# Cardinality of Two Sets with Empty Sets

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?

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.

• So for the first one $\{ \emptyset, \{ \emptyset\}\}$, the cardnality would be 2 because it only has two elements, right? – user750949 Sep 20 '20 at 18:50
• @M__: Yes, that’s right. – Brian M. Scott Sep 20 '20 at 18:52
• Wow ! What an explanation ! – user710290 Sep 20 '20 at 18:53
• 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 $2$. – Ilmari Karonen Sep 21 '20 at 2:48
• @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. – Brian M. Scott Sep 21 '20 at 2:54