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If the identity of a set depends on what the set contains, then wouldn't a set represented as A contain the same entities a set represented as {A} contains?

Hal
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3 Answers3

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Let's see if I can make this clearer. Consider the set $\Bbb Z=\{0,\pm 1,\pm 2,\pm 3,\dots\}.$ This is a set of infinitely-many numbers. However, the set $\{\Bbb Z\}$ is a set with exactly one element, and this element is not a number. It is certainly true that the element $\Bbb Z$ contains numbers, but $\{\Bbb Z\}$ only contains $\Bbb Z$ as an element. A further distinction to be made is that, while $\{\Bbb Z\}$ contains (only) $\Bbb Z,$ the set $\Bbb Z$ does not contain $\Bbb Z$ as an element.

Or what if we considered $\emptyset$ and $\{\emptyset\},$ instead? The first has no elements, while the second has one, so they are certainly not the same set.

Cameron Buie
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As given:

$A$ is a set containing an unknown number of elements, perhaps none.

$\{A\}$ is a set containing an element which happens to be a set, and in particularly, it is a set containing $A$ as its only element. In this case, we have that $A \in \{A\}$.

amWhy
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$\{A\}$ is a set with only one element, the set $A$.

Julián Aguirre
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  • So say the set A contains the elements 1,2,3 - then doesn't {A} also contain 1,2,3 - so they become the same? – Hal Jan 18 '14 at 15:19
  • Sets can be elements of another set. The set ${A}={{1,2,3}}$ contains only one element, the set $A={1,2,3}$. $$ \bigl{{1,2,3}\bigr}\ne{1,2,3}.$$ – Julián Aguirre Jan 18 '14 at 19:41
  • one of the questions in my text provides B={a,b} G={{a,b},{c,2}} and asks {B}⊆G? {B} is {{a,b}}- so I answered that {B} is not a subset of G. My text says that it is. Who's wrong? – Hal Jan 18 '14 at 20:15
  • $B$ is an element of $G$. ${B}$ is a subset of $G$. – Julián Aguirre Jan 18 '14 at 23:17