Consider the following sets:

$$ A = \{1, 2, \{1,2\}, \emptyset \} $$

$$ B = \emptyset $$

My book says that $|A| = 4$ and $|B| = 0$. Why is $\emptyset$ considered an element if it's a subset, but not when it's on its own?

  • $\begingroup$ @AsafKaragila I made a small mistake, I edited it to display what I actually meant. $\endgroup$ – Phaptitude Sep 14 '14 at 12:04
  • 5
    $\begingroup$ How many elements does $\{\{1,2\}\}$ contain? Hint: Not two. $\endgroup$ – Thomas Andrews Sep 14 '14 at 12:52
  • $\begingroup$ Now tell,what is the power set of an empty set? $\endgroup$ – rah4927 Sep 14 '14 at 16:18
  • 6
    $\begingroup$ "Why is $\emptyset$ considered an element if it's a subset, but not when it's on its own?" shows you haven't understood the notation (nor the terminology). In the first line the symbol $\emptyset$ designates an element (not a subset) of $A$. In the second line it designates the value of $B$. If it were $B=\{\emptyset\}$ instead it would be an element of $B$, and one would have $|B|=1$. $\endgroup$ – Marc van Leeuwen Sep 14 '14 at 16:26
  • $\begingroup$ So $∅ ≡ \{\}\space ∴\space A=\{1,2,\{1,2\},\{\}\}$ ? $\endgroup$ – Zaz Apr 29 '15 at 19:30

The "philosophical" issue behind this which in the beginning confuses many people is that in everyday mathematics you're almost always dealing with "typed" sets - meaning that the elements of the sets you'll encounter are always of the same kind: You might have sets of natural numbers like $\{1,2,3\}$ or sets of reals like the interval $[0,\pi)$. Later you'll maybe encounter sets of vectors or sets of functions and so on. Still, the "roles" are always kind of clear, sets are "containers" for "other" objects - the ones you are "really" dealing with. [Things get muddy once you start with topology, though.]

But in (axiomatic) set theory there are no "other" objects. Everything you'll ever encounter are sets - which entails that all sets have to be able to play both roles, the "container role" as well as the "element role".

So, your $A$ above is a set which is a container for four other things - and these four other things are also sets and one of them is $\emptyset$. (In other words, $\emptyset$ here plays the "role" of an element of a set.) But for the cardinality of $A$ the only thing that "counts" is how many objects it contains, so it is $4$. It doesn't matter whether one of its elements - $\emptyset$ - has cardinality $0$ or whether another element - $\{1,2\}$ - has cardinality $2$. Each element of a set has the same "right" to be counted - no matter whether it's the tiny empty set or a huge uncountable bouncer like $\mathbb R$.

But $\emptyset$ can also be viewed in its "role" as a set containing objects. And as a set it is a container for zero elements (by definition). So its cardinality is $0$.

[In case you're wondering: Yes, $1$ and $2$ are also sets as far as set theory is concerned. More about this can e.g. be found in other answers on this site, e.g. here.]

  • $\begingroup$ The untyped nature of the universe isn't the heart of the questioner's problem I think and in any case it is possible to have axiomatic set theories with urelements. $\endgroup$ – Francis Davey Sep 14 '14 at 20:33

Sets can be elements of other sets. They can also be subsets. That is irrelevant for defining cardinality of a set.

The cardinality of a set is "the number of elements in a set". $\varnothing$ has no elements. It has zero elements. So its cardinality $0$. Much like that $\{1,2\}$ has cardinality $2$, regardless to the fact it is an element of $A$.

  • $\begingroup$ Ah, so it has no elements but can still be an element? $\endgroup$ – Phaptitude Sep 14 '14 at 12:07
  • 5
    $\begingroup$ Yes. $\varnothing$ is an element of every power set. $\endgroup$ – Asaf Karagila Sep 14 '14 at 12:08
  • 1
    $\begingroup$ @Phaptitude: To have or to be. The empty set never has any element, but it can be one. $\endgroup$ – Marc van Leeuwen Sep 14 '14 at 16:28
  • 2
    $\begingroup$ @Phaptitude: I have no husband, but I still can be a husband. On the other hand, my car has wheels but isn't a wheel. It sounds that you are at a stage where you are looking at combinations of words without understanding their meaning. $\endgroup$ – gnasher729 Sep 14 '14 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.