2

Could anyone explain the difference between the following:

  1. $\varnothing$

  2. $\begin{Bmatrix} \varnothing \end{Bmatrix}$

  3. $\begin{Bmatrix} \varnothing & \begin{Bmatrix} \varnothing \end{Bmatrix} \end{Bmatrix}$

I know number 1 is an empty set. So if S is an empty set, can denote it like this: S = {}. Correct?

Now, I'm having difficultly with perceiving what number 2 is. Well it looks like it's a set of an empty set.

Confused with 3 as well. Could anyone explain to me the difference between the three? (Note: I've just started discrete maths this week.)

Martin Sleziak
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Bobby
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3 Answers3

15

As you say, $\varnothing$ is the empty set; you can indeed represent it as $\{\}$.

$\{\varnothing\}$ is a set with one member; that member is the empty set. If you think informally of a set as a box, $\varnothing$ is an empty box, and $\{\varnothing\}$ is a box that contains an empty box and nothing else. You could write it $\{\{\}\}$.

$\{\varnothing,\{\varnothing\}\}$ is a set with two elements; one of those elements is the empty set, and the other one is the set whose only element is the empty set. In the box metaphor $\{\varnothing,\{\varnothing\}\}$ is a box that contains two other boxes; one of those boxes is empty, and the other one contains an empty box. You could write this set $$\bigg\{\{\},\Big\{\{\}\Big\}\bigg\}\;,$$ where I’ve used different sizes of braces to make it easier to see which ones match.

Brian M. Scott
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  • Thank you so much! The analogy really helped! :) I just thought that an empty set means simply nothing. But I was clearly wrong. – Bobby Jul 29 '13 at 12:40
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    @Bobby it helps (me, at least) to remember the name: the empty set is not "nothing" because it is still a set. All sets contain n other things; the empty set is a set, which contains 0 other things. – KutuluMike Jul 29 '13 at 14:02
  • @Bobby: You’re very welcome; I’m glad that it worked for you. – Brian M. Scott Jul 29 '13 at 18:15
  • Of course, the place where your box analogy gets confusing is that the larger box and the smaller box inside it both contain the same empty box... – MartianInvader Jul 29 '13 at 19:59
4

The first listed item is the empty set.

The second is not the empty set, because it has one member. That member is the empty set.

The third (if you meant what I'd denote by $\{\emptyset,\{\emptyset\}\}$) is a set with two members, one of which is the empty set.

Michael Hardy
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3

Yes, the first is the empty set. Notation $S = \{\}$ would be correct, but it might be better to keep to $S = \emptyset$, to maintain better readability.

As for 2, this is a set that contains one object (which itself is an empty set). This means that, if $S = \{ \emptyset \}$, cardinality of $S$ is $1$, i.e., $\#S = 1$. It contains one element, and it is irrelevant that this element itself is an empty set.

Imagine an empty room. The number of items in it is zero, right? Now, put an empty cupboard in it. How many items are in that room? Obviously one, regardless of the fact that the cupboard is empty.

The third makes no sense to me. However, $S = \{ \emptyset, \{ \emptyset \} \}$ (note the comma) is a set containing two elements: an empty set and a set that itself contains an empty set.

This is a standard way to define natural numbers in the set theory. You start with the empty set, which you can axiomatize (put nothing in the bag... it's easy to accept that this exists) and then you a create new set $\{ \emptyset \}$ with one element. Now, you have two objects: $\emptyset$ and $\{ \emptyset \}$, so you use them to build a set with two elements: $\{ \emptyset, \{ \emptyset \}\}$. Now you have three different objects (since they are sets with different number of elements, they must be different), and you go on like that.

Vedran Šego
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