Show that there's a unique minimum spanning tree if all edges have different costs

Question

Show that there's a unique minimum spanning tree (MST) in case the edges' weights are pairwise different $(w(e)\neq w(f) \text{ for } e\neq f)$.

I thought that the proof can be done for example by contradiction, saying that we have $2$ different MST meaning that somewhere was possible to pick from more edges, so $w(e) = w(f)$ for $e\neq f$, contradiction. Apparently this is not correct.

How would you show that a graph has a unique MST if all edges have distinct weights?

Answer 1

If $T_1$ and $T_2$ are distinct minimum spanning trees, then consider the edge of minimum weight among all the edges that are contained in exactly one of $T_1$ or $T_2$. Without loss of generality, this edge appears only in $T_1$, and we can call it $e_1$.

Then $T_2 \cup \{ e_1 \}$ must contain a cycle, and one of the edges of this cycle, call it $e_2$, is not in $T_1$.

Since $e_2$ is a edge different from $e_1$ and is contained in exactly one of $T_1$ or $T_2$, it must be that $w ( e_1 ) < w ( e_2 )$. Note that $T = T_2 \cup \{ e_1 \} \setminus \{ e_2 \}$ is a spanning tree. The total weight of $T$ is smaller than the total weight of $T_2$, but this is a contradiction, since we have supposed that $T_2$ is a minimum spanning tree.

Answer 2

The best explanation I found was on wikipedia, Proof:

1. Assume the contrary, that there are two different MSTs $A$ and $B$.
2. Since $A$ and $B$ differ despite containing the same nodes, there is at least one edge that belongs to one but not the other. Among such edges, let e1 be the one with least weight; this choice is unique because the edge weights are all distinct. Without loss of generality, assume $e1$ is in $A$.
3. As $B$ is a MST, $\{e1\}\cup B$ must contain a cycle $C$.
4. As a tree, $A$ contains no cycles, therefore $C$ must have an edge $e2$ that is not in $A$.
5. Since $e1$ was chosen as the unique lowest-weight edge among those belonging to exactly one of $A$ and $B$, the weight of $e2$ must be greater than the weight of $e1$.
6. Replacing $e2$ with $e1$ in $B$ therefore yields a spanning tree with a smaller weight.
7. This contradicts the assumption that $B$ is a MST.

More generally, if the edge weights are not all distinct then only the (multi-)set of weights in minimum spanning trees is certain to be unique; it is the same for all minimum spanning trees [1].

Answer 3

This is a (slightly) more detailed version of the currently accepted answer

(For the sake of contradiction) Let $T_1 = (V, E_1)$ and $T_2 = (V, E_2)$ be two distinct MSTs of the graph $G = (V, E)$
Note that both have the same vertex set $V$ since both are spanning trees of $G$

Consider the set $E_{\Delta} = E_1 \triangle E_2$
Let $e = (u, v)$ be the edge in $E_{\Delta}$ having the least cost (or weight)
Note that since all costs are unique, and $E_{\Delta}$ is non-empty, $e$ must be unique.
Without loss of generality, assume $e \in E_1$

Now, there must be a path $P$, with $e \notin P$, in $T_2$ connecting $u$ and $v$, since trees are by definition, connected.
Note that at least one edge (say $e'$) that occurs in $P$ must not be in $E_1$, thus, $e' \in E_{\Delta}$
This is because, if $P \subset E_1$, $T_1$ will contain a cycle formed by the path $P$ and the edge $e$, this leads to a contradiction, since trees by definition are acyclic.
Note that by definition of $e$ and the fact that all costs are distinct, $\text{cost}(e) < \text{cost}(e')$

Now, consider $T_2' = (V, E_2')$, where $E_2' = (E_2 \backslash\{e'\})\cup\{e\}$, this has a strictly lesser total cost than $T_2$
As $T_2$ was a MST, this leads to a contradiction.