Talk:Cantor-Bernstein-Schröder Theorem/Proof 6

Notation
If you don't like that notation, I suppose I can use $fS$ for the image of $S$ under $f$.... S &amp; F write $f''(S)$, but that seems most bletcherous of all. I like the square brackets; I don't think using round brackets here for two very different things is a good idea. What do you actually want? --Dfeuer (talk) 06:43, 1 April 2013 (UTC)


 * The square bracket notation is used in Kelley 1955. --Dfeuer (talk) 06:51, 1 April 2013 (UTC)


 * Feel free to add the square bracket notation into the Definition:Mapping page in the section discussing notation. Also note that the notation used in S&F is also to be added there.


 * Round brackets are used for exactly the same thing: to define the thing upon which the mapping acts. Hence the notation $f(x)$, whatever $x$ may happen to be (an element or a set of elements) covers everything. If your elements themselves happen to be sets (as is the case when discussing power sets) the headache arising as to whether you are discussing elements or sets and determining which of the notations to use is just too much like hard work.


 * Ultimately it's a matter of house style. Round brackets are used consistently for indicating the image of a mapping, and it is a good idea to keep that convention.


 * As for Kelley using square brackets, note that that was written in 1955 and you are on record as considering works dating back to the 1960s and earlier as less worthy than more modern ones. --prime mover (talk) 08:35, 1 April 2013 (UTC)


 * I am on record not thinking they are better for choice of terminology. I do not understand your hard work comment. Please explain how to tell whether $f(x)$ means the element $y$ such that $(x,y) \in f$ or whether it means $\{y:\exists a \in x: (a,y) \in f \}$. Yes, in most cases of analysis, topology, and algebra it's obvious, but in set theory it's not. --Dfeuer (talk) 13:29, 1 April 2013 (UTC)


 * Depends on how $f$ is defined. If it is defined so that $x$ is an element of the domain then it means the element $y$ such that $(x,y) \in f$. If it is defined so that $x$ is a subset of the domain then it means $\{y:\exists a \in x: (a,y) \in f \}$. It's the responsibility of the person writing the exposition to ensure that which one is meant at any one time is crystal clear. From the context of how the proof is structured, this is not a problem; the lowercase letters relate to mappings on the elements, while the uppercase one ($E$) is a mapping on the powerset.


 * I point you towards Definition:Induced Mapping which you can of course resort to when it is necessary to distinguish between the mapping on elements of the domain and mapping on elements of the powerset (i.e. subsets of the domain). I still contend that in this context it is not necessary to distinguish between the two.


 * If you still disagree, then I recommend one of the following courses of action:


 * a) Use one of the notational options as specified on the Definition:Induced Mapping page, together with a link towards that page so as to indicate what is actually meant by the notation you have selected, or:


 * b) Add your preferred notation (e.g. the one out of Kelley) onto the Definition:Induced Mapping page, together with a link etc., remembering of course to add a note to the Sources section of that page.


 * I appreciate that there may be a few words still needed to be added to the "Image of Subset" page so as to explain this confusion and in specific cases where it may apply.


 * Your joke is indeed appropriate for the paradoxes category, BTW (especially appropriate given the date), but please do not ascribe it to me, however much you believe I may deserve the ridicule. --prime mover (talk) 15:18, 1 April 2013 (UTC)


 * This is a minor point. One of the more interesting questions is: is the "furthermore" statement actually part of the original statement of the theorem? If so, this needs to be demonstrated. If not, it should be separated out into a separate theorem page. --prime mover (talk) 14:51, 1 April 2013 (UTC)


 * Smullyan and Fitting seem very taken by that fact (which is probably true of most/all of the bijections demonstrated in the proofs). I can't really say why. They don't include it in any theorem statement, but they're rather informal about such things, and they mention the fact in connection with both of their proofs of the theorem. I can add a few words to demonstrate it. Making it a separate theorem seems like overkill&mdash;said theorem would have to either refer to the construction in the proof of this theorem, repeat its proof, or make this theorem a corollary. --Dfeuer (talk) 15:08, 1 April 2013 (UTC)

Annoyance
The vast majority of this proof is spent on the simplest part, after the Knaster-Tarski lemma has been applied. Can anyone find some way to shorten this without handwaving? --Dfeuer (talk) 19:10, 1 April 2013 (UTC)