The Axiom of Choice: History, Intuition, and Conflict

This video is about the Axiom of Choice, a tool in mathematics that is as contentious as it is useful.

Nevada City- Huma Huma

[nightspore]

Cantor was not a Russian mathematician. He was German.

[ismiregalichkochdasjetztso]

I think this deserves a sequel that explains intuitionistic logic.

[PureExile]

If the two socks in a pair are identical then each set of pairs will only contain one element e.g. {6, 6}={6}. So you choose the only element in each set of pairs for the choice function. No AC required.

[yanntal]

4:26 You don't need choice for that actually. You can prove it with much weaker additional axioms that for some reason are more acceptable than Choice. One example is just using ZF + Hahn Banach theorem.

[David-idjw]

So it sounds like the Axiom of Choice isn't that you can pick an element from a set, it's that you can pick an element from a set and _know what it is._ For example, using the well-ordered principal, you can always find the smallest value in a set, so you can always choose the smallest element. Without the well-ordered principal, it's like trying to choose a sock when you can't tell which one is the left or right.

That doesn't work with a bag of red marbles, because there's no order to them, so you can't pick the 'smallest'. If you could distinguish them by the number of atoms in each one, you could order them. If you had a bag of marbles of all different colors, you might order them by the wavelength of light that they reflect, and thus choose the one reflecting the shortest or longest wavelengths.

But without some means of ordering (which requires additional information about the marbles other than "red"), whatever gets picked is arbitrary, not defined by a rule, and thus seems to fail the assertion because there is no function to get you the result. Or more accurately, a specific, repeatable result, because a function with a given input must always produce the same output. (This assumes that the different red marbles are in fact different entities, which is required by another comment describing how probability works.)

[Necrozene]

Ever since I learned out it when I was 12, I was in love with it. It makes perfect sense, and no-one could adequately explain the problem.

[soupisfornoobs]

Pretty cool video! A little class presentation-y with the historical profiles but enjoyable nonetheless

[Gailon]

Do you need the AoC to proof the last statement? I think I saw a proof of this statement using one of Cantos Diagonal Arguments and afaik this is independent of AoC?

[rajpanchal]

How can you jump to equal cardinality from choosing exactly one element from each set?

[Necrozene]

The Banach-Tarski construction produces sets without measure, so it is not a problem at all.

[tobuslieven]

The axiom of choice doesn't sound like it places any constraint on accidentally choosing the same element more than once.

[Bethos-Arne]

Georg Cantor was German. Also if you use German names, could you pronounce them German?

[SkillUpMobileGaming]

At 0:50, you said that the portion in green "there are *infinitely many* lines through P that are parallel to L" is the negation of "there is *exactly one* line through P that is parallel to L", but this doesn't make sense to me. Wouldn't the negation of (for all X, X = 1) be (for at least one X, X ≠ 1)? So, wouldn't the negation be that "For at least one line L, there exists at least one point P not on L where the number of lines through P that is parallel to L is not equal to 1" (for example, 0, 2, 3, 4, ...)?

So, the minimum we would need to negate this statement would be to find just 1 line where 1 point not on that line can create a number of parallel lines not equal to 1.

[MobiusCoin]

I was mostly getting it until the proof at the end. Then I got completely lost lol.

[pukaman]

You cannot find the first N just as you cannot find the last real number closes to the origin.

[worldnotworld]

I was looking forward to this video, but the awful melting music drove me away... I pray you, please repost without it!

[SunShine-xcdh]

Axiom of choice says all sets are countable...