\magnification 2100 \parindent 0pt \parskip 8pt \nopagenumbers \hsize 7.5truein \vsize 9.9 truein \hoffset -0.45truein \def\u{\vskip -7pt} \def\v{\vskip -5pt} \input ../../../PAPERS/GOLD/psfig Thm 19.1, 2: Comparison of box and product topologies. Let ${\cal B}_\alpha$ be a basis for $X_\alpha$ Basis for the box topology: $\{\Pi U_\alpha ~|~ U_\alpha$ open in $X_\alpha \}$ \centerline{or $\{\Pi B_\alpha ~|~ B_\alpha \in {\cal B}_\alpha \}$} Basis for the product topology: $\{\Pi U_\alpha ~|~ U_\alpha$ open in $X_\alpha$, \rightline{$U_\alpha = X_\alpha$ for all but finitely many $\alpha \}$} or $\{\Pi B_\alpha ~|~ B_{\alpha_i} \in {\cal B}_{\alpha_i}, ~i = 1, ..., n,$ \rightline{$B_\alpha = X_\alpha$ for $\alpha \not= {\alpha_i}, ~i = 1, ..., n \}$} Hence box topology is finer then the product topology Thm 19.3: Let $A_\alpha$ be a subspace of $X_\alpha$. Then $\Pi A_\alpha$ is a subspace of $\Pi X_\alpha$ if both products are given the box topology or if both products are given the product topology. Thm 19.4: If $X_\alpha$ is Hausdorff for all $\alpha$ then $\Pi X_\alpha$ is Hausdorff in both the box and product topologies. HW p. 118: 3, 5, 6, 7 Thm 19.5: $\Pi \overline{A_\alpha} = \overline{\Pi A_\alpha}$ in both the box and product topologies. Thm 19.6: Suppose $f_\alpha: X \rightarrow Y_\alpha$. Define $f: X \rightarrow \Pi_{\alpha \in A} Y_\alpha$ by $f(x) = (f_\alpha(x))_{\alpha \in A}$. Let $\Pi X_\alpha$ have the product topology. Then $f$ is continuous is and only if $f_\alpha$ is continuous $\forall$ $\alpha$ \end Suppose $D^*$ is the condensation of $D= (V, A)$. Let $K_1, ..., K_p$ be the strong components of $D$. Then $\{K_1, ..., K_p\}$ are the vertices of $D^*$. Thm 2.7: $D^*$ is acyclic. Suppose $D*$ contains a cycle $K_{i_1}, ..., K_{i_m}$ where $K_{i_1} = K_{i_m}$. Then there exists $v_{j,b} \in K_{i_j}$, $v_{j+ 1,e} \in K_{i_{j+1}$ such that $(v_{j,b}, v_{j+ 1,e}) \in A$. Since $v_{j, e}, v_{j,b}$ are in the same strong component, $K_{i_j}$, there exists a path $v_{j, e}, u_{j, 1}, ..., u_{j, n_j}, v_{j,b}$ Similarly $v_{1, b}, v_{m, e} are in the same strong component, $K_{i_1}$, there exists a path $v_{m, e}, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$ Thus $v_{1, b}, v_{2, e}, u_{2, 1}, ..., u_{2, n_2}, v_{2, b}, v_{3, e}, u_{3, 1}, ..., u_{3, n_2}, v_{3, b}, ..., v_{m-1, b}, v_{m, e}, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$ is a closed chain in $D$. Hence there exists a path from $v_{1, b}$ to $v_{2, e}$ (namely $v_{1, b}, v_{2, e}$) and there is a path from $v_{2, e}, v_{1, b}$ (namely $v_{2, e}, u_{2, 1}, ..., u_{2, n_2}, v_{2, b}, v_{3, e}, u_{3, 1}, ..., u_{3, n_2}, v_{3, b}, ..., v_{m-1, b}, v_{m, e}, u_{1, 1}, ..., u_{1, n_1}, v_{1,b}$). Thus $v_{2, e} \in K_{i_1}$. But this is a contradiction by thm 2.6 since $v_{2, e} \in K_{i_2}$, $K_{i_1} \not= K_{i_2}$ (since $K_{i_1}, ..., K_{i_m}$ is a cycle). Note: using alot of notation helped us to write the above proof. Often notation can help us in writing a proof and can also help as understand . But proofs using less notation and more English are also valid and sometimes are easier to understand. For example: Thm 2.8: An acyclic digraph, $D$ has a unique vertex basis consisting of all vertices with no incoming arcs. Let $B$ be the set of all vertices with no incoming arcs. If $v \in B$. Then since $v$ has no incoming arcs, $v$ is only reachable from itself. Thus $B$ must be a subset of any vertex base. Suppose $u \not in B$. Let $u_1 = u$. Then $u_1$ has an incoming arc $(u_2, u_1)$. If $u_2 \in B$, then $u_1$ is reachable from a vertex in $B$. If $u_2 \not\in B$ then $u_2$ has an incoming arc $(u_3, u_2)$. Suppose the path $u_n, ..., u_1$ is defined such that all vertices are distinct. If $u_n \in B$, then $u$ is reachable from a vertex in $B$. If $u_n \not\in B$ then $u_n$ has an incoming arc $(u_{n+1}, u_n)$. If $u_{n+1} = u_i$ for some $i = 1, ..., n$, then $u_{n+1}, u_n, ..., u_i$ is a cycle, a contradiction. Hence all the vertices of $u_{n+1}, u_n, ..., u_1$ are distinct. Since the number of vertices of $D$ is finite, this process must eventually end, say with the path $u_t, ..., u_1$. Since we cannot continue this process, $u_t$ must not have any incoming arcs. Hence $u_t \in B$, and hence $u$ is reachable from a vertex in $B$. Thus any vertex basis must be contained in $B$. Hence $B$ is the unique vertex basis of $D$. \end