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enter image description here I understood that $T$ is one of the spanning trees which has disjoint edges compared to other $k-1$ edges. then, $T$ has $|V(G)|-1$ edges.

What does the author say in the underlined statement? Isn't it a trivial thing?we know that $|\mathscr P-1|\leq|V(G)|-1$. Is it follow from this inequality and pigeonhole principle? How do I prove the converse?Where will I get the converse? I draw the example for the illustration of the theorem,enter image description here

dark blue edged and narrow red edged trees are two edge disjoint trees, what does the theorem say?

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  • $\begingroup$ What book is this from? $\endgroup$ – Shaun Sep 22 '18 at 9:28
  • $\begingroup$ The author meant that there are at least $|\mathscr{P}|-1=p-1$ edges of $T$ that connects vertices from different classes in $\mathscr{P}$. This is obvious if you collapse each $V_i$ to a point. $\endgroup$ – user10354138 Sep 22 '18 at 9:49
  • $\begingroup$ @shaun Textbook of graph theory Balakrishnan and Ranganathan $\endgroup$ – Math geek Sep 22 '18 at 13:10
  • $\begingroup$ T is a spanning tree, so graph uses every verices of G, each vertices are connected, so, it must have n-1 edges. each edge, one end is in any one of the $V_j$. so it must have p-1 edges of T . right?@user10354138 $\endgroup$ – Math geek Sep 23 '18 at 10:32
  • $\begingroup$ Where can I find the converse? $\endgroup$ – Math geek Sep 23 '18 at 10:34
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The highlighted sentence would be better if it said: whenever $V_1, V_2, \dots, V_p$ is a partition of $G$, and $T$ is a spanning tree of $G$, $T$ must contain at least $p-1$ edges joining distinct parts.

One way to prove this claim is: if we delete all edges of $T$ that join distinct parts among the $V_1, V_2, \dots, V_p$, then $T$ becomes a forest with $p$ components (one on each $V_i$). But deleting one edge can only increase the number of components of a graph by at most $1$, so at least $p-1$ edges must be deleted to get from $1$ to $p$ components.

Another way, as suggested in the comments: consider the graph $T^*$ on $p$ vertices $\{v_1, v_2, \dots, v_p\}$ with an edge $v_iv_j$ whenever $T$ has an edge between $V_i$ and $V_j$. $T^*$ must be connected: if there were no way to get from $\{v_1, \dots, v_k\}$ to $\{v_{k+1}, \dots, v_p\}$ in $T^*$, there would be no way to get from $V_1 \cup \dots \cup V_k$ to $V_{k+1} \cup \dots\cup V_p$ in $T$. So $T^*$ has at least $p-1$ edges.

Once the claim is done, take the $k$ edge-disjoint spanning trees $T_1, T_2, \dots, T_k$ and apply the claim to them. Each $T_i$ has $p-1$ edges joining distinct parts of the partition $V_1, V_2, \dots, V_p$, and these must be different edges for each $T_i$ (because they're edge-disjoint). Altogether, we get $k(p-1)$ edges joining distinct parts, which was what we wanted.

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  • $\begingroup$ How do I prove the converse? $\endgroup$ – Math geek Sep 23 '18 at 16:11
  • $\begingroup$ @Mathgeek I don't know; it seems hard. Have you tried looking up the citations Tutte [181] or Nash-Williams [145] in your book? They should correspond to papers cited at the end of the book or at the end of the chapter. $\endgroup$ – Misha Lavrov Sep 23 '18 at 16:22
  • $\begingroup$ proof is not given in the textbook. from where will i get this paper? $\endgroup$ – Math geek Sep 24 '18 at 0:51
  • $\begingroup$ Tutte [181] or Nash-Williams [145], what does that mean? $\endgroup$ – Math geek Sep 24 '18 at 0:52
  • $\begingroup$ It means that somewhere in your textbook, there is a list of papers cited, and number 181 on this list is a paper by Tutte that proves this theorem. The paper may or may not be available online, but without knowing what the citation is, nobody else can say. $\endgroup$ – Misha Lavrov Sep 24 '18 at 1:15

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