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Given a set $S = \{A,B\}$ such that $A = \{1,2,3\}$ and $B = \{1,2,3\}$, what is the cardinality of S?

I know this may seem very trivial, and I am inclined to believe that the answer is $2$; my question comes from the fact that $A = B$. Since both $A$ and $B$ are the same, would that not be the same as having a repeated element, therefore making the answer $1$?

Crosby
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Carlos Romero
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3 Answers3

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$S$ has one element, so has cardinality $1$. (Although we have two labels for $\{1,2,3\}$, there is still only one such set.) Sets don't have duplicates. (Multisets can have duplicates. If $S$ were a multiset, it would have cardinality $2$.) The set $\{x, x, x, x, x, x, x, x\}$ has one element, $x$.

Eric Towers
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The cardinality of your the set $S$ would be $1$. Since $A$=$B$, the cardinality is $1$.

Also, sets that are subsets of a set are just items if you will, and can be counted as elements. The only reason the cardinality is not $2$ is because the sets are equal

Prime
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  • The word ''subset'' here has a different meaning, right? –  Jul 01 '18 at 01:22
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    Thanks so much for the clarification! – Carlos Romero Jul 01 '18 at 01:26
  • If $S={1, 2, 3}$, then $2$ is an element of $S$, which is denoted by $2 \in S$. $S$ is a subset of $S$, which is denoted by $S \subset S$. –  Jul 01 '18 at 01:36
  • I coincide with you, @user529760. That $x\in A$ for some set $A$ does not mean that $s\subseteq S$. By definition, $A\subseteq B$ if and only if $x\in A\implies x\in B$ for all objects $x$. So this should mean that the inclusion “$\subseteq$” is not defined if either $A$ or $B$ is not a set. To see this, set $A=1$ and $B={1,2,3}$. Clearly, $A\in B$. But is it true that $\color{red}{x\in 1}\implies x\in{1,2,3}$? No, as there is no reasonable way to define the sentence $x\in 1$! Moreover, it is possible that $A\subseteq B$ but $B\notin A$ (e.g. take $A={1,2}$ and $B={1,2,3}$). – Crosby Jul 01 '18 at 12:31
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By the vanishing property, since $A=B$, $S=\{A\}=\{B\}$, hence $|S|=1$.

aidangallagher4
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  • Ok, but how does it relate to ${A,B}$? – peterh Jul 01 '18 at 01:36
  • The vanishing property (I feel like I’ve made up the name in hindsight but anyway) says that repeated elements in a set don’t matter; ie ${1,2,1}={1,2}$ or in your case ${A,B}=({A,A}={B,B})={A}={B}$ since $A=B$ – aidangallagher4 Jul 01 '18 at 01:40