Talk:Cantor–Bernstein–Schroeder theorem

Translation of parts from german article

In the german article de:Cantor-Bernstein-Schröder-Theorem, which I translated from the english version, I added a visualization of the map h.

Someone might want to translate the image displayed at that article into english and look over this trial of a translation, and then integrate it into the english article.

Here comes my translation of the section "Veranschaulichung" in the german article.

---

(Before that, I added a little remark on the map h.)

Note, that this definition of h is nonconstructive, in the sense that there exists no general method to decide in finitely many steps for any given sets A, B and injections f, g, if an element x of A lies in C or not. For special sets and maps this might of course be possible.

Visualization

The definition of h can be visualized with the following diagram.

(Then the diagram) [1]

Displayed are parts of the (disjoint) sets A and B together with parts of the mappings f and g. If you interpret the set A union B together with the two maps as a directed graph, then this bipartite graph has several connected components.

These can be divided into four types: pathes extending infinitely to both directions, finite cycles of even length, infinite paths starting in the set A, and infinite paths starting in the set B (the path passing through the element a in the diagram is infinitely long into both directions, so the diagram contains one path of every type). In general it is not possible to decide in finitely many steps, what type of path a given element of A or B belongs to.

The set C defined above contains precisely the elements of A which are part of an infinite path starting in A. The map h is then defined in such a way, that for every path it yields a bijection of the elements of A onto the elements of B direct before or after it in the path. For the both-side infinite path and for the finite cycles, we choose to map every element to its predecessor in the path.

---

Hope this will be of use. --SirJective 10:54, 29 Oct 2003 (UTC)

Still hope someone might take a look at it... :-( --SirJective 19:36, 13 Jun 2004 (UTC)

I now copied this section into the article (and corrected some minor mistakes). The image still needs to be translated, and I will do it as soon as someone tells me the correct english words in the context of graph theory. --SirJective 15:45, 1 Aug 2004 (UTC)

---

imho, "One can then check that h : A → B is the desired bijection." is not really clear. Don't you think that more details may be necessary to understand this proof ? What do you think about the french article ?

Alternative presentation

The Cantor-Berstein-Schröder theorem is one that many people find difficult to understand intuitively. At present, I think the sketch proof is not quite as clear as it could be, and the helpful diagram is not used as effectively as it could be. How about this presentation, formalising the intuitive approach which provides a more complete proof that is well illustrated by the diagram?

Assume without loss of generality that A and B are disjoint. For any a in A or b in B we can form a unique two-sided sequence of elements that are alternatively in A and B, by repeatedly applying $f$ and $g$ to go right and $g^{-1}$ and $f^{-1}$ to go left (where defined).

$\cdots \rightarrow f^{-1}(g^{-1}(a)) \rightarrow g^{-1}(a) \rightarrow a \rightarrow f(a) \rightarrow g(f(a)) \rightarrow \cdots$

For any particular a, this sequence may terminate to the left or not, at a point where $f^{-1}$ or $g^{-1}$ is not defined.

Call such a sequence (and all its elements) an A-stopper, if it stops at an element of A, or a B-stopper if it stops at an element of B. Otherwise, call it doubly-infinite if all the elements are distinct or cyclic if it repeats.

By the fact that $f$ and $g$ are injective functions, each a in A and b in B is in exactly one such sequence to within identity, (as if an element occurs in two sequences, all elements to the left and to the right must be the same in both, by definition).

By the above observation, the sequences form a partition of the whole of the disjoint union of A and B, hence it suffices to produce a bijection between the elements of A and B in each of the sequences separately.

For an A-stopper, the function $f$ is a bijection between its elements in A and its elements in B.

For a B-stopper, the function $g$ is a bijection between its elements in B and its elements in A.

For a doubly infinite sequence or a cyclic sequence, either $f$ or $g$ will do. (unsigned post from User:Elroch circa Feb 2006).

Great presentation! Finally I understood this proof. 88.152.251.205 16:34, 18 March 2006 (UTC)

I've used your proof in the Hebrew Wikipedia. 88.152.251.205 12:29, 25 March 2006 (UTC)

I am very pleased that you found this of use, and appreciate your feedback. Elroch 01:22, 27 March 2006 (UTC)

this argument is actually from Julius König.Kope 16:06, 2 September 2006 (UTC)

I would be surprised indeed if no-one had taken this route before! Thanks for identifying to whom credit is due for discovering it. Elroch (talk) 20:34, 4 April 2010 (UTC)

I am now pretty sure that my proof is based on what I was taught in 1979, but more complete. It's been rather a long time since I took or looked at the lecture notes though! Elroch (talk) 09:41, 13 April 2010 (UTC)

Why can we assume the sets are disjoint? What if they aren't? - Sikon (talk) 12:55, 25 May 2011 (UTC)

If A and B were identical, then the identity map on A would be a bijection between them. Thus we may assume they are distinct. Clearly

$\{ \langle a, \langle a, A \rangle \rangle \vert a \in A \}$

is a bijection between A and A×{A}. Similarly, there is a bijection between B and B×{B}. Notice that the sets A×{A} and B×{B} are disjoint because

$\langle a, A \rangle \neq \langle b, B \rangle$

follows from A≠B. f and g can be modified to go between these disjoint sets and then one proceeds as above. Finally, the resulting bijection h can be modified to go between the initial sets A and B. JRSpriggs (talk) 18:28, 25 May 2011 (UTC)

A supershort proof hiding infinite steps

There is a supershort proof that goes like this:

Let A and B be sets and f:A->B and g:B->A be functions. Then there is a subset A0 of B such that g(B-f[A0])=A-A0. If f and g are injective, it follows immediatly that h(x)=f(x) for x in A0 and else h(x)=g−1(x) is a bijection.

