Kruskal's algorithm worked example

Consider this original connected graph on vertices $A,B,C,D,E$. Its undirected edges, already in increasing weight order, areHere $AB$ means the unordered pair $\{A,B\}$; the weights are dimensionless costs. Accept $AB$ and $BC$, with total $3$. Reject $AC$: its endpoints already have the graph path $A,B,C$, so it would close the graph cycle $A,B,C,A$. Accept $CD$ and $DE$, reaching all five vertices with four edges. The resulting minimum spanning tree isStop here: $CE$ and $BE$ need not be examined. If processing continued, both would be rejected because their endpoints are already in the same connected component of a graph.

Taking the three cheapest edges $AB,BC,AC$ instead gives a graph cycle on $A,B,C$, leaving $D,E$ isolated: it is neither a forest nor a spanning tree. Even taking the four cheapest edges adds only $CD$ and leaves $E$ isolated. A cheap infeasible selection cannot compete with a minimum spanning tree. For this small graph, checking all $\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$ four-edge subsets gives a direct certificate: among those that are spanning trees, exactly one has minimum cost $12$, namely $T$. The general algorithm avoids this exhaustive search.