A minimum spanning tree of an edge-weighted connected graph is a subset of the edges which connects all the vertices without any cycles and with the minimum possible edge weight.
Question: How is the matter of a minimum spanning tree handled for infinite graphs? Especially when the sum of weights is infinite?
Taking as an example the positive integers with the euclidean topology from the real line, and defining the weight of the edge connecting any two integers to be the distance between them, it is clear that any spanning tree will contain infinitely many edges of at least length one and therefore the total weight of any connected graph will always be infinite. On a prima facie basis one might conclude the weights of all spanning trees are incomparable and it's impossible to choose a minimum.
But a more compelling argument is that the minimum spanning tree is obviously the tree $M=(1,2),(2,3),(3,4),\ldots$
One can easily formalise this into a proof: pick any spanning tree $O$ other than $M$. Now $O$ will have at least two edges which overlap each other. Proceed by replacing each pair of overlapping edges in turn with either two or three consecutive edges which span the same points until you will arrive at $M$ by a proces of removing segments, hence $M<O$
So how is the matter of spanning weights typically handled when the sum of weights is infinite?
I can see that in the example above, a reasonable approach might be to form a function from the spanning graphs into the ordinals with $\omega$ representing the length of the real line and the length of the smallest spanning tree. Then $\omega+1$ is the length of the same tree but adjusted to have a single line segment of length one overlapping, e.g. by replacing $(1,2),(2,3)$ with $(1,3),(2,3)$.
Best Answer
For finite graphs, minimum spanning trees are characterized by the following property: tree $T$ is an MST of $G$ if and only if, for every edge $vw \in E(G) \setminus E(T)$, no edge of the unique $v-w$ path in $T$ has a higher weight than $vw$. This is a property that we could easily generalize to infinite graphs; it only requires comparing edge costs one at a time.
There might not be an MST by this definition. Consider the complete graph on vertex set $\mathbb N$, and give each edge $ij$ the weight $\frac1{\max\{i,j\}}$. Then no spanning tree has the property described above. However, I also feel comfortable saying that this graph simply does not have an MST; no matter which spanning tree you pick, there is a better spanning tree.