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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    $\begingroup$ Hint: What would it mean for it not to be convex? $\endgroup$ Jan 7, 2016 at 11:54
  • $\begingroup$ @AkivaWeinberger It would mean that the segment that joints two generic points of the set is not contained in the set itself $\endgroup$
    – PhDing
    Jan 7, 2016 at 12:04
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    $\begingroup$ It would mean that there exists two points in it, $a$ and $b$, such that the line connecting $a$ and $b$ isn't contained in it. This is false, because there doesn't exist two points in it at all. (In other words, $\emptyset$ is convex because there's no counterexample to the convexity condition.) $\endgroup$ Jan 7, 2016 at 12:05
  • $\begingroup$ The statement that the empy set is convex is vacuously true, as @Akiva pointed out. $\endgroup$
    – drhab
    Jan 7, 2016 at 12:17

3 Answers 3


$\forall$ 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.

  • $\begingroup$ How about a statement of the type "there exists ..." $\endgroup$
    – BigbearZzz
    Jan 7, 2016 at 11:54
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    $\begingroup$ More specifically: $\forall x\in\emptyset,P(x)$ is true no matter what $P(x)$ is. $\endgroup$ Jan 7, 2016 at 11:54
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    $\begingroup$ Statements of existence are false on empty domains, because they require you to produce an example, for which there is none. $\endgroup$ Jan 7, 2016 at 11:56
  • $\begingroup$ @BigbearZzz This is why the answer begins "For all statements," that is, statements of the form $\forall x . . .$. $\endgroup$ Jan 7, 2016 at 11:57
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    $\begingroup$ $\forall x\in\emptyset \,\, \exists y \,\, y=x$ is one of the logically true statements which seem deliberately misleading $\endgroup$
    – Henry
    Jan 7, 2016 at 12:03

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?


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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