I can't understand how to use Brooks' Theorem. I only succeed to derive the maximum degree of $G$ is greater than 3. How to derive that the minimum degree of $G$ is greater than 3 from Brooks' Theorem? Chvatal and Harary's proof is wrong as it is? or not? (If you have other elegant proof of $r(C_4,K_4)=10$, then It also help me.)

For upper bound, we think about above graph $G$. Using $r(C_4,K_3)=7$, easily we have there is no vertex with degree $\le2$. By lemma, we have at least one vertex (say $u$) whose degree is 3.

Claim. The subgraph induced by vertices non-adjacent to $u$ contains $2K_3$.Then

The subgraph induced by vertices non-adjacent to $u$ has 6 vertices. So we easily have two triangles $T_1,T_2$ in this subgraph. If $T_1,T_2$ has two common vertex, we get $C_4$. If $T_1,T_2$ has only one common point, let $T_j=v_0v_1^jv_2^j$. Let $w$ be the other vertex non-adjacent to $u$. Then $v_2^1w$ isn't an edge by symmetry and avoiding $C_4$, namely $v_0v_1^1wv_2^1$. We also have edge $v_1^2w$ by avoiding 4 independent vertices $v_2^1wv_1^2u$.By symmetry, we have $C_4$, namely $v_0v_1^2wv_2^2$ and it's a contradiction.

Claim. The neighborhood subgraph $N(u)$ of $u$ is $\bar{K_3}$.At now, it's

If notso difficult to see this is contradiction by thinking about , the connection neighborhood subgraph of above $2K_3$ u$is an isolated vertex$v_1$and an edge$\bar{K_3}$.v_2v_3$. $v_2$ and $v_3$ has at least one edge to $T_1\cup T_2$, since their degree $\ge3$. If $v_2$ and $v_3$ has edges to common triangle, we get $C_4$. So if we let $T_j=w_1^jw_2^jw_3^j$, we can assume there are edges $v_2w_1^1, v_3w_1^2$. ($v_1,v_2$ has no other edges to $T_1\cup T_2$.) Then both edges $v_1w_2^1,v_1w_3^1$ cannnot be exist. So we can assume there isn't edge $v_1w_3^1$, then we have edge $w_3^1w_1^2$, since otherwise we have 4 independent vertices $v_1v_2w_3^1w_1^2$. By symmetry, we also have edge $w_1^1w_3^2$. So we have $C_4$, $w_1^1w_3^2w_1^2w_3^1$. It's a contradiction.

Now, we have $T_j=w_1^jw_2^jw_3^j$ and 6 edges $v_iw_i^j$. Then we have edge $w_1^1w_1^2$, since otherwise $w_1^1w_1^2v_2v_3$ form 4 independent vertices. By symmetry, we have $C_4$, namely $w_1^1w_1^2w_2^2w_2^1$. It's a contradiction.

3 Capitalization of names

I'm studying about graph ramsey theory Graph Ramsey Theory now. Starting this study, I'm reading chvatal Chvatal and harary's Harary's series of papers. In the second paper (V.Chvatal, F.Harary, Generalized ramsey theory for graphs,Ⅲ. Small off-diagonal numbers, Pacific Journal of Mathematics 41, No.2, 1972, pp.335-345), I can't understand the proof of $r(C_4,K_4)=10$. Their proof is like following.

Let $G$ is arbitrary simple graph of order 10 with point independence number $<4$. It is sufficient to prove $G$ contains $C_4$. From $G$'s point independence number is $<4$, $G$'s (point) chromatic number is $\ge4$. Hence by Brooks' theorem either $K_4$ (and hence $C_4$) is contained in $G$, or the degree of each point of $G$ is at least four. If the first case occur, we have done. If the second case occur, we also have $C_4$ in $G$ by the following lemma (I omit the proof of this lemma but it's not so difficult).

Lemma. If a graph $G$ with $p$ points has minimum degree $d$ and $d(d-1)>p-1$, then $G$ contains $C_4$.

I can't understand how to use Brooks' Theorem. I only succeed to derive the maximum degree of $G$ is greater than 3. How to derive that the minimum degree of $G$ is greater than 3 from Brooks' Theorem? (If you have other elegant proof of $r(C_4,K_4)=10$, then It also help me.)

supplementation:I got a (awkward?) proof of $r(C_4,K_4)=10$. The proof is like following.

For lower bound, we use chvatal-harary Chvatal-Harary theorem.

For upper bound, we think about above graph $G$. Using $r(C_4,K_3)=7$, easily we have there is no vertex with degree $\le2$. By lemma, we have at least one vertex (say $u$) whose degree is 3. The subgraph induced by vertices non-adjacent to $u$ contains $2K_3$. Then we easily have the neighborhood subgraph of $u$ is $\bar{K_3}$. At now, it's not so difficult to see this is contradiction by thinking about the connection of above $2K_3$ and $\bar{K_3}$.

I'm studying about graph ramsey theory now. Starting this study, I'm reading chvatal and harary's series of papers. In the second paper (V.Chvatal, F.Harary, Generalized ramsey theory for graphs,Ⅲ. Small off-diagonal numbers, Pacific Journal of Mathematics 41, No.2, 1972, pp.335-345), I can't understand the proof of $r(C_4,K_4)=10$. Their proof is like following.

Let $G$ is arbitrary simple graph of order 10 with point independence number $<4$. It is sufficient to prove $G$ contains $C_4$. From $G$'s point independence number is $<4$, $G$'s (point) chromatic number is $\ge4$. Hence by Brooks' theorem either $K_4$ (and hence $C_4$) is contained in $G$, or the degree of each point of $G$ is at least four. If the first case occur, we have done. If the second case occur, we also have $C_4$ in $G$ by the following lemma (I omit the proof of this lemma but it's not so difficult).

Lemma. If a graph $G$ with $p$ points has minimum degree $d$ and $d(d-1)>p-1$, then $G$ contains $C_4$.

I can't understand how to use Brooks' Theorem. I only succeed to derive the maximum degree of $G$ is greater than 3. How to derive that the minimum degree of $G$ is greater than 3 from Brooks' Theorem? (If you have other elegant proof of $r(C_4,K_4)=10$, then It also help me.)

supplementation:I got a (awkward?) proof of $r(C_4,K_4)=10$. The proof is like following.

For lower bound, we use chvatal-harary theorem.

For upper bound, we think about above graph $G$. Using $r(C_4,K_3)=7$, easily we have there is no vertex with degree $\le2$. By lemma, we have at least one vertex (say $u$) whose degree is 3. The subgraph induced by vertices non-adjacent to $u$ contains $2K_3$. Then we easily have the neighborhood subgraph of $u$ is $\bar{K_3}$. At now, it's not so difficult to see this is contradiction by thinking about the connection of above $2K_3$ and $\bar{K_3}$.

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