Can a set be a subset of another set but not be contained it the other set

elementary-set-theory

Say I have the set $A=\{\{1,2,\{3\}\}\}$ , and so the set cardinality of $A$ is $1$ from my understanding.

If $B = \{\{1,2,\{3\}\}, \{\{1,2,\{3\}\}\}$, then the set cardinality of B is $2$ too?

My question is

Suppose we have another set $C=\{ \emptyset \} \cdot A$，and another set $D=\{ \emptyset \} \cdot B $. Indeed, $ C \subseteq D$, but why isn't $C \in D$？

Because from my understanding the set $C$ can be written as $ \{(\emptyset , \{1,2,\{3\})\}$ and $D$ written as $ \{(\emptyset , \{1,2,\{3\}), (\emptyset , \{\{1,2,\{3\}\})\}$.

Sorry if my question isn't clear! Thanks for the help 🙂

Best Answer

  • Yes, the cardinality of $A$ equals $1$, and the cardinality of $B$ equals $2$. This just means that $A$ has one element, and $B$ has two.
  • The notation $\{\emptyset\}\cdot A$ does not have a clear meaning according to ordinary mathematical convention, and so it is up to you to define it. Assuming you mean the cartesian product of $\{\emptyset\}$ and $A$, this ought to be denoted with the $\times$ symbol. It is indeed true that $\{\emptyset\}\times A=\big\{(\emptyset,\{1,2,\{3\}\})\big\}$.
  • Yes, $C\subseteq D$, even though it is not the case that $C\in D$. The symbols $\in$ and $\subseteq$ simply have different meanings: $A\in B$ means that $A$ is an element of the set $B$, whereas $A\subseteq B$ means that every element in $A$ is an element in $B$. To give a simpler example, $\{1,3\}\subseteq\{1,2,3\}$, even though $\{1,3\}$ is not an element of $\{1,2,3\}$. The only elements of $\{1,2,3\}$ are $1$, $2$, and $3$.
  • The word "contained" is often used ambiguously. Sometimes, "$A$ is contained in $B$" means "$A$ is a member of $B$", whereas other times it means "$A$ is a subset of $B$". While the intended meaning of "contained" can sometimes be guessed from context, is safest to avoid using this word entirely.
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