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I know cardinality is counting the number of elements in a set.

$\{ \emptyset, \{ \emptyset\}\}$ - I said that the cardinality of the set above was $2$ because $\emptyset$ is one element, and $\{\emptyset\}$ is another.

$\{ \emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$ - With this set I said the cardinality was $4$ because there is four elements. $2$ from what I previously stated and the other $2$ from $\{\emptyset, \{\emptyset\}\}$, it being two elements.

I don't understand cardinality with $\emptyset$ well. I know an empty set can be a set of its self and that the first one is the power set of a power set of an empty set, $P(P(\emptyset))$. Is my understanding flawed with cardinality? My logic for coming to my conclusions valid?

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Your understanding is flawed, I’m afraid. When you count the elements of a set, you count just the elements of that set: you don’t count the elements of those elements separately, and there is nothing special about $\varnothing$ in this context.

Let’s consider the set $x=\Big\{a,\{b\},\big\{c,\{d\}\big\}\Big\}$. It has $3$ elements: $a$, $\{b\}$, and $\big\{c,\{d\}\big\}$, so its cardinality is $3$. That last element happens to be a set with $2$ elements of its own, but it’s still just one member of $x$. We could replace it with the infinite set $\Bbb Z$ of all integers, getting the set $\big\{a,\{b\},\Bbb Z\big\}$, and we’d still have a set of cardinality $3$.

Now the cardinality of $x$ is $3$ no matter what $a,b,c$, and $d$ are.1 In particular, it’s $3$ even if $a=b=c=d=\varnothing$, so that $x=\Big\{\varnothing,\{\varnothing\},\big\{\varnothing,\{\varnothing\}\big\}\Big\}$. It’s also $3$ if $a=b=c=d=\Bbb Z$, and $x=\Big\{\Bbb Z,\{\Bbb Z\},\big\{\Bbb Z,\{\Bbb Z\}\big\}\Big\}$. In the first case the $3$ elements of $x$ are $\varnothing$, $\{\varnothing\}$, and $\big\{\varnothing,\{\varnothing\}\big\}$; in the second they are $\Bbb Z$, $\{\Bbb Z\}$, and $\big\{\Bbb Z,\{\Bbb Z\}\big\}$.

1 That’s not quite true, but the two exceptions involve a technicality that beginners sometimes find confusing. Specifically, if $a=\{b\}$, then $$x=\Big\{a,a,\big\{c,\{d\}\big\}\Big\}=\Big\{a,\big\{c,\{d\}\big\}\Big\}$$ and has only $2$ elements. Similarly, if $a=\big\{c,\{d\}\big\}$, then $$x=\big\{a,\{b\},a\big\}=\big\{a,\{b\}\big\}$$ and again has only $2$ elements. In no case, however, does $x$ have $4$ elements.

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  • $\begingroup$ So for the first one $\{ \emptyset, \{ \emptyset\}\}$, the cardnality would be 2 because it only has two elements, right? $\endgroup$ – user750949 Sep 20 '20 at 18:50
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    $\begingroup$ @M__: Yes, that’s right. $\endgroup$ – Brian M. Scott Sep 20 '20 at 18:52
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    $\begingroup$ Wow ! What an explanation ! $\endgroup$ – user710290 Sep 20 '20 at 18:53
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    $\begingroup$ Technically, your statement that "the cardinality of $x$ is $3$ no matter what $a,b,c$, and $d$ are" isn't quite true. For example, if $a = \{b\}$, then the cardinality of $x$ is at most $2$. $\endgroup$ – Ilmari Karonen Sep 21 '20 at 2:48
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    $\begingroup$ @IlmariKaronen: True. I don’t want to overcomplicate matters, but it might be worth adding a note in the answer itself; I’ll think about it for a little. $\endgroup$ – Brian M. Scott Sep 21 '20 at 2:54

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