\magnification 2200 \parindent 0pt \parskip 5pt \nopagenumbers \hsize 7.5truein \vsize 9.9 truein \hoffset -0.45truein \def\u{\vskip -7pt} \def\v{\vskip -5pt} \input ../../../PAPERS/GOLD/psfig 2.4: Digraphs and matrices Recall if $A = (a_{ij}) $ and $B = (b_{ij}) $ are two $n \times n$ matrices, then \centerline{$A + B = ( a_{ij} + b_{ij} )$ } \centerline{$A \times B = ( a_{ij} b_{ij} )$ } \centerline{$AB = ( c_{ij} ) $ where $c_{ij} = \Sigma_{k=1}^n a_{ik}b_{kj}$ = $row(i) \cdot column(j)$} The following refer to the digraph $D = (V, A)$ where $V = (v_1, ..., v_n)$ where the vertices have not been given a particular order. Defn: The adjacency matrix of $D = (V, A) = A(D) = (a_{ij})$ where $a_{ij} = \cases{1 & if $(v_i, v_j) \in A$ \cr 0 & otherwise}$. Thm 2.11: If $A$ is the adjacency matrix of $D$, then the $i, j$th entry of $A^k$ is the number of paths of length $k$ from $v_i$ to $v_j$. Defn: The reachability matrix of $D = R(D) = (r_{ij})$ where $r_{ij} = \cases{1 & if $v_j$ is reachable from $v_i$ \cr 0 & otherwise}$. Defn: The Boolean function $b: {\cal M} \rightarrow \{0, 1\}$ where ${\cal M} = \{0, 1, 2, ... \}$ is defined by $b(x) = \cases{0 & if $x = 0$ \cr 1 & otherwise}$. $B: $ set of matrices with nonnegative integer entries $ \rightarrow $ set of matrices with entries 0 and 1 is defined by $B(A) = ( b(a_{ij} ) )$. Thm 2.12: Suppose $A$ is the adjacency matrix and $R$ is the reachability matrix of the digraph $D$. If $D$ has $n$ vertices, then \centerline{$R = B(I + A + A^2 + ... _ A^{n-1}) = B( (I + A)^{n-1} )$} Defn: $I =$ the identiy matrix = $(i_{kp})$ where $i_{kp} = \cases{0 & if $k \not= p$ \cr 1 & if $k = p$}$ $J = (j_{kp})$ where $j_{kp} = 1$ for all $k, p$. $R' =(r_{ji})$ is the transpose of $R = (r_{ij})$. Thm 2.13 (Ross and Harary 1959). Suppose $A$ is the adjacency matrix and $R$ is the reachability matrix of the digraph $D$. Let $n =$ the number of vertices of $D$. Then (a.) $D$ is strongly connected if and only if $R = J$. (b.) $D$ is unilaterally connected if and only if $B(R + R') = J$. (c.) $D$ is weakly connected if and only if $B( I + A + A' )^{n-1}) = J$. Thm 2.14. Suppose $R$ is the reachability matrix of the digraph $D$. Let $R \times R' = (t_{ij})$ Let $R^2 = (s_{ij})$. Then (a.) $v_j$ is in the strong component containing the vertex $v_i$ if and only if $t_{ij} = 1$. (b.) The number of vertices in the strong component $u_i$ is $s_{ii}$. Defn: The distance matrix of $D$ the matrix $(d_{ij})$ where $d_{ij}$ is the distance from vertex $v_i$ to $v_j$. Thm 2.15: $d_{ij}$ is the smallest number $k$ such that $B(A^k) = 1$. HW: 2.3: 1, 2, 3, 4, 6, 10, 11, 18; 2.4: 1, 2, 3, 4, 5, 7, 8, 12 (13 or 14 or 15). --------------------------------------------------------------------------------- Thm: If $X$ is Hausdorff, then the diagonal, $D$, is closed in $X \times X$. Need to prove $D$ is closed in $X \times X$. Question: How do you prove a set is closed? a.) Show $X \times X - D$ is open. b.) Show $D' \subset D$. c.) Show $D = \overline{D}$. d.) Show $D$ is a finite union of closed sets or an arbitrary union of open sets. e.) etc. Question: What is given? If $X$ is Hausdorff. Hence a.) for every pair of points in $X$ there exists a pair of disjoint open sets containing them (on exams give the precise definition). b.) Every finite point set in $X$ is closed. c.) 17.9 involving limit points, 17.10 involving sequences, etc. Question: How can we relate what is given to what we need to prove. Which defintions/theorems involving what we need to prove are most related to which definitions/theorems of what is given. The defintion of Hausdorff gives open sets (in X). Hence we will try focusing on the definition $D$ closed in $X \times X$ iff $X \times X - D$ is open in $X \times X$. Concern: $X$ Hausdorff gives open sets in $X$, but we need open set in $X \times X$. We will ignore this concern for either of the following two reasons: 1.) We have to try something. Might as well give this a try. Working blindly with little motivation as to why this might work may be disconcerting, BUT if often results in the correct answer. If you don't know what to do, try something (hopefully relating what is given to what you need to prove even if weakly related). When successful, look back over your proof and see if you can figure out why it worked and how you could have come up with it with more motivation. 2.) $X \times X$ has the product topology, thus we can create the needed open set in $X \times X$ by taking the product of two open sets in $X$. Since $X$ Hausdorff gives us two open sets in $X$, this is excellent motivation. Thus we will try to show $X \times X - D$ is open in $X \times X$. There are many ways to show a set is open. We'll try the definition. Take $(x, y) \in X \times X - D$ (note since it often helps to be specific, I didn't take a point $p$ in $X \times X - D$, but instead took an ordered pair $(x, y)$). We need to find a set $W$ open in $X \times X$ such that $(x, y) \in W \subset X \times X - D$ OR we could look for a basis element $U \times V$ where $U$ and $V$ are open in $X$ such that $(x, y) \in U \times V \subset X \times X - D$ (note usually it is easier to find a basis element because you know what basis elements look like, but keep in mind that not all open sets are basis elements). Looking for a basis element is particularly useful for this problem since Hausforff gives us open sets in $X$ and we are now looking for $U$, $V$ open in $X$ such that $x \in U$, $y \in V$, and $U \times V \subset X \times X - D$. We will now try to apply the hypothesis that $X$ is Hausdorff. To apply Hausforff, we need two distinct points in $X$, preferably ones related to what we are tyring to prove. Since $(x, y) \in X \times X - D$, $x, y$ are points in $X$. Are they distinct? Since $(x, y) \not\in D = \{(a, a) ~|~ a \in X \}$, $x \not= y$. Hence $x, y$ are two distinct points in $X$. Hence there exists $U, V$ open in $X$ such that $x \in U$, $u \in V$, $U \cap V = \emptyset$. Thus $(x, y) \subset U \times V$. Is $U \times V \subset X \times X - D$. Suppose $(u, v) \in U \times V$. We have not yet used $U \cap V = \emptyset$. $u \in U, ~v \in V$, $U \cap V = \emptyset$ implies $u \not= v$. Hence $(u, v) \not\in D$. Thus $(u, v) \in X \times X - D$. Now you know how to relate $X$ Hausdorff to $D$ is closed in $X \times X$ to prove $X$ Hausdorff implies $D$ is closed in $X \times X$. Perhaps this relationship can also be used to prove $D$ closed in $X \times X$ implies $X$ Hausdorff. Sometimes a different relationship is needed to prove the other direction. Note this proof was actually an "easy" proof. All it used was definitions. We used the definition of closed to start the proof and give us the two distinct points $x, y \in X$ so we could apply Hausdorff. We used the definition of Hausdorff to find the $U \times V$ that we needed. Thus the proof for this direction is: Take $(x, y) \in X \times X - D$. Since $(x, y) \not\in D$, $x \not= y$. Thus there exists $U, V$ open in $X$ such that $x \in U$, $u \in V$, $U \cap V = \emptyset$. Thus $(x, y) \subset U \times V \subset X \times X - D$. Hence $X \times X - D$ is open in $X \times X$. Thus $D$ is closed in $X \times X$. -------------------------------------------------------------------------------- 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 $X_\alpha \}$ or $\{\Pi B_\alpha ~|~ B_\alpha \in {\cal B}_\alpha \}$ Basis for the box topology: $\{\Pi U_\alpha ~|~ U_\alpha$ open in $X_\alpha$, $U_\alpha = X_\alpha$ for all but finitely many $\alpha \}$ or $\{\Pi B_\alpha ~|~ B_{\alpha_i} \in {\cal B}_{\alpha_i}, ~i = 1, ..., n , B_\alpha = X_\alpha$ 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 subspace 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 $f(x) = (f_\alpha(x))_{\alpha \in A}$ \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 Dear Viveca, Just got your letter and am excited that you and your family are planning to visit Iowa. We would be happy to have you stay with us for as long as your time allows. I will probably be visiting Texas in June and perhaps the beginning of July, so after July ? would be best for me. If you can only make it before July , mother will stay in Iowa, so you are welcome anytime. My schedule is not yet set, but one of my students is defending her thesis this summer, so it will not be very flexible. But I will definintely be free after July . Last summer we bought a badmitton set, but have not had time to use it as we have spent our time unpacking boxes and replacing the upstairs carpet with laminate flooring. Mother and I are doing the work ourselves so it is taking longer than expected. But I think it will look good in a couple of months when we have finished the upstairs. Halsinger