If we have an (arbitrary) connected undirected graph G, whose edges have distinct weights,
Also, I'm more thankful if someone can give a hint of the key things one must keep in mind when dealing with such MST questions.
This is a homework problem. Thanks.
I see you too are studying for CSC263 through the 2009 test? (Same here!)
Another way to see that the minimum is always in the MST is to look simply at this minimum edge (call it e):
e v1 ---------------- v2
(Assume this has connections to other verticies). Now, for e NOT to be included in the final MST means at one point we have, without loss of generality, v1 in the MST but not v2. However, the only way to add v2 without adding e would be to say that the addition of v1 didn't add e to the queue (because by definition, e would be at the top of the queue because it has lowest priority) but this contradicts the MST construction theorem.
So essentially, it is impossible to have an edge with minimum weight not get to the queue which means that any MST constructed would have it.