For two graphs $G,H$, define $G\otimes H$: it has vertex $V(G)\times V(H)$, $(v_1,v_2)(v_1',v_2')\in E(G\otimes H)$ if it satisfies all the following three requirments
(i) $(v_1,v_2)\neq (v_1',v_2')$
(ii) $v_1=v_1'$ or $v_1v_1'\in E(G)$
(iii) $v_2=v_2'$ or $v_2v_2'\in E(H)$
Show that $\alpha(G\otimes H)\le R_2(\alpha(G)+1,\alpha(H)+1)-1$, where $\alpha(G) $ is the order of maximal independent set in $G$.
As we know, $\alpha(G\otimes H)\ge\alpha(G)\alpha(H)$, since if $A,B$ are independent sets in $G,H$ respectively, then $A\times B$ is an independent set in $G\otimes H$. I want to show by contradiction: if there is an independent set $T$ in $G\otimes H$ has order at least $max\{R_2(\alpha(G)+1,\alpha(H)+1), \alpha(G)\alpha(H)\}$, then get a contradiction. But from the constraint of $R_2(\alpha(G)+1,\alpha(H)+1)$, of course there is no $K_{\alpha(G)+1}$ in $T$, then there is a independent $(\alpha(H)+1)$-set in $T$. I want to project $T$ to first or second coordinate. But I think it is not strong enough. May I get some help?
Suppose that $T$ is an independent set in $G\otimes H$ of cardinality $n=R_2\big(\alpha(G)+1,\alpha(H)+1\big)$. Let $T=\{\langle u_k,v_k\rangle:k=1,\ldots,n\}$. Let
$$P=\{\langle k,\ell\rangle:1\le k<\ell\le n\text{ and }u_k\ne u_\ell\text{ and }u_ku_\ell\notin E(G)\}$$
and
$$Q=\{\langle k,\ell\rangle:1\le k<\ell\le n\text{ and }v_k\ne v_\ell\text{ and }v_kv_\ell\notin E(H)\}\;;$$
since $T$ is independent, $P\cup Q=\{\langle k,\ell\rangle:1\le k<\ell\le n\}$. The definition of $n$ ensures that either there is $N\subseteq[n]=\{1,\ldots,n\}$ such that $|N|=\alpha(G)+1$ and
$$\{\langle k,\ell\rangle\in N\times N:k<\ell\}\subseteq P\;,$$
or there is $N\subseteq[n]$ such that $|N|=\alpha(H)+1$ and
$$\{\langle k,\ell\rangle\in N\times N:k<\ell\}\subseteq Q\;.$$
Without loss of generality we may assume the former. Now verify that $\{u_k:k\in N\}$ is an independent set of cardinality $\alpha(G)+1$ in $G$, which is of course impossible.