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

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.

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.)

This is incredibly helpful, thank you

Is this some topology hoodoo?

Great video, thanks for making it! By chance, can you let us know which books to find the Stolz-Cesàro Theorem in? I look forward to the day when asymptotics and perturbation theory are all over CZcams math channels. Carl Bender has close to 1 million views on his first lecture video now.

In 8:40 you say we're allowed to take that radius r_1 such that 0<r_1<1. Is it so because as B(x,r) \cap U_1 is open then we can find a radius l>0 where B(y_1, l) is contained in B(x,r) \cap U_1 but if we take a real number r_1 such that 0<r_1<min(1,L) small enough (the density of the rationals or the Archimedean property can help us control its smallness) such that the closure of B(y_1, r_1) (the ball itself with the point z in X such that d(z, y_1) = r_1 ) is contained in B(x,r) \cap U_1? And is it the same reasoning for 9:26, right?

Amazing!

Why the creepy music lol

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

If the directional derivative is the scalar product with a turning unitary vector, it's mean that 2 points which have the same gradient on 2 different curve will have necessarly the same directional derivative in each direction at that given point ? Probably because the 2 different curve are the same LOCALLY ? -- altough intuitively I would think it could be different and depend on the law of evolution of it's own curve.

super cool! love your videos 😊

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

YOOO this is such a good video thank you!!!

Love your videos! Thank you so much!!

The content of group theory is really interesting ( very simple ) in english, I used to study it in french which have that kind of complex abstraction even in introduction ( Fratini group and its représentation , association with galois…) I really needed this kind of fresh and clear explanation Good job !

That is super cool!

The background sound is very annoying

gracias

That is a real "real example" to explain the topic. Thanks Andrew!

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

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.

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

For the 6 x 6 grid first shown at 7:20, how does your answer account for the straight line which contains the points (0,5), (1,3) and (2,1) for instance?

At around 4:40, it should be C'_n+1(x) = -S_n(x)

I found this video is really helpful for my studying about calculus 2 in uni. Could you share that geogebra animation, I really want to learn more about animating in geogebra! After all, thank you so much for your work!

It's amazing how I am watching a video about something so interesting, and yet I get bombarded with an advert half way through with that useless idiot Greg Secker waving 150 quid at me

Really nice visuals btw

Very helpful video

Thank you

thanks!

Incredible explanation. Thanks so much!!!

Mr. Andrew thank for explanation

Nice visuals

Type "Imaginary number" with your eyes closed

You need to work on your audio.

💯

missing the case x=z ≠ y? (which is obvs also true, 0≤1+1, but exhaustive proof requires this)

Are you kidding? Only 2k subscribers?! Wonderful. Thank you for sharing!

Axiom of choice says all sets are countable...

bro this video is just awesome! helped me a lot, thanks

thanks for the explanation

thanks!❤❤❤

Great video

thank you!!!!!

thank you so much for this explanation, im doing my IB extended essay in mathematics and this was a giant help :)

for what you need B` ? hmm?

Finally someone explaining the 1-z part! I was pulling my hair out because I was looking at the wrong similar triangles and it's hard to look at something differently after your brain already decided how to see it. Your video was the only one after about a dozen that explained it! And every source online said things like "easily shown by similar triangles" or didn't explain it at all! Thank you!

Thank you so much 🙏🙏🙏🙏🙏

Thank you, very easy to follow.