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I have a problem that asks me this:

If $A = \{a, b, c\}$ and $B = \{b, \{c\}\}$, is $B$ a subset of $A$?

What I'm confused about is if you treat $\{c\}$ and $c$ as the same elements.

I'm sort of confident that $\{c\}$ is distinct from $c$ and this would not be a subset, but I want to make sure.

rogerl
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mymemesarespiciest
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    ${c}$ is a set containing the element $c$. So no, $B = {b, {c}}$ is not a subset of $A = {a, b, c}.$. – amWhy Feb 09 '18 at 00:22
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    $;c;$ is an element, $;{c};$ is a set containing one single element, namely $;c;$ , so no: they are not the same at all and thus $;{c}\in B\setminus A\implies B\not\subset A;$ ...and it is not a very good idea to use the exact same symbol for two different things... – DonAntonio Feb 09 '18 at 00:24
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    Thanks for the quick responses! I appreciate it. – mymemesarespiciest Feb 09 '18 at 00:26
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    Similar: https://math.stackexchange.com/questions/723377/why-a-unit-set-is-not-the-same-as-its-element-x-ne-x https://math.stackexchange.com/questions/2620616/what-is-the-difference-between-x-and-x-when-x-itself-is-a-set – Chris Culter Feb 09 '18 at 00:26
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    Beware this problem may be more subtle... What I mean is, of course $c\ne {c}$, but what if $a={c}$? So, without knowledge about $a,b,c$ we cannot tell. –  Feb 09 '18 at 00:27
  • c is a cookie, {c} is a cookie jar with one cookie. – William Elliot Feb 09 '18 at 01:24
  • @mymemesarespiciest Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png – user Feb 10 '18 at 23:00

3 Answers3

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I'm sort of confident that $\{c\}$ is distinct from $c$ and this would not be a subset, but I want to make sure.

Be more confident.   It is so in axiomatic set theory that a set cannot be a member of itself.

As others have pointed out, it is possible that $a=\{c\}$, or even $b=\{c\}$; so we should make that cavat.$$\{b,\{c\}\}\nsubseteq \{a,b,c\}\text{ unless }a=\{c\}\text{ or }b=\{c\}$$

Graham Kemp
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$\{c\}$ is a set containing the element $c$, one is a subset of $A$: $\{c\} \subset A$, and the other is an element in $A: c\in A.$

So no, $B = \{b, \{c\}\}$ is not a subset of $A = \{a, b, c\}.$ $B$ is a set which contains an element of $A$ and a subset of $A$. But since $\{c\} \notin A$, $B$ cannot be a subset of A.

amWhy
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    Unless $a={c}$... I cannot tell if that was envisaged by the author of the problem... –  Feb 09 '18 at 00:29
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    That's a good point you raise. If in fact $a = {c}$, then indeed $B = {{x}, a}$, which means – amWhy Feb 09 '18 at 00:34
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No it is not, since "$c$" in A is an element which different from $\{c\}$ in B where it is a set.

max_zorn
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user
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