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Show that the two interpretations of the components of a graph are equivalent: (1) a connected subgraph of G that is not a proper subgraph
Show that the two interpretations of the components of a graph are equivalent: (1) a connected subgraph of G that is not a proper subgraph of any other connected subgraph of G (2) a subgraph of G induced by the vertices in an equivalence class resulting from the equivlance relation *defined in attached image*
or a graph G, a component of G has been defined as (1) a connected sub- graph of G that is not a proper subgraph of any other connected subgraph or G and has been described as (2) a subgraph of G induced by the ver- tices in an equivalence class resulting from the equivalence relation defined in Theorem 1.7. Show that these two interpretations of components are equivalent. Theorem 1.7 Let R be the relation defined on the verter set of a graph G by u R v, where u,v E V(G), if u is connected to v, that is, if G contains au-v path. Then R is an equivalence relation. Proof. It is immediate that R is reflexive and symmetric. It remains therefore only to show that R is transitive. Let u, v, w E V(G) such that u R v and v R w. Hence G contains a u v path P and a v+ w path P". As we have seen earlier, following P' by P" produces a u - w walk W. By Theorem 1.6, G contains a u- w path and so u R w. or a graph G, a component of G has been defined as (1) a connected sub- graph of G that is not a proper subgraph of any other connected subgraph or G and has been described as (2) a subgraph of G induced by the ver- tices in an equivalence class resulting from the equivalence relation defined in Theorem 1.7. Show that these two interpretations of components are equivalent. Theorem 1.7 Let R be the relation defined on the verter set of a graph G by u R v, where u,v E V(G), if u is connected to v, that is, if G contains au-v path. Then R is an equivalence relation. Proof. It is immediate that R is reflexive and symmetric. It remains therefore only to show that R is transitive. Let u, v, w E V(G) such that u R v and v R w. Hence G contains a u v path P and a v+ w path P". As we have seen earlier, following P' by P" produces a u - w walk W. By Theorem 1.6, G contains a u- w path and so u R wStep by Step Solution
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