Why is it the empty set, trivially convex? I see this results stated into a proof as something known, but I do not understand what's the idea idea behind it. How could I reason about convex combinations if the set has no elements?
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$\begingroup$$\forall$ statements are true on empty domains. The definition of convex requires that for every two points in the set the line connecting them is also in the set. In order to show that this didn't work you would need to produce a counterexample, for which there is none, because there's nothing to choose from.
$\endgroup$ 11 $\begingroup$Here's another approach, without thinking about the formal quantifier logic of it.
The intersection of any two convex sets is convex, yes? Well, what's the intersection of two disjoint circles?
$\endgroup$ $\begingroup$You have to consider the definition of convexity. It says that for every pair of points in the set the line connecting the points are in the set.
Now this boils down to quantifier logic. The statement is of the form "for every x in the set ...". The quantifier logic says that such a statement is true for the empty set.
One could avoid this by explicitly excluding the empty set, but that would complicate things rather than simplify. For example there's a result that the intersection of two convex sets is convex which is complicated if you exclude the empty set in the definition.
Another trivial example of convex sets are single-point sets as it would imply that any two points in the set would be the same and the line segment connecting them would only be the point itself.
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