Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?

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Currently learning about spanning trees and using Kruskal's algorithm and I was wondering whether a minimum weight spanning tree of a weighted graph must contain one of the least weight edges of every vertex.

Is it the case?

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Yes.

Let's assume that's not true, i.e. there exists a vertex $v$ such that MST does not use any of its smallest weight edges (there may be more than one). Let $e$ be any of such edges, then you can add $e$ to MST and then remove the other edge of $v$ on that cycle, which by definition was of strictly greater weight. We reach a contradiction with the weight of MST.

I hope this helps $\ddot\smile$