What is the product of an empty family of similiar algebras, that is $\prod\langle \mathbf{A}_i \mid i \in I \rangle $, where $I = \emptyset$?

The family $\langle \mathbf{A}_i \mid i \in \emptyset \rangle $ equals $\emptyset$ beacuse the function $i \colon \emptyset \mapsto X$ (whatever the $X$ is) is the empty function $\emptyset$.

So the question is: what is $\prod \emptyset$? Is it $\{ \emptyset \}$?

Am I right? I believe it is, but I haven't a step-by-step explanation. What is the universum of such product? How does a function interpretation look like (and more important: why?)


  • $\begingroup$ To be honest I've never seen the notation $\prod x$ without indices. Can you provide the definition of $\prod x$, for an arbitrary $x$? $\endgroup$ – Git Gud Oct 6 '14 at 18:27
  • $\begingroup$ Or the definition of $\prod\langle \mathbf{A}_i \ | \ i \in I \rangle$. $\endgroup$ – Git Gud Oct 6 '14 at 18:37
  • 1
    $\begingroup$ Yes. As a set, the product is the collection of all functions $f$ with domain $I$ such that for all $i\in I$ we have $f(i)\in|\mathbf A_i|$, where $|\mathbf M|$ is the universe of the algebra $\mathbf M$. If $I=\emptyset$, the only such function is the empty function. $\endgroup$ – Andrés E. Caicedo Oct 6 '14 at 18:46
  • $\begingroup$ @Git Gut. $\prod \langle \mathbf{A}_i \ | \ i \in I \rangle$ denotes the same object as $\prod_{i \in I} \mathbf{A}_i$. $\endgroup$ – Roy Oct 6 '14 at 18:52
  • $\begingroup$ Then it's as Andres said. See the wikipedia entry on this. Which isn't surprising considering that the empty sum yields its neutral element, so does the empty product (since $\{\varnothing \}$=1). $\endgroup$ – Git Gud Oct 6 '14 at 19:26

To cut down on the number of unanswered questions, here we go.

Given any set of sets $\{B_i\}_{i\in I},$ we have by definition that $$\prod_{i\in I}B_i:=\left\{f:I\to\bigcup_{i\in I}B_i\mid\forall i\in I,f(i)\in B_i\right\},$$ so when $I=\emptyset,$ this is clearly just the set containing the empty function $\emptyset,$ and nothing else.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.