If all edge weights are distinct, then the MST is unique

[CGandC]

Homework Statement

This is not a homework question, I thought about it myself and was wondering if the uniqueness of an MST is defined in terms of edge uniqueness or weight uniqueness

Relevant Equations

- Assume graph is connected and undirected with weight function w: E ->R
- MST = minimum spanning tree

I was having trouble understanding what "uniqueness" means when we talk about MSTs, here's the theorem and its proof:

( Source: Jeff Erickson, Algorithms. https://jeffe.cs.illinois.edu/teaching/algorithms/book/07-mst.pdf )

I don't fully understand what "unique" stands for when talking about MSTs. I thought it meant that all the edges of the different MSTs must be the same, not only the weights.

To clarify:
I know that the multiset of weights of all MSTs of a graph must be equal, so if there's an MST with all weights unique then it must be so in every other MST; the question is - are the edges also unique?
If the answer's 'yes' , then how do we conclude from the proof above that the edges must be equal in $T_1$ and $T_2$?
If the answer's 'no', then what is a necessary and sufficient condition for all the edges of MSTs to be equal? ( i.e. the MSTs to be unique in terms of the edges themselves )

From the proof above it seems uniqueness is defined in terms of weight of edges, but it feels to me a little bit odd, hence I'm asking the question to have a second opinion on this.

Thanks in advance for any kind of help!

 

[FactChecker]

The weights are defined by the graph, not by the spanning tree, so it does not make sense to talk about the weights of the spanning trees being different.
Furthermore, the second paragraph of the proof states "Each of our spanning trees must contain an edge that the other tree omits." This confirms that an edge difference is required for two spanning trees to be different.

 

[CGandC]

The weights are defined by the graph, not by the spanning tree, so it does not make sense to talk about the weights of the spanning trees being different.

Agreed, it's known that the multiset of edge weights of every MST are the same, hence if all the weights of one MST are unique, so are for the other MST.

Furthermore, the second paragraph of the proof states "Each of our spanning trees must contain an edge that the other tree omits." This confirms that an edge difference is required for two spanning trees to be different.

That's correct, if we'll assume otherwise we'll have that all edges in $T$ and $T'$ are the same, hence $T = T'$ which is a contradiction to the assumption that $T \neq T'$
But I don't understand where's the contradiction in the above proof if we consider the only case left which is "Each of our spanning trees must contain an edge that the other tree omits."

It seems to me the proof above only showed that if we have two different msts where the edge weights are unique, then we can swap an edge from both trees that have the same weight; where's the contradiction here to the assumption that $T \neq T'$?

 

[FactChecker]

But I don't understand where's the contradiction in the above proof if we consider the only case left which is "Each of our spanning trees must contain an edge that the other tree omits."

This must lead to a contradiction, therefore proving that there is no edge in one that is not in the other.
At the end of the proof, it says "$w(e)=w(e'')$". But $e \ne e''$ and the weights are all distinct, so that is a contradiction. The MSTs must be identical.

 

[CGandC]

This must lead to a contradiction, therefore proving that there is no edge in one that is not in the other.
At the end of the proof, it says "$w(e)=w(e'')$". But $e \ne e''$ and the weights are all distinct, so that is a contradiction. The MSTs must be identical.

Thanks for putting the time and helping me. I overthought about it and after taking a break and coming back to the proof I understand it.