Talk:Product of Summations is Summation Over Cartesian Product of Products

Happy for it to be merged with Product of Summations. It is similar.

Peter Driscoll (talk) 03:46, 10 September 2017 (EDT)


 * I need to analyse it to see whether this approach can genuinely be considered a different proof from the existing one. If they are essentially the same, we may regrettably have to ditch this, as the effort to bring it to house style may be considerable. --prime mover (talk) 04:00, 10 September 2017 (EDT)


 * Note that the statements of the two are not identical. In Product of Summations, $B = \{1\cdots m\}$ does not depend on $a \in A$. So it is a proper generalisation and ought to be treated as such. &mdash; Lord_Farin (talk) 05:12, 10 September 2017 (EDT)

Product of Summations talks about standard summation re-ordering. The cartesian product version, although related has a single sum and a cartesian product to iterate over. This is useful in one particular derivation of the Eulers Product formula for the zeta function. It allows the construction of the set of positive natural numbers from the cartesian product of powers of primes.

The result is not rocket science. Just a simple proof to get started with. Learn the rules of the game here.

Peter Driscoll (talk) 10:19, 10 September 2017 (EDT)


 * I had a look at house style. It is going to take a while to get used to. As I understand your structure, the proof should have its own page but be referenced by Product of Summations, with the result listed there. Can you rename this page to, Product of Summations is Summation Over Cartesian Product of Products. If renaming is not supported, then I will create the new page and you can delete this one.


 * Your house style doesn't seem to give examples. Examples are incredibly usefull.


 * Peter Driscoll (talk) 21:02, 10 September 2017 (EDT)


 * You should have the ability to rename: hover over More and click on Move (top right). I believe standard auths should get you this.


 * For house style there's actually fairly copious notes, but the suggested technique is by example. But it's a good start to note that all "big" operators like Union, Summation, Product etc. need to be on a line with "\displaystyle" in them somewhere before the invocation of the command, and that brackets around them need to be prefaced with \left and \right as appropriate, with braces for ease of making sure of the correct scope. (You will immediately notice the vast improvement of readability and presentational aesthetics.) Also, check out the use of the eqn template, into which all (for a given value of "all") multi-row equations and logical arguments are to go. --prime mover (talk) 01:33, 11 September 2017 (EDT)


 * Oh yes, and note that the eqn template automatically places the \displaystyle into every $\LaTeX$ string, so you don't need to add them there. I notice that you have already started experimenting with this.


 * Also note the "\mathop" that goes before the operator in the limits of e.g. \sum and \prod -- it adds an extra space around the operator. This was a kludge we learned early on that was needed for MathJax, usually excellent but has occasional clumsiness like this. --prime mover (talk) 01:37, 11 September 2017 (EDT)
 * I have no "more". Unless I am going blind. Top right? Not seeing.
 * Next to "Read", "Edit", "Add topic", "View history" there should be a drop-down menu. As I say, I don't know whether this works for non-admin contributors, it may no longer be available. recent MediaWiki upgrades may have changed this.
 * Not there. Only Read Edit Add topic View history and watch list. Peter Driscoll (talk) 03:48, 11 September 2017 (EDT)
 * I have been playing around with equation templates. Cool. Yes I have seen the \left and \right. I will take note of the \mathop. I wondered what that was for.
 * There is something about it in the help pages. --prime mover (talk) 03:25, 11 September 2017 (EDT)
 * Peter Driscoll (talk) 03:13, 11 September 2017 (EDT)


 * Thanks for the move. I will clean this up, as I learn house style.
 * Peter Driscoll (talk) 20:39, 11 September 2017 (EDT)

Indexed Families of Sets
I am reading indexed families of sets.
 * Cartesian Product
 * Indexing Set/Family of Sets

It seems to provide some of the the right mechanism, but the terminology is unfamiliar to me. Roughly speaking,
 * $A$ is the indexing set
 * $B$ is the set of sets

I need a cartesian product of n of the sets. The I make a cartesian product of that product with another set to form a catesian productof n+1 sets. I don't see all the mechanisms I need here. The constructions needed for the proof are, The whole set for the next element $B_a$.
 * $(\prod_{a \in \{1..n\}} B_a) \times B_{n+1}$

A single element,
 * $(\prod_{a \in \{1..n\}} B_a) \times \{k\}$

A subset.
 * $(\prod_{a \in \{1..n\}} B_a) \times X_{n+1}$

I am investigating. Any help appreciated.

