Points with shortest distance always form an edge of a delaunay triangulation.

Question

So I have to proof the following:

Points with shortest distance to each other always form an edge of a delaunay triangulation/minimal spanning tree.

Since minimal spanning tree is a subtree of a delaunay triangulation I guess it is enough to proof it for delauny triangulation alone.

I do not really know where I should start with this problem.

Answer 1

The "closest pair property" is a consequence of the "empty circle property" as shown for example in http://www.tcs.fudan.edu.cn/rudolf/Courses/Algorithms/Alg_ss_07w/Webprojects/Lu_delaunay/delaunay.html