Russell's paradox and classes

We have already talked about properties, but we didn't define them. I'm not going to that right now either, but I remind you that a property (or a predicate) is some definite way to separate sets to those which have that property and to those which do not. For example, if $P(x)$ is a property and $a$ is a set, then $a$ might have the property $P$, written as $P(a)$, or it might not have the property $P$, written as $\mathrm{\neg }P(a)$. Now, for any set $A$, we may look at the property $S_A(x)$ asserting "$x$ belongs to $A$". In this way, every set can be seen as a property, but the other way around is not true.

An example of the property which doesn't come from a set is given by so-called Russell's paradox. Namely, consider the property $R(x)$ asserting that "$x$ does not belong to $x$". I claim that this property does not come from a set, or more precisely, that there is no set $r$ such that the property $R$ is the same as the property $S_r$. Just a side note, you might be asking yourself right now what set could satisfy $x\in x$. In fact, one of the future axioms will exclude this possibility, so all sets will have the property $R$. However, we don't know and we don't need to know this right now.

Okay, how do we see that $R$ does not come from a set? Since we are convinced that this is true, we may try assuming that it is not and then conclude an absurd statement from this assumption. In this way, we will know that our assumption is itself absurd and that $R$ indeed does not come from a set. Well then, let us assume that there exists a set $r$ for which the properties $R$ and $S_r$ are one and the same. That these properties are the same simply means that for all sets $a$, it is true that $R(x)$ holds if and only if it is true that $S_r(x)$ holds. (A side note: "bla if and only if yada" is just another way of saying "if bla, then yada, and if yada, then bla".) Now we ask whether $R(r)$ is true or false (and it must be either one or the other).

• Case 1. $R(r)$ is true. By the definition of $R$ this means that $r\notin r$. However, we also know that $R$ is the same property as $S_r$, so $S_r(r)$ must also be true. Looking up the definition of $S_r$, we see that this means that $r\in r$. Well, well, we somehow have that $r\notin r$, but also that $r\in r$! This is of course absurd.
• Case 2. $R(r)$ is false. By the definition of $R$, this means that $r\in r$. And then again, $R$ is the same thing as $S_r$, so we have that $S_r(r)$ is false as well. Reading the definition of $S_r$ once again, we get that $r\notin r$ and we arrive and the same kind of absurd as in the first case!

There were two possible cases and both of them turn out to be absurd! This can only mean that our assumption was absurd and that $R$ can't possibly come from a set.

We now have an idea in our mind: all sets are properties, but not all properties are sets. This is kind of an abuse of language and we don't like it, so we introduce a new word for a property, a class. Formally, there is no distinction between a class and a property, but conceptually, a property is something that sets have or don't have, whereas a class is something to which sets may or may not belong. So, if $P$ is a property/class, then the expressions "$P(x)$ is true", "$x$ has the property $P$", and "$x\in P$" are all saying the same thing, with maybe a difference in intuition behind them in the third case. 

As you can see, we use the set-like notion for classes, as in "$x\in P$". Our observation from the beginning of the post can now be safely expressed as "every set is a class" and we shall think of classes as generalisations of sets. Russell's paradox shows that there are classes which are not sets and we call these classes proper. If $P$ is a property, we might write

$$\{x:P(x)\}$$

for the corresponding class (which is just a matter of notation and not a new object). Thus, an example of a proper class is the class

$$\{x:x\notin x\}.$$

Following the analogy between sets and classes, we can define the intersection and the union of two classes $P$ and $Q$:

$$P\cap Q:=\{x:P(x)\text{ and }Q(x)\}$$

$$P\cup Q:=\{x:P(x)\text{ or }Q(x)\}.$$

Now, observe this: if $a$ is a set, then $a$ is also a class, so $a\cap P$ is a well defined class. However, we have that

$$a\cap P=\{x:x\in a\text{ and }P(x)\}=\{x\in a:P(x)\}$$

and this last thing is a set by the Axiom of Comprehension! In fact, the Axiom of Comprehension can be reformulated by simply stating that the intersection of a set and a class is a set.

An important example of a class is the class $V$ of all sets, also called the universal class or the universe. This class corresponds to the property $x=x$, which is true for all sets $x$. It is easy to see now that $V$ is a proper class: if it were a set, then $V\cap R$ would also be a set (where $R$ is from Russell's paradox), but $V\cap R=R$ and $R$ is not a set. Another important example of a class is the class $\mathcal{P}(a)$ of all subsets of a given set $a$. This class is called the powerset of $a$ and it corresponds to the property "$x\subseteq a$". However, the question of whether or not this class is proper is more delicate than in the case of $V$ and I leave it to you for now to contemplate on it. 😁