0
$\begingroup$

I've asked a couple of pretty stupid questions, and I think I've hit on my basic confusion that will allow me to ask a question that helps me move forward without saying "please do my homework for me." (It's not actually homework.) (Though I obviously need to take some classes.)

Say you have a category of posets all over the set of, let's just say, English letters of the alphabet. It seems to me there are two possibilities.

Possibility one: the set of objects of this category, which in some way includes the set of English letters of the alphabet, would repeat elements since each each poset contains the same set of letters. I.e., you have multiple posets, each of which contains the letters a, b, c, etc., and therefore repeats elements. Repeating elements is a no-no, so this has to be wrong.

Possibility two: the posets themselves are the objects, and the elements that the posets order are not themselves objects of the category of posets. But I believe I've seen mathematicians act like they can directly reference the elements of the posets in a category like this one, so I do't understand how that works. For example, you might want to have a functor from the category of posets over the set of English letters to a category of posets over the set of English letters where the letter E is terminal. It seems like there's an obvious logical way to do said functor, but I'm confused on how the mapping actually works. How do you reference the elements inside a particular poset when the posets themselves are the elements of the set of objects of the category?

I guess I don't even need to ask about posets. The category of sets would include the sets $\{a\}, \{a, b\}, \{a, c\}$, and therefore be repeating elements, unless those elements are "inaccessible."

One way I've tried to approach this is to tell myself that, when (po)sets are objects, the list of elements in any particular set within the larger category is really just the name of the set. So when you ask for a functor from the category of posets over the English alphabet to the category of posets that specifically end at E, you could see this as referencing the posets whose names end with E, without actually getting into their elements. That doesn't immediately strike me as contradictory, but it also doesn't really seem like a proper solution.

I've realized that I don't even know how products of sets work. I feel like, if I'm going to say that I have two sets to take a product from, then those sets have to be in the same set, or how am I counting them in the first place? But then my same question arises: how do I reference the individual elements in each set?

Maybe related: Are elements of sets always and necessarily points?

I'm probably confused about basic set theory, not category theory, but since I am trying to work with category theory, I'd appreciate an answer that ties back into category theory after helping resolve whatever incredibly basic misunderstandings about sets I apparently have....

$\endgroup$
6
  • 1
    $\begingroup$ If the objects in a category are sets, then you can refer to their elements. The fact that there are categories where the objects are not sets doesn't mean you aren't allowed to treat the sets in a concrete category like sets. You're not restricted to using only category theory axioms. $\endgroup$ Apr 7, 2019 at 17:29
  • 1
    $\begingroup$ Some set theories allow “ur-elements” (elements that are not themselves sets); others do not allow ur-elements, and “it’s sets all the way down”. In fact, standard Zermelo-Fraenkel Set Theory does not have any elements that are not themselves sets. But in pretty much every set theory, there are always sets whose elements are not “always and necessarily points”. Every set theory I know has a notion of “power set”: if $X$ is a set, the power set of $X$ is a set whose elements are precisely the subsets of $X$; that is, its elements are all sets. $\endgroup$ Apr 7, 2019 at 17:40
  • $\begingroup$ I'm an idiot. Elements are sets. I already knew that. So I'm already referencing "the elements inside the sets inside the sets" when I'm doing anything at all. $\endgroup$
    – MalDLittle
    Apr 7, 2019 at 17:43
  • $\begingroup$ If $X$ is a set, $\cup X$ is the set $\cup X=\{a\mid \exists y\in X\text{ s.t. }a\in y\}$. So, for example, if $X=\{\{1,2,3\}, \{3,4,5\}\}$, then $\cup X = \{1,2,3,4,5\}$ (“erase the outermost set of brackets inside $X$“). If $X=\{\{1,\{2\}\}, \{3,4\}\}$, then $\cup X=\{ 1,\{2\}, 3, 4\}$. If $X=\{1,2,3,4\}$, and these are ur-elements, then $\cup X=\varnothing$. You can iterate this operator (sometimes called “amalgamated [unary] union”) get at elements inside the elements inside a set. $\endgroup$ Apr 7, 2019 at 17:49
  • 1
    $\begingroup$ Note that in a category, an arrow does not have to be described in formal terms in a particular language. Arrows need not even be functions in the usual sense (for example, the category of a single poset $(P,\leq)$ has the elements of $P$ as objects, and for each $x,y\in P$, an arrow set $\mathbf{P}(x,y)$ which has a single object if $x\leq y$, and is empty otherwise. And symmetrically, you can use whatever language you want to describe the arrows of the category, you don’t have to stick to just what the category “sees”. $\endgroup$ Apr 7, 2019 at 17:51

0

You must log in to answer this question.

Browse other questions tagged .