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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.

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    $\begingroup$ A has one element but B has two. $\endgroup$ – John Douma Jul 16 '16 at 2:02
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    $\begingroup$ Are you asking if A is a subset of B or if A is equal to B? $\endgroup$ – John Douma Jul 16 '16 at 2:06
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    $\begingroup$ If being a subset is the same as being equal, then all sets are empty. $\endgroup$ – Asaf Karagila Jul 16 '16 at 2:11
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    $\begingroup$ 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 $. $\endgroup$ – fleablood Jul 16 '16 at 2:35
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    $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen Jul 16 '16 at 8:04
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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.

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    $\begingroup$ And now think about this: does the box of all the boxes that don't contain themselves contain itself? Hahaha $\endgroup$ – Pierre Arlaud Jul 16 '16 at 11:15
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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, $A\neq B$

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