In “An element of a set can never be a subset of itself”, what does ‘itself’ stand for?

I have just begun learning about sets. My first language isn't English. I'm in high school.

Here's an example problem I found in my textbook:

Example 11: Let $A, B$ and $C$ be three sets. If $A∈B$ and $B⊂C$, is it true that $A⊂C$? If not, give an example.

Solution: No. Let $A=\{1\}, B=\{\{1\}, 2\}$ and $C=\{\{1\}, 2, 3\}$. Here $A∈B$ as $A=\{1\}$ and $B⊂C$. But $A⊄C$ as $1∈A$ and $1∉C$.

Note that an element of a set can never be a subset of itself.

The link to the textbook's chapter.

What does “itself” stand for here? Does it mean an element of a set can't be it's own (the element's) subset?

Or does that mean an element cannot be both an element and a subset of a set at the same time?

If $P=\{p\}, Q=\{\{p\}, q\}$, and $R=\{\{p\}, q, r\}$, we can say that $P∈Q$. But, can we say that both $Q∈R$ and $Q⊂R$ are true? Is it so that $Q$ cannot be both an element and a subset of $R$? Is $\{\{p\}, q, r\}$ not the same as $\{p, q, r\}$?

Solution 1:

Both interpretations are sensible. Unfortunately, both interpretations are false statements! That comment is just misguided. (It's not your fault; it's the author's fault)

For instance: your first interpretation is:

If $A$ is a set, and $x\in A$ is an element of $A$, then $x$ cannot be a subset of $x$.

But that is false. In Set Theory, sets can be elements of other sets, and every set is a subset of itself. So $x$ can certainly be a subset of itself. For example, if $A=\{\{1\},\{2\}\}$, then $x=\{1\}$ is an element of $A$, and $x$ is a subset of itself.

Your second interpretation is:

If $A$ is a set, and $x\in A$, then $x$ cannot be a subset of $A$.

But that is also false. In fact, there is a whole class of sets, known as "transitive set", with the property that every element is also a subset. For instance, the set $A=\{\varnothing,\{\varnothing\}\}$, whose elements are (i) the empty set and (ii) the set whose only element is the empty set; has the property that each of its elements is, in addition to being an element of $A$, also a subset of $A$.

In short: I'm not sure what the author meant to say with that comment, but both natural interpretations of it are false.

What is true is that, in general, if $A$ is a set and $x\in A$ is an element of $A$, then you cannot say, from these facts alone, whether $x$ is a subset of $A$ or not; and if your set theory allows for objects that are not sets ("ur-elements"), then you may not know whether $x$ is a subset of itself or not.

It is also true that in many set theories, one cannot have a set be an element of itself: that is, you can never have $A\in A$. (But there are set theories where this is valid, however...)

Solution 2:

It helps to think of the braces $\{\}$ as being quite literal. So if $P=\{p\}$, $Q=\{\{p\},q\}$, and $R=\{\{p\},q,r\}$, then:

• When we write $Q=\{\{p\},q\}$, it means that the set $Q$ contains the two elements $\{p\}$ and $q$. In symbols, $\{p\}\in Q$ and $q\in Q$. Since $P=\{p\}$, we can interchange those two things, so we can also write $P\in Q$.
• The statement "$P\subset Q$" means "any element of $P$ is an element of $Q$." Well, $p$ is an element of $P$, but not of $Q$.
• The set $R$ contains $\{p\}$, $q$, and $r$, and $Q$ contains $\{p\}$ and $q$. Thus, $Q\subset R$. However, $Q\notin R$, because $R$ does not contain the element $Q=\{\{p\},q\}$.

As someone pointed out, it is not true that if $x\in S$, then $x$ is not a subset of $S$, nor the similar statement that if $x\subset S$, then $x\notin S$. The set $\{1,\{1\}\}$ gives a counterexample to both. The only true statement I can think of here is that a set $S$ is never an element in itself. We can never have $S\in S$.