The meat begins at 12:00.05:30 Errr, no, the square is not "made of" points. It *contains* those points. A point just marks a certain location inside that (pre-existing) square. One _cannot_ build the area of the square from points, because POINTS DON'T HAVE ANY AREA BY THEMSELVES!!! Areas could be made only from smaller (e.g. unit) areas. Otherwise there would be no difference between a scalar, a vector, a pseudovector, and a pseudoscalar in Clifford Algebras etc. There's a reason we distinguish those dimensions, right? :PThe ridiculousness of that statement comes out nicely at 06:10 where he writes the sum of the probabilities for these points. He said in the previous lecture that the probability of a point is 0 (because it has no area, and we assumed that area = probability in this model), and he restated this assumption again in this lecture here. So we're actually *adding up 0s here*! And we _cannot_ get a positive result (in this case: 1) from summing up zeros, no matter how much of them we add up (be it 10, 100 or infinitely many). So the right side of this equation cannot be equal to the left side, and the entire equation is false.06:40 No, the mistake is not in this point. The union is fine, as long as we're adding up actual areas. The mistake is in assuming that a point can have an area :P07:20 Of course I can! Cantor's diagonalization argument is one example. Another one is Hilbert's space-filling curve. Long story short: yes, there are ways of ordering points on a surface, as well as in space, by threading them on a line. In other words, there exist functions which map points on a surface (or in space) onto a line. And this is another proof that this is not what is wrong with this argument. The "points have area" assumption is.17:20 If the probability of B is 0, then we're contradicting ourselves, because the "|B" in "P(A|B)" was supposed to mean that B occurred *for sure*. So it _cannot_ be 0! The very statement that "B occurred" means that this probability is _not_ 0, and we are OK to put it in the denominator. Logic, Mr. Titskills :P####[]