}\left\vert G_{u, v} \right\vert\mathclose{} - \mathopen{}\left\vert \Gamma(u, v, u_{1}) \cup \Gamma(u, v, u_{6}) \right\vert\mathclose{}) + 9(\mathopen{}\left\vert G \right\vert\mathclose{} - \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{}) \\& = 9 \mathopen{}\left\vert G \right\vert\mathclose{} - 2 \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{} - 2 \mathopen{}\left\vert \Gamma(u, v, u_{1}) \cup \Gamma(u, v, u_{6}) \right\vert\mathclose{} \leqslant 9 \mathopen{}\left\vert G \right\vert\mathclose{} - 2 \mathopen{}\left\vert G_{u, v} \right\vert\mathclose{} - 2 \mathopen{}\left\vert G_{u, v, u_{1}} \right\vert\mathclose{} \\& \leqslant 9 \mathopen{}\left\vert G \right\vert\mathclose{} - 2(2 \delta(G) - \mathopen{}\left\vert G \right\vert\mathclose{}) - 2(3 \delta(G) - 2 \mathopen{}\left\vert G \right\vert\mathclose{}) = 15 \mathopen{}\left\vert G \right\vert\mathclose{} - 10 \delta(G),\end{align*}

which contradicts $\delta(G) > 3/4 \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ .

Now let $b \geqslant 5$ and suppose $G + uv$ does contain a $K_{b - 2} + T_{0}$ where both of u, v are in the $(b - 2)$ -clique. Let L be the $(b - 2)$ -clique without u, v: L is a $(b - 4)$ -clique in G and so $G_{L}$ is 5-colourable and so locally 4-partite. Also, by Lemma 4.2, $\delta(G_{L}) > 22/29 \cdot \mathopen{}\left\vert G_{L} \right\vert\mathclose{}$ and, by Remark 4.3, the pair u, v is 4-sparse in $G_{L}$ . Finally, $G_{L} + uv$ contains $uv + T_{0}$ , which contradicts the result just proved for $b = 4$ .

Proof. Take an edge-maximal locally b-partite graph G with $\delta(G) > (1 - 1/(b + 1/7)) \cdot \mathopen{}\left\vert G \right\vert\mathclose{}$ . We need to show that G is $(b + 1)$ -colourable. We may assume by induction that the theorem holds for all b $^{\prime}$ with $3 \leqslant b' < b$ (if there are any).

We first show that for any b-sparse pair u, v of G, $G' = G + uv$ contains a $(b - 2)$ -clique K with ${G^{\prime}_{K}}$ not 3-colourable. Indeed, for $b = 3$ , G $^{\prime}$ is not locally tripartite (by edge-maximality) so there is a vertex in G $^{\prime}$ whose neighbourhood is not 3-colourable. Take K to be this vertex. For $b > 3$ , G $^{\prime}$ is not locally b-partite so contains a vertex $w_{1}$ with ${G^{\prime}_{w_{1}}}$ not b-colourable. Applying Lemma 4.2 with $X = \mathopen{}\left\{w_{1}\right\}\mathclose{}$ and $\gamma = 1/7$ gives

\begin{equation*}\delta({G^{\prime}_{w_{1}}}) > \biggl(1 - \frac{1}{b - 1 + 1/7}\biggr) \cdot \lvert {G^{\prime}_{w_{1}}} \rvert.\end{equation*}

By the induction hypothesis, if ${G^{\prime}_{w_{1}}}$ was locally $(b - 1)$ -partite, then it would be b-colourable. In particular, ${G^{\prime}_{w_{1}}}$ is not locally $(b - 1)$ -partite and so there is a vertex $w_{2}$ in ${G^{\prime}_{w_{1}}}$ with ${G^{\prime}_{w_{1}, w_{2}}}$ not $(b - 1)$ -colourable. Repeating this argument gives a $(b - 2)$ -clique K with ${G^{\prime}_{K}}$ not 3-colourable.