We know that if $T_1 = \{e_1, \dots, e_n\}$ and $T_2=\{f_1, \dots, f_n\}$ are MST then if we order the edges such as $$w(e_1) \le w(e_2) \le \dots \le w(e_n)$$ $$w(f_1) \le w(f_2) \le \dots \le w(f_n)$$

then $w(e_i)=w(f_i)$ for all $i\in [n]$

This also (trivially) implies that $w(e_1) \le w(f_i)$ for all $i\in [n]$

Does this equality also holds when $T_2$ is a spanning tree but not minimal? I have been trying to prove it for a while and are not succeeding.

Assume that it is not and let $k$ be the first index where $w(f_i) < w(e_i)$ and denote $f=(u,v)$. Let us look at the unique path connecting $f=(u,v)$ in $T_1$ if that path contains any of the $\{e_{k}, \dots, e_n\}$ edges then we can replace that edge with $f_i$ and we get that $T_1$ is not a MST, otherwise that path contains at-least one of the edges $\{e_1, \dots, e_k\}$.

There is where I am stuck, I am trying to say that $\{f_1, \dots, f_{k}\}$ must contain an edge that is safe to add to $T_1$, my idea is that one of the edges in $\{f_1, \dots, f_{k}\}$ must close a loop in the edges $e_{k}, \dots, e_n$ and therefore it is safe to replace it but I can't get it to work.

Appreciate any help.

This also proves that MSTs are invariant under increasing functions (that is, if $f(x)$ is an increasing function that an MST w.r.t $w(E)$ would remain an MST w.r.t $f(w(E))$.

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  • $\begingroup$ Yes, I assume that $T_2$ is still a spanning tree (I tried the approach where I assume that none of the $f_i$ close a loop and show a contradiction; that is $T_2$ is not spanning but also failed) $\endgroup$
    – Rab
    Sep 21 at 12:03

1 Answer 1


The property you speak of is indeed true.

Denote $E = \{e_1, e_2, …, e_{k-1}\}$ and $F = \{f_1, f_2, …, f_k\}$. Note $V'$ the vertices covered by $E$: $V' = \{v\in V\mid \exists e\in E\text{ such that } v\in e\}$.

I will prove that there exists $f\in F\setminus E$ such that $(V, E\cup\{f\})$ is acyclic. That would mean that $w(e_k) \leqslant w(f) \leqslant w(f_k)$.

Suppose that $(V, E\cup\{f\})$ contains a cycle for each $f\in F$.

Denote $C_1, …, C_m$ the connected components of $(V, E)$. Since $(V, E\cup \{f\})$ contains a cycle for all $f\in F$, that means that each $f$ is an edge between two vertices of a certain $C_i$. If we define $E_i = \{e\in E\mid e \text{ is an edge between two vertices of }C_i\}$ and $F_i = \{f\in F\mid f \text{ is an edge between two vertices of }C_i\}$, the pigeonhole principle says that there exists $i\in \{1, …, m\}$ such that $|F_i| > |E_i|$. But since $(C_i, E_i)$ is a tree, that means that $|E_i| = |C_i| - 1$. That implies that $(C_i, F_i)$ contains a cycle, and that $T_2$ contains a cycle.

We conclude by contradiction that $w(e_k) \leqslant w(f_k)$.


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