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Determine whether each of these pairs of sets are equal$$A = \{\{1\}\} \qquad \qquad B = \{1, \{1\}\}$$

I believe $A$ is equal to $B$ because all elements in $A$ are in $B$, but the answer says that it's not.

Rodrigo de Azevedo
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jame_smith
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    A has one element but B has two. – John Douma Jul 16 '16 at 02:02
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    Are you asking if A is a subset of B or if A is equal to B? – John Douma Jul 16 '16 at 02:06
  • @JohnDouma Aren't A is a subset of B and A is equal to B the same thing? – jame_smith Jul 16 '16 at 02:09
  • 1 is not the same as the set containing 1. – Doug M Jul 16 '16 at 02:10
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    If being a subset is the same as being equal, then all sets are empty. – Asaf Karagila Jul 16 '16 at 02:11
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    Aren't A a subset of B the same as A equal B? What ?!?!?! absolutely of course not! A equal B means A and B have the same elements. A subset B means all of the elements of A are in B but not nescessarily are all elements of B in A. In the is case they can't be equal because they don't have the same elements. B has 1 as an element but A does not. $B \not \subset A $. For $A=B $ then both $A\subset B $ and $B \subset A $. – fleablood Jul 16 '16 at 02:35
  • @fleablood Thx for clearing out the confusion :) – jame_smith Jul 16 '16 at 02:43
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    The answer actually depends on whether $1={1}$ or $1\ne{1}$. The former would make $1$ a Quine atom, which is an entity you will not want to encounter in the set theory you learn. – Hagen von Eitzen Jul 16 '16 at 08:04
  • @Hagen: Unless you learn about permutation models and ZFA. In that case, Quine atoms are very useful. – Asaf Karagila Jul 16 '16 at 14:12
  • @AsafKaragila I did not want to imply that you might not be interested in them. Rather, that the OP might lead a much happier life by postponing this encouner a bit :) – Hagen von Eitzen Jul 17 '16 at 19:40

2 Answers2

47

Think of $A$ as a bag which contains within it another smaller bag with a one in it.

$A=\underbrace{\{~~~~~~~\overbrace{\{1\}}^{\text{second bag}}~~~~~~~~\}}_{\text{first bag}}$

On the other hand, $B$ is a bag which contains in it not only a second bag with a one in it, but also a one which is loose.

$B=\underbrace{\{~~~~~~~~\overbrace{\{1\}}^{\text{second bag}}~~~~~\overbrace{1}^{\text{this too}}~~~~~~~\}}_{\text{first bag}}$

$1\in B$ but $1\not\in A$. There is no "loose 1" in $A$, there is only a bag with a one in it in $A$.

Thus, $A\neq B$

JMoravitz
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28

You're correct that all elements in $A$ are in $B$, but not the other way around - $B$ includes the element $1$, but $A$ only has $\{1\}$. Think of it like boxes - $B$ is a box that includes one item and also a box that itself contains one item; $A$ is just a box containing a box containing an item.

Reese Johnston
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