User:IssaRice/Schröder–Bernstein theorem

questions:

• are the ancestor proof and the iterated partition proof actually different, or do they just look different on the surface? btw this is a very nice writeup of the ancestor proof (much clearer than wikipedia's version); HT Satira. UPDATE: these are actually basically the same proof, just written in different notation.
• what is the proof that uses axiom of choice, and how does choice simplify the proof? It's described here, but I don't know enough about ordinals/cardinals to understand it: https://math.stackexchange.com/questions/223587/looking-for-cantors-original-proof-of-the-cantor-bernstein-theorem-that-relies
• why does the wikipedia proof (and some of the other proofs) assume that A and B are disjoint? what does this buy us?
• what is the point of having the set C in exercise 8.3.2? i don't see how we need this extra level of generality when applying this result in the following exercise, and the picture is simpler to draw in 8.3.2 itself.

In my proof at https://taoanalysis.wordpress.com/2020/09/01/exercise-8-3-2/ i think the proof of injectivity can be improved. consider dictinct x,y, and then split cases depending whether both are green, both are white, or one is green and the other is white. Also in the surjectivity proof, i am already doing this green/white split, so it would be good to mention those colors in the proof itself (just to help the reader).

Unified notation

Let $A,B$ be sets. (In Tao's book we assume $A\subseteq B$ but this isn't the case in the other proofs.)

$f:A\to B$ (In Tao's book we assume that $f=\iota _{A\to B}$ is the inclusion map that sends each $x\in A$ to itself.)

$g:B\to A$

$C_{0}=A\setminus g(B)$

$C_{n+1}=(g\circ f)(C_{n})$ for all $n\geq 0$

$D_{0}=B\setminus f(A)$

$D_{n+1}=(f\circ g)(D_{n})$ for all $n\geq 0$ (In Tao's book, since $f$ is just the inclusion map and $g$ maps into $A$, we have $f\circ g=g$ which is why Tao can write $D_{n+1}=g(D_{n})$. Note that Tao uses "f" instead of "g" because his notation is different from everybody else's.)

Chain types

Type 1: Loop

Type 2: infinitely goes backwards without repeating

Type 3: Chain stops in A, with $g^{-1}$ eventually undefined.

• 3a: Elements of $C_{0}$ ($g^{-1}$ immediately undefined)
• 3b: Elements of $\bigcup _{n=1}^{\infty }C_{n}$ ($g^{-1}$ is defined, but eventually if you keep going $g^{-1}$ will be undefined)

Type 4: Chain stops in B, with $f^{-1}$ eventually undefined.

• 4a: Elements of $D_{0}$ ($f^{-1}$ immediately undefined)
• 4b: Elements of $\bigcup _{n=1}^{\infty }D_{n}$ ($f^{-1}$ is defined, but eventually if you keep going $f^{-1}$ will be undefined)

Definitions of the bijection in various books

• Book of Proof: h(x) = f(x) if x is of type 3; $h(x)=g^{-1}(x)$ if x is of type 1,2,4.
• Cornell page: h(x) = f(x) if x is of type 1,2,3; $h(x)=g^{-1}(x)$ if x is of type 4.
• Tao: h(x) = f(x) if x is of type 1,2,3,4a (actually case 4a doesn't even exist in the Tao book since A is a subset of B); $h(x)=g^{-1}(x)$ if x is of type 4b
• Wikipedia: h(x) = f(x) if x is of type 3; $h(x)=g^{-1}(x)$ if x is of type 4; h(x) can be defined either way if x is of type 1 or 2 (I think this is the key insight, and explains a lot of the variation between different proofs, since they just pick either f or g^-1 for types 1 and 2)

Viewing the theorem as a fixed point result

Is the result a fixed point result? (if so how can we phrase it as such?) or is it just that some of the proofs make use of the fixed point ideas? Yes, this can be viewed as a fixed point result. This page cleared things up for me: https://ncatlab.org/nlab/show/Cantor-Schroeder-Bernstein+theorem#proof

basically when we write the definition of the bijection h, we have two cases, and if we call one of those cases X (specifically, the case where we just apply f), then we can write X = g(f(X)^c)^c where ^c is set complementation, so X is a fixed point of the operation x -> g(f(x)^c)^c

This page explains the relevance of knaster-tarski better (HT Satira): https://proofwiki.org/wiki/Cantor-Bernstein-Schr%C3%B6der_Theorem/Proof_6

So it's something like:

• first we notice that SB can be stated as a fixed point theorem
• the operation that has a fixed point happens to fit some preconditions of knaster-tarski lemma, so we can use this lemma to produce a fixed point
• given the fixed point, we can "unwrap" the operations to produce the subsets needed to define the bijection h

Exercise 9 here https://www.greaterwrong.com/posts/9a2asxypuNjCmga3p/iteration-fixed-point-exercises writes the same fixed point instead as B' = f(A') and A-A' = g(B-B') (but if you substitute in B' into the second equation and then take the complement of both sides, you get exactly the same fixed point.)

References

• https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem
• https://www.cs.cornell.edu/courses/cs2800/2017sp/lectures/lec05-cantor.html
• http://www.people.vcu.edu/~rhammack/BookOfProof/BookOfProof.pdf#page=242
• https://math.stackexchange.com/questions/3323133/questions-about-the-schroeder-bernstein-theorem
• https://math.stackexchange.com/questions/1726578/understanding-a-proof-of-schr%C3%B6der-bernstein-theorem
• https://proofwiki.org/wiki/Cantor-Bernstein-Schr%C3%B6der_Theorem/Lemma
• https://proofwiki.org/wiki/Cantor-Bernstein-Schr%C3%B6der_Theorem