existence of spanning trees in complete graphs implies choice?

202 Views Asked by At

it is known that the existence of spanning trees in arbitrary (connected) graphs implies the Axiom of Choice. I was wondering if this result still holds if we restrict ourselves to spanning trees of complete graphs. Does the Axiom of Choice still follow?

I would be very happy if you can provide a reference.

1

There are 1 best solutions below

4
On BEST ANSWER

I wouldn't think so. A complete graph should always have a spanning tree.

Pick your favorite vertex $u$ from the vertex set $V$, infinite or not, and construct a giant star tree with $u$ as the center. That is, construct the tree $T = (V, F)$ where the edge set is $F = \{uv : v \in V \setminus \{u\} \}$. This has to be a spanning tree (and as pointed out in comments,note that you don't need the axiom of choice to pick an element from a single set $V$).

So denote by $S$ the statement "every complete graph has a spanning tree". If the statement "$S$ implies the axiom of choice" is true, then the axiom of choice is true since $S$ is true. So this implication cannot be - or otherwise you'd have a proof of the axiom of choice.