2.3: Communication Network Question: What (and who) are the minimum number of people you need to give a message in order for

the message to reach everyone one else in the network.

Answer: Find a vertex basis. Defn: A vertex basis of G = (V, A) is a sertof vertices B ⊂ V such that

(1) Every vertex in V is reachable form some vertex in B.

(2) If C ⊂ B then there exists a v ∈ V such that v is not reachable from any vertex in C. Defn: A strong component of G is a strongly connected subgraph D generated by a set of vertices such

that this set of vertices is maximal with respect to strong connectedness.

Thm 2.6: Every vertex is in exactly one strong component. Defn: G∗ = (V ∗, A∗)isthecondensationofG = (V, A)if

(1) V ∗ = {K_{1}, ..., K_{p}} = thesetofstrongcomponentsofG.

(2) There exists and arc (K_{i}, K_{j}) iff there exists u ∈ K_{i} and v ∈ K_{j} such that (u, v) ∈ A. Thm 2.7: G∗ is acyclic (i.e. G does not contain any cycles). Thm 2.8: An acyclic digraph has a unique vertex base consistin of all vertices with no incoming arcs. Cor: In an acyclic digraph, there exists a vertex with no incoming arcs. Thm 2.9: Let B∗ = {K_{i_{1}}, ..., K_{i_{n}}} be the vertex basis of G∗ where G∗ is the condensation of G. Then

the set {v_{1}, ..., v_{n}} is a vertex basis of G if v_{j} ∈ K_{i_{j}}. Furthermore, every vertex basis of G can be formed in this way.

2.4: Digraphs and matrices Recall if A = (a_{ij}) and B = (b_{ij}) are two n × n matrices, then

A + B = (a_{ij} + b_{ij})

A × B = (a_{ij}b_{ij})
AB = (c_{ij}) where c_{ij} = Σ_{k=1}^{n}a_{ik}b_{kj} = row(i) · 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} = \{ \begin{matrix}1 & if (v_{i}, v_{j}) ∈ A \\ 0 & otherwise \end{matrix} .

Thm 2.11: If A is the adjacency matrix of D, then the i, jth 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} = \{ \begin{matrix}1 & if v_{j} is reachable from v_{i} \\ 0 & otherwise \end{matrix} .

Defn: The Boolean function b : M → \{0, 1\} where M = \{0, 1, 2, ...\} is defined by b(x) = \{ \begin{matrix}0 & if x = 0 \\ 1 & otherwise \end{matrix}. B : set of matrices with nonnegative integer entries → set of matrices with entries 0 and 1isdefinedbyB(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

R = B(I + A + A^{2} + ..._{A}^{n−1}) = B((I + A)^{n−1})

Defn: I = the identiy matrix = (i_{kp}) where i_{kp} = \{ \begin{matrix}0 & if k ≠ p \\ 1 & if k = p \end{matrix}

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 × 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 × X. Need to prove D is closed in X × X. Question: How do you prove a set is closed? a.) Show X × X − D is open. b.) Show D′ ⊂ 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 × X iff X × X − D is open in X × X.

Concern: X Hausdorff gives open sets in X, but we need open set in X × 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 × X has the product topology, thus we can create the needed open set in X × 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 × X − D is open in X × X. There are many ways to show a set is open. We’ll try the definition.

Take (x, y) ∈ X × X − D (note since it often helps to be specific, I didn’t take a point p in X × X − D, but instead took an ordered pair (x, y)).

We need to find a set W open in X × X such that (x, y) ∈ W ⊂ X × X − D OR we could look for a basis element U × V where U and V are open in X such that (x, y) ∈ U × V ⊂ X × 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 ∈ U, y ∈ V, and U × V ⊂ X × 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) ∈ X ×X −D, x, y are points in X. Are they distinct? Since (x, y) ∉ D = {(a, a) | a ∈ X}, x ≠ y. Hence x, y are two distinct points in X. Hence there exists U, V open in X such that x ∈ U, u ∈ V, U ∩ V = ∅.

Thus (x, y) ⊂ U × V. Is U × V ⊂ X × X − D. Suppose (u, v) ∈ U × V. We have not yet used U ∩ V = ∅. u ∈ U, v ∈ V, U ∩ V = ∅ implies u ≠ v. Hence (u, v) ∉ D. Thus (u, v) ∈ X × X − D.

Now you know how to relate X Hausdorff to D is closed in X × X to prove X Hausdorff implies D is closed in X ×X. Perhaps this relationship can also be used to prove D closed in X ×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 ∈ X so we could apply Hausdorff. We used the definition of Hausdorff to find the U × V that we needed.

Thus the proof for this direction is:

Take (x, y) ∈ X × X − D. Since (x, y) ∉ D, x ≠ y. Thus there exists U, V open in X such that x ∈ U, u ∈ V, U ∩ V = ∅. Thus (x, y) ⊂ U × V ⊂ X × X − D. Hence X × X − D is open in XX. Thus D is closed in X × X.

——————————————————————————– Thm 19.1, 2: Comparison of box and product topologies. Let B_{α} be a basis for X_{α} Basis for the box topology: {ΠU_{α} | U_{α} open in X_{α}} or {ΠB_{α} | B_{α} ∈ B_{α}} Basis for the box topology: {ΠU_{α} | U_{α} open in X_{α}, U_{α} = X_{α} for all but finitely many α} or {ΠB_{α} | B_{α_{i}} ∈ B_{α_{i}}, i = 1, ..., n, B_{α} = X_{α} for α ≠ α_{i}, i = 1, ..., n} Hence box topology is finer then the product topology Thm 19.3: Let A_{α} be a subspace of X_{α}. Then ΠA_{α} is a subspace of ΠX_{α} if both products are given

the box topology or if both products are given the subspace topology.

Thm 19.4: If X_{α} is Hausdorff for all α then ΠX_{α} is Hausdorff in both the box and product topologies. HW p. 118: 3, 5, 6, 7 Thm 19.5: Π\overline{A_{α}} = \overline{ΠA_{α}}inboththeboxandproducttopologies. Thm 19.6: Suppose f_{α} : X → Y_{α}. Define f : X → Π_{α∈A}Y_{α} by f(x) = (f_{α}(x))_{α∈A}