Set Theory and Beyond

Everything I said about $n$-tuples can also be said about "$I$-tuples", where $I$ is now an arbitrary set. What I mean by this is that we can define an $I$-tuple to be a function with domain $I$. If we set $I=n$, we just get the notion of an $n$-tuple! However, in this case we usually don't use the term "tuple", but rather a  family.  So, a family  indexed by set $I$  is just a function with domain $I$. An $n$-tuple is then a family indexed by $n$. A family of elements of $X$ (indexed by $I$) is a function $$s:I⟶X.$$ We sometimes write $(a_i:i\in I)$ to denote the family $s$ indexed by $I$ such that $s(i)=a_i$ for all $i\in I$.  If $(A_i:i\in I)$ is a family (of sets), we define its product $$\prod _{i\in I}{A}_i$$ to be the class consisting of all families $s$ indexed by $I$ which satisfy that $s(i)\in A_i$ for all $i\in I$. If we have that $A_i=X$ for all $i\in I$, we use the notation \[X^I:=\pr...

I defined the ordered pair as $$⟨x,y⟩:=\{\{x\},\{x,y\}\}.$$ I might've used the notation $(x,y)$ in some places by accident, but let's be particularly careful of our notation in this post. Having defined an ordered pair $⟨x,y⟩$, we might wonder what an ordered triple $⟨x,y,z⟩$ would be. One way to approach this is to define $n$-tuple by recursion on $n$. So we say: $$⟨x_0⟩:=x_0,$$ $$⟨x_0,\dots ,x_n⟩:=⟨⟨x_0,\dots ,x_{n-1}⟩,x_n⟩.$$ In this definition, we know what is a 1-tuple and we can easily see that a 2-tuple exactly corresponds to an ordered pair. From there, we can see that $⟨x,y,z⟩$ is actually $⟨⟨x,y⟩,z⟩$, and so on. However, there is a subtle issue here and it has to do with the rigor that I'm suppressing from my posts. Very briefly, our definition would here go something like: "given sets $x_0,\dots ,x_{n-1}$,...

So, a binary relation is a set of ordered pairs, any such set. A binary relation on a set $X$ is a binary relation which is a subset of $X\times X$. A binary relation between sets $X$ and $Y$ is a binary relation which is a subset of $X\times Y$. Now, I want to say what is a function: a function will be a special kind of a binary relation. Let us fix a binary relation $R$. This just means that $R$ is a set of ordered pairs. Two natural sets assigned to $R$ are its domain  and its range :  $$\mathrm{dom}(R):=\{x:\mathrm{\exists }y,⟨x,y⟩\in R\},$$ $$\mathrm{ran}(R):=\{y:\mathrm{\exists }x,⟨x,y⟩\in R\}.$$ Of course, it's not immediate that these are sets, but there is no doubt that they are classes. In order for a class to be a set, it suffices for it to be contained in a set. More precisely, if a class $C$ is contained in a set $X$, then $C$ is equal to the intersection $X\cap C$. The intersection of a set and a class is a set by Axiom of...

 [I continue writing in the form of a dialogue between Master (M) and Apprentice (A).] M: As I promised last time we met, I'll explain to you today what is an ordering. A: I've been waiting so eagerly for this! Please, go on! M: Let's look at some set $X$. You now know what a binary relation on $X$ is. Let $<$ be a binary relation on $X$. By the way, since $<\subseteq X\times X$, we can have that $(x,y)\in <$ for some elements $x$ and $y$ of $X$. However, we usually write "$x<y$" instead of "$(x,y)\in <$". A: Oh, yes, that's a standard mathematical notation. Now I understand what it formally means. M: Good. Okay, so $<$ is a binary relation on $X$, but in order for it to be an ordering, it needs to have some additional properties. In fact, I'll first tell you what is a strict ordering. Binary relation $<$ will be a strict ordering just in case it is irreflexive and transitive. Being irreflexive mea...

[I continue writing in the form of a dialogue between Master (M) and Apprentice (A).] A: O Noble Teacher, last time we talked, you said that a unary relation is just an arbitrary set. But you didn't explain why would you introduce another word just to mean "set"! M: O My Dear Apprentice, that is because when we say $R$ is a unary relation, we usually have context in mind and we usually say more. For example, we might be thinking about some set $X$ and then we might say that $R$ is a unary relation on set $X$ . In this way, we want to say more then just "$R$ is a set". Formally, we mean that $R$ is a subset of $X$, but intuitively, we separate elements $x$ of $X$ into those for which the relation $R(x)$ holds and those for which the relation $R(x)$ does not hold. Here, "$R(x)$ holds" just means "$x\in R$" and "$R(x)$ does not hold" means $x\in R-X$. A: Okay, that makes more sense. But what about binary r...

I'll write this post in form of a dialogue between Master (M) and Apprentice (A). A: Say, O Noble Teacher, what are those "relations" that mathematicians always talk about? M: Why, my dear Apprentice, you have already seen one kind of relations! A: How so, O Exalted One? M: Well, $\in$ is a relation and $V$ is also a relation! These are class-sized relations. A: Oh, I see! But how does their "relation-ness" manifest? M: In the case of $\in$, it divides pairs of sets $(x,y)$ into those for which $x$ belongs to $y$ and into those for which $x$ does not belong to $y$. So, it tells us whether the statement ''$x\in y$" is true or false for any particular pair of sets $(x,y)$. In this sense, they are no different from predicates or classes. Class-sized relations are the same thing as predicates. A: I seem to understand now! $\in$ is a binary predicate, so a class-sized relation. $V$ is a class, which is the same thing as a unary pr...