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So I want to understand how induction proves that Kruskal's Algorithm is correct in terms of giving us a minimum spanning tree. I understand why the algorithm gives us a spanning tree, but I don't understand how it gives us a minimum.

So I was reading my lecture notes from class, and it tries to explain it using induction on the algorithm stage number. So the proof is as follows:

Let F$_i$ be the set of edges chosen at stage i of the algorithm. Let P$_i$ be the proposition: "There is some minimum spanning tree that contains F$_i$." Clearly any minimum spanning tree contains F$_0$ which is just the empty set of edges, so P$_0$ is true. Now suppose that P$_i$ is true. Then there is a minimum spanning tree T that contains F$_i$. Let e be the edge considered next by the Algorithm. If e is in T, then F$_i$ $\cup$ {e} $\in$ T, so P$_{i+1}$ is true. Now assume that e is not in T. Then F$_i$ $\cup$ {e} contains a cycle with e in it.

So P$_{i+1}$ cannot be true in this case since it's impossible to form a new tree that contains this cycle, so I don't see how induction holds. Can some please explain this to me so that it makes sense?

Thank you in advance.

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  • $\begingroup$ You have to prove that there is a minimum spanning tree containing the edges chosen so far. You haven't chosen a cycle yet. $\endgroup$ – saulspatz Mar 31 at 14:26
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We have to prove that that there is some minimum spanning tree containing the edges chosen so far. The easy case is when $e$ is in $T,$ and we have to deal with the case when $e$ is not in $T.$ $T\cup\{e\}$ contains a cycle $C,$ and obviously $e$ is one of the edges of $C.$ No edge $e'$ of $C$ can have greater weight than that of $e,$ for then we could delete $e'$ and obtain a spanning tree of lesser weight than $T.$ $e$ cannot have been the last edge of $C$ to be chosen, for the algorithm never creates a cycle. Therefore, some edge $e'$ of $C$ was chosen after $e,$ and we must have $w(e)=w(e').$ Deleting $e'$ gives a spanning tree $T'$ of the same weight as $T,$ containing all edges chosen so far.

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