2
$\begingroup$

Is there a difference between $\{1,\emptyset\}$ and $\{1\}$? Does the former have $2$ elements and the latter $1$ element? Does adding a string of $\emptyset$'s increase the number of elements of a set?

$\endgroup$
  • 2
    $\begingroup$ The empty set can be an element of a set. Consider for example the power-set always containing the empty set. $\endgroup$ – Peter Apr 23 '17 at 22:58
  • 1
    $\begingroup$ Adding "a string of" any element won't increase the number of elements. For instance, $$ \{1,2\} = \{1,2,1,1,1\} = \{1,2,1,2,2,1,1\} $$ $\endgroup$ – Omnomnomnom Apr 23 '17 at 23:00
  • 2
    $\begingroup$ Note also that $\emptyset = \{\}$, and $\{\emptyset\} = \{\{\}\} \neq \emptyset$ $\endgroup$ – Omnomnomnom Apr 23 '17 at 23:03
  • 1
    $\begingroup$ It is true, as other noted in answers and comments, that adding a string won't increase the number of elements. This is the usual and widely accepted convention, or rather the convention for sets. But, it might be worth noting that there are so-called multisets, see en.wikipedia.org/wiki/Multisets (a different thing, perhaps unimportant in the context of your question) for which a string does increase the number of elements. $\endgroup$ – Mirko Apr 24 '17 at 0:52
3
$\begingroup$

The main problem here is that $1\in\mathbb N$ and $\varnothing$ are different in nature. $1$ is an integer, while $\varnothing$ is a set.

You can't have a set mixing sets and non-sets, $\{1,\varnothing\}$ is simply wrong in general.

Remark : emphasis on "in general", because as in Unix we say "everything is file", for set theoretists "everything is set", naturals, reals, functions, etc... See below for instance the construction of naturals, for a possible meaning of $\{\varnothing,1\}=2$.


$\{\{1\},\varnothing\}$ would be fine and it has $2$ elements which are sets.

The sets $\{\varnothing\}$ and $\{\varnothing,\varnothing\}$ are equal, in the same way that $\{1\}=\{1,1\}$ because in a set, elements appear only once.

So repeating an element won't increase the cardinality, it's just the same set.


Now, the fact that for a set $S$ we have $S\neq\{S\}$ is the base of the definition of the axioms of the naturals. There are two main constructions :

  • Zermelo

$\begin{array}{l} 0=\varnothing\\ 1=\{0\}=\{\varnothing\}\\ 2=\{1\}=\{\{\varnothing\}\}\\ 3=\{2\}=\{\{\{\varnothing\}\}\}\\ n+1=\{n\}=\{\{..\{\varnothing\}..\}\}\} \end{array}$

  • Von Neumann

$\begin{array}{l} 0=\varnothing\\ 1=0\cup\{0\}=\{0\}=\{\varnothing\}\\ 2=1\cup\{1\}=\{0,1\}=\{\varnothing,\{\varnothing\}\}\\ 3=2\cup\{2\}=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}\\ n+1=n\cup\{n\}=\{0,1,2,..,n\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\},\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\},...\}\\ \end{array}$

In the Zermelo case, the cardinal of a number is always $1$ but elements are themselves sets of cardinal $1$ nested like russian dolls.

In the Von Neumann case, the cardinal of the number, is the number itself, but the nesting is more complex. Every natural is the set of all its predecessors.

$\endgroup$
  • $\begingroup$ I prefer $|n| = n$ $\endgroup$ – Henno Brandsma Apr 24 '17 at 17:22
  • $\begingroup$ Since everything is a set, I see nothing wrong in $\{1,\emptyset\}$. And in the second part you're saying that $1=\{\emptyset\}$ and $2=\{1,\emptyset\}$, contradicting yourself. $\endgroup$ – egreg Apr 26 '17 at 22:45
  • $\begingroup$ Nothing wrong under the point of view of sets, but OP asks if there is a difference between $\{1\}$ which I assume he considers a subset of $\mathbb N$ and $\{1,\varnothing\}$ which this time is not a subset of $\mathbb N$ since $\varnothing$ is not an element of $\mathbb N$, unless we use the canonical mapping described above. This is why it is dangerous to write such "unstructured" sets, not because it is forbidden, but because there is a risk a confusion between things whose nature is different. Am I wrong ? $\endgroup$ – zwim Apr 26 '17 at 23:09
  • $\begingroup$ -1. You wrote: "You can't have a set mixing sets and non-sets, $\{1,\emptyset\}$ is simply wrong in general." This statement disagrees with the standard mathematical conception of set. According to this conception, it is certainly true that given any finite list of mathematical objects (like $1$ and $\emptyset$), there is a set containing exactly these objects. There is an attractive philosophical position that mathematics should be done in a typed system, in which every set has type Set of T, where T is another type, e.g. Set of Natural Numbers, Set of (Sets of Real Numbers)... $\endgroup$ – Alex Kruckman May 4 '17 at 13:36
  • $\begingroup$ But even in typed systems, you can often form "sum"/"disjunction"/"union" types, so if $\emptyset$ has type Set of Natural Numbers and $1$ has type Natural Number, I could form the type Set of (Natural Numbers OR Sets of Natural Numbers), and $\{1,\emptyset\}$ would have this type. $\endgroup$ – Alex Kruckman May 4 '17 at 13:38
3
$\begingroup$

To your first question - Yes, the two sets are different. In fact $\{1\}\subset\{1,\emptyset\}$.

Adding a string of anything doesn't increase elements in a set - $\{1,2,3,4\}=\{1,1,2,3,3,3,3,4,4\},$ so no this doesn't change anything for the empty set either.

$\endgroup$
2
$\begingroup$

Yes they are different sets. The empty set is not nothing. It is the set which contains nothing, an empty box, if you will. Thus the set which contains both one and an empty box is different than the set which contains just one. Incidentally, the first set you mentioned is what we define the ordinal 2 to be.

$\endgroup$

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.