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I have been reading Naive Set theory book by Holmes and it is stated that

$A\in A$

is not true of any reasonable set and hence it isn't reflexive.

Why isn't belonging ($\in$) reflexive ? I cannot convince myself why it isn't reflexive. Can anybody explain it with some simple examples.

Sibi
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  • You might be getting confused between $\in$ and $\subseteq$. –  Apr 21 '14 at 13:44
  • Naive set theory does not disallow a set being an element of itself. Most modern set theories do disallow this possibility though, because Russell's Paradox implies that naive set theory is inconsistent. – Dan Rust Apr 21 '14 at 13:48
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    @Daniel: That's not the reason. The reason is that we benefit well from the well-foundedness of the $\in$ relation. But it was once very useful to have many sets satisfying $x={x}$. – Asaf Karagila Apr 21 '14 at 14:30

1 Answers1

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In order for the relation $\in$ to be reflexive, it must be true that $A\in A$ for each set $A$. As Halmos points out, this is not true. Just about any set $A$ you choose will give a counterexample. For example, take $A = \mathbb N = \{0,1,2,\ldots\},$ the set of natural numbers. Then $A\notin A$, because $\mathbb N$ is not itself a natural number.

For an even simpler example, take $A = \{0,1\}$. Then $A\notin A$, as $\{0,1\}\neq 0$ and $\{0,1\}\neq 1$.

And for the smallest example (which is not the same as the simplest example, I suppose), take $A$ to be the empty set $\emptyset$. Then $A$ has no elements, so it is certainly not an element of itself.

bradhd
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  • what if A is "set of sets"? – athos Apr 22 '14 at 01:14
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    @athos: then $A \in A$, and you're one step away from Russell's paradox: define $S = {x \in A: x \notin x}$ and consider whether or not $S \in S$. Anyway, the meaning of "$\in$ is reflexive" is not, "there is some set $A$ for which $A \in A$". "Reflexive" means "for every set $A$, $A \in A$. So to show $\in$ is not reflexive we only need one set that is not an element of itself. Once you have one that isn't you don't care about the rest for the purpose of this question. – Steve Jessop Apr 22 '14 at 01:40
  • @SteveJessop thanks for the answer. I totally agree. – athos Apr 22 '14 at 01:57