This proof has the virtue of avoiding infinite steps. The infinite steps are needed though in proving the first assertion. (Sry bout my mathematical syntax. I have only notepad ;) —The preceding unsigned comment was added by YohanN7 (talk • contribs) 18:59, 30 March 2007 (UTC).

alternative proof

I have an alternative proof which does not involve the use of inverse functions. I don't know whether it is worth adding to the article. I'd also need some help attributing the proof to the right source. I think it is from Boolos and Jeffrey's Computability and Logic... —Preceding unsigned comment added by 137.205.8.2 (talk • contribs)

Variants with surjectivity

What about variants of the theorem with injectivity replaced by surjectivity? There are two such variants: one with an injection and a surjection in the same direction, and one with surjections in opposite direction. Can they proved without AC as well? —Preceding unsigned comment added by 131.174.22.39 (talk) 17:28, 19 February 2009 (UTC)

I do not know, but I suspect that AC is necessary in the case where both are surjections. JRSpriggs (talk) 13:25, 20 February 2009 (UTC)

Indeed, you need to "invert" functions, so that mere surjectivity entails choices for the inverse : you have to choose one element in the inverse image of the element you're interested in, hence you need to do infinite choices, hence you need AC.

I don't know if one can easily prove that AC is needed, given that this form of "Cantor-Bernstein" doesn't seem to be equivalent to AC (I don't see how to build surjections without using AC), therefore one would need to exhibit a model of ZF where it is not satisfied, which can be hard, since it is not trivial for AC. Samuel Bach

Lead with the König proof?

I tried to follow the first proof and my head started to spin, but I found the second proof very straightforward and pleasing. Is there any reason not to lead with the simpler proof, and move the harder one into a subsection? ciphergoth (talk) 22:08, 28 April 2009 (UTC)

Title of theorem

Is it really true that most sources use Cantor's name in the title? Most of the sources I've seen personally (and most of the first few Google hits for "Schroeder Bernstein") and most of the math people with whom I've spoken have actually just said Schroeder-Bernstein. I don't have any source that proclaims one to be more common than the other, though. But maybe the title should be changed to reflect what appears to me to be common practice. 207.207.126.206 (talk) 00:06, 5 August 2009 (UTC)

General idea

I've tried to make the proof "natural", which happens to repeat the second proof, and shows the link with the first (although perhaps not clearly in what I've written). The difficulty is that it is quite clear on a picture (there are chains going to infinity, starting from "lonely" points, which are points with no arrow ending at them, and chains infinite on both sides, for which one can choose any orientation of arrows), but less so in English. Also, my English might be slightly unsatisfactory... If someone could draw a picture, or/and make the explanations clearer, I think it could help visualize the proof quite easily. --Samuel Bach (talk) 16:28, 23 April 2010 (UTC)

I agree my first attempted contribution was quite unnecessary and poorly written. However, I think it could be useful to point out the link between the two proofs of the article : elements of C_n are the nth elements of A-stoppers. What do you think? Furthermore, I would like to know how to write "A" the same way as in the article. Thank you.--Samuel Bach (talk) 16:10, 24 April 2010 (UTC)

Generally, I use the editor to look at the source in order to see how to generate a particular symbol or version of a letter (then use the back-up button to exit). Find an example of what you want to write and look at the source for it. If you are talking about how to italicize a letter, it is done by putting doubled single-quotes around the letter. Tripled single-quotes generate bold-face. See Help:Displaying a formula for additional information. JRSpriggs (talk) 23:45, 24 April 2010 (UTC)

Thank you. That is what I tried, but I saw double quotation marks instead of two single quotation marks, hence my mistake. What do you think of my proposal to explain why the two proofs of the article are the same (simply with a sentence like that above)?--Samuel Bach (talk) 16:20, 25 April 2010 (UTC)

I think it might be useful for you to show how the two proofs are related.

To maximize the chance of your changes being accepted (without an edit-war), I would suggest that: (1) you make small (minimal) changes; and (2) you wait a day or two after each change for it to be accepted (or modified or rejected) before making another change. JRSpriggs (talk) 18:41, 25 April 2010 (UTC)

Set-based proof?

This is from Willard (General Topology, problem 1J, 1970).

Let A be a set, P(A) the power set of A.

If F:P(A) → P(A) is a function such that whenever C ⊆ D ⇒ F(C) ⊆ F(D), then there exists E such that E = F(E). Proof: Let E = ∪{ C ⊆ A : C ⊆ F(C) }. Claim: E = F(E). Let x ∈ E. Then x ∈ C for some C such that C ⊆ F(C) ⊆ F(E). Hence, E ⊆ F(E). E ⊆ F(E) ⇒ F(E) ⊆ F(F(E)) ⇒ F(E) = C such that C ⊆ F(C) for some C ⇒ F(E) ⊆ E.

Cantor-Bernstein: If A and B are sets and f:A → B and g:B → A are injective functions, then there exists h:A → B such that h is bijective. Proof: Define F:P(A) → P(A) as F(C) = A - g(B - f(C)). F preserves subsets from above, so there exists E such that E = F(E). Define h:A → B by h(x) = f(x) for x ∈ E and h(x) = g^-1(x) for x ∉ E —Preceding unsigned comment added by Quagmirede (talk • contribs) 07:31, 30 April 2011 (UTC)