Since $D_{+}(f) = \text{Proj}(S) \cap D(f)$, these sets are open. This proves (1). Also (2) follows as $D(ff') = D(f) \cap D(f')$. Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(I)$ are closed. This proves (8). \medskip\noindent Suppose that $T \subset \text{Proj}(S)$ is closed. Then we can write $T = \text{Proj}(S) \cap V(J)$ for some ideal $J \subset S$. By definition of a homogeneous ideal if $g \in J$, $g = g_0 + \ldots + g_m$ with $g_d \in S_d$ then $g_d \in \mathfrak p$ for all $\mathfrak p \in T$. Thus, letting $I \subset S$ be the ideal generated by the homogeneous parts of the elements of $J$ we have $T = V_{+}(I)$. This proves (9). \medskip\noindent The formula for $\text{Proj}(S) \cap D(g)$, with $g \in S$ is direct from the definitions. This proves (3). Consider the formula for $\text{Proj}(S) \cap D(g_0)$. The inclusion of the right hand side in the left hand side is obvious. For the other inclusion, suppose $g_0 \not \in \mathfrak p$ with $\mathfrak p \in \text{Proj}(S)$. If all $g_0f \in \mathfrak p$ for all homogeneous $f$ of positive degree, then we see that $S_{+} \subset \mathfrak p$ which is a contradiction. This gives the other inclusion. This proves (4). \medskip\noindent The collection of opens $D(g) \cap \text{Proj}(S)$ forms a basis for the topology since the standard opens $D(g) \subset \Spec(S)$ form a basis for the topology on $\Spec(S)$. By the formulas above we can express $D(g) \cap \text{Proj}(S)$ as a union of opens $D_{+}(f)$. Hence the collection of opens $D_{+}(f)$ forms a basis for the topology also. This proves (5). \medskip\noindent Proof of (6). First we note that $D_{+}(f)$ may be identified with a subset (with induced topology) of $D(f) = \Spec(S_f)$ via Lemma \ref{lemma-standard-open}. Note that the ring $S_f$ has a $\mathbf{Z}$-grading. The homogeneous elements are of the form $r/f^n$ with $r \in S$ homogeneous and have degree $\deg(r/f^n) = \deg(r) - n\deg(f)$. The subset $D_{+}(f)$ corresponds exactly to those prime ideals $\mathfrak p \subset S_f$ which are $\mathbf{Z}$-graded ideals (i.e., generated by homogeneous elements). Hence we have to show that the set of $\mathbf{Z}$-graded prime ideals of $S_f$ maps homeomorphically to $\Spec(S_{(f)})$. This follows from Lemma \ref{lemma-Z-graded}. \medskip\noindent Let $S = \mathbf{Z}[X_1, X_2, X_3, \ldots]$ with grading such that each $X_i$ has degree $1$. Then it is easy to see that $$ \text{Proj}(S) = \bigcup\nolimits_{i = 1}^\infty D_{+}(X_i) $$ does not have a finite refinement. This proves (7). \medskip\noindent Let $I \subset S$ be a graded ideal. If $\sqrt{I} \supset S_{+}$ then $V_{+}(I) = \emptyset$ since every prime $\mathfrak p \in \text{Proj}(S)$ does not contain $S_{+}$ by definition. Conversely, suppose that $S_{+} \not \subset \sqrt{I}$. Then we can find an element $f \in S_{+}$ such that $f$ is not nilpotent modulo $I$. Clearly this means that one of the homogeneous parts of $f$ is not nilpotent modulo $I$, in other words we may (and do) assume that $f$ is homogeneous. This implies that $I S_f \not = S_f$, in other words that $(S/I)_f$ is not zero. Hence $(S/I)_{(f)} \not = 0$ since it is a ring which maps into $(S/I)_f$. Pick a prime $\mathfrak q \subset (S/I)_{(f)}$. This corresponds to a graded prime of $S/I$, not containing the irrelevant ideal $(S/I)_{+}$. And this in turn corresponds to a graded prime ideal $\mathfrak p$ of $S$, containing $I$ but not containing $S_{+}$ as desired. This proves (10) and finishes the proof.