0
$\begingroup$

On page 3/(4) of Dummit&Foote's Abstract Algebra (3rd ed.):

(Let $A$ be a nonempty set) A partition of $A$ is any collection $\{A_i\lvert i\in I\}$ of nonempty subsets of $A$ ($I$ some indexing set) such that (a) $A=\bigcup_{i\in I}A_i$, and (b) $A_i \cap A_j=\emptyset$, for all $i,j\in I$ with $i\neq j$. i.e., $A$ is the disjoint union of the sets in the partition.

  • May I ask, by the word indexing in this context, if the indexing set can contain infinitely many elements? In particular, can it contain uncountably many elements? (The answer to this question is probably "yes")

  • Furthermore, can the function from $I$ to those nonempty subsets of $A$ (i.e. the process of indexing) not be injective? More importantly, are there cases where we intentionally want it not to be injective? And why? [For this question we shall modify the definition (b) to "$A_i \cap A_j=\emptyset$, for all $A_i,A_j\in A$ with $A_i\neq A_j$"]

Some stranger questions:

  • Am I allowed to have $I=A$? If so, is it meaningful? (A kind of self-indexing?)

  • Is that possible, by any chance, to have some partition of some set being unable to be indexed?

$\endgroup$
2
$\begingroup$

In general an indexing set can be any set at all. Yes, one member of a partition can have more than one index, though the default would be an injective map. Yes, it’s possible to index a partition of $A$ by $A$ itself: if $\mathscr{P}$ is a partition of $A$, for each $a\in A$ let $A_a$ be the unique member of $\mathscr{P}$ containing $a$. In this case the indexing is clearly not injective unless $\mathscr{P}$ is the partition of $A$ into singletons. This of course shows that every partition can be indexed. In fact, every partition can be injectively indexed, since it can trivially be indexed by itself: the function is $$f:\mathscr{P}\to\mathscr{P}:P\mapsto P\,,$$ or, in the notation of Dummit & Foote’s definition, $A_P=P$ for each $P\in\mathscr{P}$. This is pretty pointless, but it is technically an indexing of the partition.

I dislike that definition, because, as some of what I’ve written may suggest, the indexing is completely unnecessary. A partition $\mathscr{P}$ of a non-empty set $A$ is simply a family of pairwise disjoint, non-empty subsets of $A$ such that $\bigcup\mathscr{P}=A$. Any surjective function $f$ from a set $I$ to $\mathscr{P}$ can then be used to index $\mathscr{P}$: for each $i\in I$ we have $P_i=f(i)$.

$\endgroup$

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.