Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Comments on Product of empty set of elements vs. product over empty indexing set — is there any difference?

Post

Product of empty set of elements vs. product over empty indexing set — is there any difference?

+4
−0

I am reading Lang's Algebra (3rd ed., Pearson, 2003). In $\S$I.1 Monoids, on page 4 the author defines the meaning of and notations for products of finitely many elements of a monoid as follows:

Let $G$ be a monoid, and $x_1, \dotsc, x_n$ elements of $G$ (where $n$ is an integer $> 1$). We define their product inductively: $$ \prod_{\nu = 1}^n x_\nu = x_1 \dotsm x_n = (x_1 \dotsm x_{\nu - 1})x_\nu. $$ We then have the following rule: $$ \prod_{\mu = 1}^m x_\mu \cdot \prod_{\nu = 1}^n x_{m + \nu} = \prod_{\nu = 1}^{m + n} x_\nu, $$ which essentially asserts that we can insert parentheses in any manner in our product without changing its value. The proof is easy by induction, and we shall leave it as an exercise.

One also writes $$ \prod_{m + 1}^{m + n} x_\nu \quad \text{instead of} \quad \prod_{\nu = 1}^n x_{m + \nu} $$ and we define $$ \prod_{\nu = 1}^0 x_\nu = e. $$

As a matter of convention, we agree that the empty product is equal to the unit element.

If I understand Lang correctly, in the last two points above he is trying to differentiate between the following two cases: Suppose that the elements of $G$ are indexed by elements from a linearly ordered set $I$. Then,

  • if $J \subset I$, $J \neq \emptyset$ such that $\{ x_\nu \in G : \nu \in J \} = \emptyset$, then we define $\prod_{\nu \in J} x_\nu = e$;
  • as a matter of convention, $\prod_{\nu \in \emptyset} x_\nu = e$.

Is there really a difference between these two cases; that is, do we need to separately define both, the product of an empty set of elements, as well as the product over an empty indexing set? Is it not possible to derive one from the other? Even better, is it not possible to derive the values of these expressions directly from the definition of products of finitely many elements itself?

One reason why Lang's exposition here doesn't feel so clean to me is that apparently many choices need to be made. I would prefer it if the definition of products of finitely many elements automatically took care of these edge cases, instead of us having to insert them in "by hand", so to speak. After all, there is only one logical choice for these values, so I expect that choice to be a consequence of the definition itself rather than it being a separate definition or convention.

History
Why does this post require attention from curators or moderators?
You might want to add some details to your flag.
Why should this post be closed?

1 comment thread

General comments (7 comments)
General comments
The Amplitwist‭ wrote almost 4 years ago

I would have preferred to have some algebra-related tags on my question, but it seems that such tags don't already exist and I cannot create them. Could users with more privileges please help me choose a better set of tags?

Derek Elkins‭ wrote almost 4 years ago · edited almost 4 years ago

Assuming $x_{(-)} : J \to G$, i.e. $x$ is a $J$-indexed collection of elements of $G$, then it can't be the case that $J \neq \varnothing$ while $\{x_\nu \in G \mid \nu \in J\} = \varnothing$. You could define a notion of product that took a finite multiset and compare that to a set-indexed notion of product to get a meaningful question. However, as r~~ points out, this is unrelated to Lang's definition which only defines a product notation for a range of naturals.

The Amplitwist‭ wrote almost 4 years ago · edited almost 4 years ago

@DerekElkins In my mind, I was thinking of a finite monoid (or group) $G$ whose elements are indexed as $\{ x_1, \dotsc, x_n \}$ and $I = \mathbb{N}$, $J = \{ n + 1\}$. But perhaps the notion of indexing already assumes a bijection, so taking $I = \mathbb{N}$ is improper?

Peter Taylor‭ wrote almost 4 years ago

To be clear: am I correct to understand that by "the last two points" you mean everything from "and we define" until the end?

The Amplitwist‭ wrote almost 4 years ago

@PeterTaylor Yes, that's right. I included more than just that part in the quote because I felt that it provided context for what Lang is trying to convey; hopefully, it isn't too confusing!

Monica Cellio‭ wrote almost 4 years ago

What algebra-related tags do you want to add? Moderators can create the tags for you. (So could I, as an admin, except that I don't know math well enough to know what to create on my own.)

The Amplitwist‭ wrote almost 4 years ago

@MonicaCellio I tried editing the tags on the post just now and was somehow able to create new tags! I'm not sure what went wrong for me earlier. Thank you for pinging, though. :-)