If I shortcut an optimal TSP solution, is still optimal?

Question

Let's have a complete undirected metric graph with k nodes; metric graph is a graph that satisfy the triangle inequality, so being w the weight function the for all nodes a, b, c it is true that w(a, c) is less or equal to w(a,b) + w(b,c).

Wlog let's say that the cycle: <1, 2, 3, ..., k, 1> is the optimal TSP solution for that graph.

My question is: if I remove one node from the graph (for example the n-th) and I shortcut the cycle just skipping n is the resulting cycle still an optimal TSP solution?

n.b., The cycle would become <1, 2, ..., n-1, n+1, ..., k, 1>

Answer 1

No, this does not hold. A rather handwaving counterexample is given below. I trust you can add the numbers, do the math, and formally verify this (I used this online solver to verify my claims).

Consider these points:

The top point is clearly far away, so it has to be connected to the closest points. The other links then follow, as shown here:

If we exclude the top point, it is more optimal to have the two top points connect to the center point, as shown below. So just short-cutting is not optimal: