Given set $ A = \{ \emptyset \}$ what is $|A|$? I just learned about sets today in class and I'm unsure how to answer this.
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$\begingroup$$A=\{\emptyset\}$ is a set containing one element. That element is itself also a set, but this is irrelevant.
Note the difference between $A=\{\emptyset\}$, $B=\emptyset$, and $C=\{\{\emptyset\}\}$. These are all different sets. $A$ is the set containing the emptyset. $B$ is the emptyset, and $C$ is the set containing the set containing the empty set.
Cardinality of a finite set is simply the number of elements contained in the set, so in this case $|A|=1$
Similarly, $|C|=1$ and $|B|=0$ for the other examples I added above.
$\endgroup$ $\begingroup$In a way it is the start of the construction of natural numbers (cardinalities of finite sets)
0=| ∅ |, 1=|{ ∅ }|, 2=|{ ∅ ,{ ∅ } }|, 3=|{ ∅ ,{ ∅ },{ ∅ ,{ ∅ } } }|,...as you see the first set (between |.|) has no elements, the second one has one element, the third one has two distinct elements as the empty set and the set whose only element is an empty set are different. The (usually confusing) ... just adds the nested braces.
$\endgroup$ $\begingroup$I'm assuming the difficulty is from the imprecision of language, it may sound like the "set of the empty set" is the same as an empty set sort of like how a double-negative in English can really mean a negative. Just call "the empty set" something else to make it less confusing. Say it's "the special set" instead, something less intuitively transitive or inheritable than emptiness, denoted by $S$ instead of $\emptyset$. Then $A = \{S\}$. How many elements does $A$ have?
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