Peter Driscoll (talk) 18:26, 14 September 2017 (EDT)


 * I haven't forgotten this.
 * Peter Driscoll (talk) 10:45, 19 September 2017 (EDT)


 * Way over my head I'm afraid. I'm going to have to leave this for someone else. --prime mover (talk) 16:13, 19 September 2017 (EDT)


 * Okies. No worries. I know what I want to say. Just need to say it your way.
 * Informally it is like this. We have a set of sets which are indexed by the indexing set can be anything, as long as it is ordered. We think of it as numbers, but it could be anything. The indexing set indentifies both sets, and elements in the tuple.
 * An example. You have two pets, Bruiser and Claude. {Bruiser, Claude} is the indexing set. They each have a set of toys. Bruisers toys are the rubber bone, the chewed over tennis ball, and your missing slipper {bone, ball, slipper}. Claude has jungle bell roller, the fluffy mouse on a string, and the table tennis ball {bell, mouse, ball}.
 * We have a tuple of sets
 * ({bone, ball, slipper}, {bell, mouse, ball})
 * This tuple indexed by Bruiser is {bone, ball, slipper}.
 * Indexed by Claude gives {bell, mouse, ball}.
 * What are all the combinations of toys that Claude and Bruiser can play with? This is the Cartesian product of Bruiser and Claude's toys.
 * { (bone, bell) (ball, bell) (slipper, bell)
 * (bone, mouse) (ball, mouse) (slipper, mouse)
 * (bone, ball) (ball, ball) (slipper, ball) }
 * Each of these tuples is indexed by {Bruiser, Claude}. So (ball, mouse) indexed by Bruiser is Ball. (ball, mouse) indexed by Claude is mouse.
 * Peter Driscoll (talk) 21:16, 19 September 2017 (EDT)
 * Peter Driscoll (talk) 21:16, 19 September 2017 (EDT)


 * I'm totally out of my depth. Perhaps I ought to be removed. --prime mover (talk) 01:56, 20 September 2017 (EDT)


 * Putting together a proof wiki such as this is a thankless job. Only those who try to achieve such things understand how difficult it is.
 * So thank you for all you have done.
 * Is that one thank? Is a thank a thing? How can it be thankless, if there is no such thing as a thank.
 * Have I contradicted myself?
 * I babble.
 * Peter Driscoll (talk) 06:33, 20 September 2017 (EDT)

Reading your requirements, it would seem that Definition:Finite Cartesian Product is all you need. Some basic theorems about it seem to be missing (of the type Cartesian Product with Finite Cartesian Product, although there is only a "natural equivalence" in the category-theoretic sense) but this needn't be limiting.

For optimal rigour, you might need to fix a bijection $\{1 \ldots n\} \to A$, guaranteed to exist by finiteness of $A$ and construct the Cartesian products using this mapping.

Does this address your needs? &mdash; Lord_Farin (talk) 14:58, 20 September 2017 (EDT)
 * I think there may be versions of the cartesian product where that bijection is built in. That would simplify the proof. So the sets B are named after the elements of A, and so are elements in the tuple. That is really just hiding the bijection inside the mechanics of the cartesian product.

Unfortunately the math notation they have used for these more advanced cartesian product definitions is not familiar to me. Peter Driscoll (talk) 20:27, 5 October 2017 (EDT)