# VC dimension of linear separator in 3D

I am confused about the Vapnik-Chervonenkis dimension of a linear separator in 3 dimensions.

In three dimensions, a linear separator would be a plane, and the classification model would be "everything on one side of a plane."

It's apparently proved that the VC dimension of linear separators is d+1, so in 3D, its VC dimension is four. That means it should be able to put any set of 1, 2, 3, or 4 points on one side of a plane.

+1    -1

-1    +1


This is the case that a line (2-dimensional linear separator) cannot handle, but the 3-dimensional linear separator is supposed to be able to shatter this. But, I can't see how you could put two corners on "one side of a plane" because all four points are coplanar.

Could someone explain how a 3-d linear separator can shatter the four points I just described?

The VC dimension is the maximal number $d$ such that there exists a set of $d$ points that is shattered by the class. It doesn't mean that every set is shattered.