If $R = k[T_0,\cdots,T_n]/I$ is integral, then $\text{dim Proj R}=\text{dim D}_+(T_i)$ for some $i$ Why is it that if $R = k[T_0,\cdots,T_n]/I$ is integral, $k$ a field, then $\text{dim Proj R}=\text{dim D}_+(T_i)$ for some $i$?
$D_+(T_i)$ is just the set of primes of $\text{Proj R}$ which don't contain $T_i$
$V_+(T_i)$ is just the set of primes of $\text{Proj R}$ which contain $T_i$
I know that the fact that $R$ is integral means $(0)$ is a prime ideal, so $V_+(0)= \text{Proj R}$ is irreducible. In particular, open sets are dense in it and also irreducible, and $\text{Proj R}= \cup_{i=0}^nD_+(T_i)$
The assertion $\text{dim Proj R}=\text{dim D}_+(T_i)$ essentially means that for some $i$, the largest ascending chain of irreducible closed sets of $\text{Proj R}$ (minus $\text{Proj R}$ itself) entirely lies in some $D_+(T_i)$, but why? How can I be sure my next-to-last irreducible closed set in the ascending chain won't just overlap with several of the $D_+(T_i)$? I know for sure that it is not going to contain any of the $D_+(T_i)$ by virtue of their density inside $\text{Proj R}$ but this does not allow me to conclude that it lies inside one
 A: $\newcommand{\Proj}{\mathrm{Proj}}$
$\newcommand{\dim}{\mathrm{dim}}$
If you take a maximal chain of closed, irreducible subsets $\Proj(R) = Z_0 \supsetneq Z_1 \supsetneq \cdots \supsetneq Z_d = \{x\}$, then $x \in D_+(T_i)$ for some $i$, since (as you've said) these open sets give a cover. Now check that $Z_j \cap D_+(T_i)$ is non-empty for all $0 \le j \le d$. What does this say about the dimension of $D_+(T_i)$?
Edit/partial solution: This does get kind of involved, working from basic definitions of the Zariski topology, so I'll provide a partial solution building off my answer.
As discussed in the comments, we want to show that the $Z_j \cap D_+(T_i)$ is irreducible as a closed subset of $D_+(T_i)$, and that this chain is still strictly decreasing, which will imply $\dim(D_+(T_i)) \ge \dim(\Proj(R))$. It's also true that $\dim(D_+(T_i)) \le \dim(\Proj(R))$, by just taking closures of a maximal chain of closed subsets of $D_+(T_i)$. To simplify notation somewhat, let's fix $D = D_+(T_i)$ for some $i$ such that $x \in D_+(T_i)$. Let $V = V_+(T_i)$ for the same $i$. 
Suppose $Z_j \cap D = Z_{j,1} \cup Z_{j,2}$ for two closed subsets of the subspace $D$, considered with the induced/subspace topology. We want to show $Z_j \cap D = Z_{j,1}$ or $Z_{j,2}$. Since the topology of $D$ is induced by that of $\Proj(R)$, there must be two closed subsets of $\Proj(R)$, call them $C_{j,1}$ and $C_{j,2}$, such that $C_{j,m} \cap D = Z_{j, m}$ for $m = 1,2$. As stated in OP's question, $D$ is dense in $\Proj(R)$, so the fact that $Z_j \cap D$ is non-empty implies this is dense in $Z_j$, meaning the closure is $Z_j$. To see this: $Z_j$ is a closed subset of $\Proj(R)$ containing $Z_j \cap D$, so it must contain the closure of this set. However, if $C := \overline{Z_j \cap D}$ is not all of $Z_j$, then $Z_j = C \cup (Z_j \cap V)$ expresses $Z_j$ as the union of two proper, non-empty closed subsets (remember $V$ is closed in $\Proj(R)$!). This contradicts that $Z_j$ was irreducible. Therefore, $Z_j \cap D$ is dense in $Z_j$. 
Thus, if $Z_j \cap D = Z_{j,1} \cup Z_{j,2}$, then by the density just proven, we see that $\overline{Z_{j,1}} \cup \overline{Z_{j,2}} = Z_j$. Since $Z_j$ is irreducible, one of the $\overline{Z_{j,m}}$ must be $Z_j$; assume $\overline{Z_{j,1}} = Z_j$. Then $Z_{j,1} = Z_j \cap D$, and $Z_j \cap D$ is proven to be irreducible. 
Showing the chain is strictly descending should be less technical, and can be done using similar ideas. 
