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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 $\{p, q, r\}$?

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    $\begingroup$ Both are sensible... and both are false! The author is saying something that either makes no sense, or else is wrong. $\endgroup$ – Arturo Magidin May 27 at 4:03
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    $\begingroup$ I bet that the author here was thinking of the statement "A set cannot be a member of itself" (which follows, in ZFC, from the axiom of foundation) and got confused between "member" and "subset." $\endgroup$ – Milo Brandt May 27 at 4:19
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    $\begingroup$ The only thing this makes me think of is the classical Barber's paradox by Bertrand Russel. "The barber shaves everyone who don't shave themselves". So does the barber shave himself? To avoid allowing self references like this which create logical paradoxes. $\endgroup$ – mathreadler May 27 at 13:52
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    $\begingroup$ @MiloBrandt: Unfortunately, that's not the reason. I know many textbooks that make all kinds of false or even nonsensical claims. The one quoted here is yet another to add to the list. Rishi, it is probably best to avoid that textbook if you want to study proper mathematics. $\endgroup$ – user21820 May 27 at 13:53
  • $\begingroup$ Note that, if $C = \{ 1, \{1\}, 2,3 \}$ then $A \in C$, $A \subset C$, and $A \subseteq A$. $\endgroup$ – steven gregory May 27 at 17:07
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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...)

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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 $S\in S$.

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