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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?

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  • $\begingroup$ @AsafKaragila I made a small mistake, I edited it to display what I actually meant. $\endgroup$ – Phaptitude Sep 14 '14 at 12:04
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    $\begingroup$ 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
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    $\begingroup$ "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
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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.]

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  • $\begingroup$ 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
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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 $A$.

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  • $\begingroup$ Ah, so it has no elements but can still be an element? $\endgroup$ – Phaptitude Sep 14 '14 at 12:07
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    $\begingroup$ Yes. $\varnothing$ is an element of every power set. $\endgroup$ – Asaf Karagila Sep 14 '14 at 12:08
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    $\begingroup$ @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
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    $\begingroup$ @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

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