The cardinality of the real numbers has been blowing my mind for the last 40 years. This proof adds yet another layer onto the mind-boggling hugeness of R. It's so big that it contains an infinite recursion of mappings onto itself! *boggle*
Just a couple of remarks. Surprisingly this proof does not use the axiom of choice. But if there exist surjections between A and B, the AC is required to prove they have same cardinality. Another surprising application of this theorem is the fact that continuous fonctions from R to R have the same cardinality than R itself (hint: a continuous function is completely determined by its values on the rationals.)
Turning a pair of surjections into a bijection is clearly a way to pick an arbitrarily element from each equivalence class so it makes sense that would require the AC. Injections have much more of a feel of all the choices having been made in advance.
g: B -> A is an injective function. So if B is non-empty there is a x in B with g(x) = y (y is in A). And since x originates from B but gets mapped to A under g, there exists a inverse image from g-¹(y) in A - which is x from B. He specifically chose a point in A which can be assigned a inv image under g. If there weren't at least as many points in A having an inv image in B under g, as there are elements in B, then (1) there would either be a element in B without an image, but that can't be since a function maps *every* element in its domain (in the case of g, its domain is B) to an element in the co-domain (which is A for g). (2) Or every element in B is assigned a image in A under g, but there are x1 and x2 in B with g(x1)=g(x2), which contradicts the assumption that g is injective. Hope that helps
And we don't know which element in V specifically has no inverse image under g. He just chose one which has, because there *has* to be one. Note that with inverse image l mean pre-image.
Very thankful for this series. Proofs were my weakness many years ago when I first started math. Good to have this as a refresher for certain topics.
as a high school student watching this series, thanks for making it :D super interesting stuff watched it all.
I listened to all this playlist and I’ve learned sooooo much. Thank you a lot! ❤
The cardinality of the real numbers has been blowing my mind for the last 40 years. This proof adds yet another layer onto the mind-boggling hugeness of R. It's so big that it contains an infinite recursion of mappings onto itself! *boggle*
I was just thinking I would love to see a proof of this in Michael Penn form. Thanks for fulfilling my unspoken wishes!
Just a couple of remarks. Surprisingly this proof does not use the axiom of choice. But if there exist surjections between A and B, the AC is required to prove they have same cardinality. Another surprising application of this theorem is the fact that continuous fonctions from R to R have the same cardinality than R itself (hint: a continuous function is completely determined by its values on the rationals.)
Turning a pair of surjections into a bijection is clearly a way to pick an arbitrarily element from each equivalence class so it makes sense that would require the AC. Injections have much more of a feel of all the choices having been made in advance.
THANKS!
Sir please start geometry lecture series after this
How can we use g inverse in definition of h ? How do we know that there is not an element in V that does not have a pre image under g inverse?
g: B -> A is an injective function. So if B is non-empty there is a x in B with g(x) = y (y is in A). And since x originates from B but gets mapped to A under g, there exists a inverse image from g-¹(y) in A - which is x from B. He specifically chose a point in A which can be assigned a inv image under g.
If there weren't at least as many points in A having an inv image in B under g, as there are elements in B, then
(1) there would either be a element in B without an image, but that can't be since a function maps *every* element in its domain (in the case of g, its domain is B) to an element in the co-domain (which is A for g).
(2) Or every element in B is assigned a image in A under g, but there are x1 and x2 in B with g(x1)=g(x2), which contradicts the assumption that g is injective.
Hope that helps
And we don't know which element in V specifically has no inverse image under g. He just chose one which has, because there *has* to be one.
Note that with inverse image l mean pre-image.
yesterday，wtf
today，amazing
13:39 Lol, RUclips labelled this section of the video "Surge Activity"
bit lost at
16:28 , how can he cancel g from both sides?
He applied g^(-1) on both sides.
g o f would be better read as "f composed with g". But we are used to hear "aluminum", so it doesn't matter much.