I've recently started studying discrete mathematics for my computer science degree, and I have a doubt regarding membership and inclusion. I'll explain with an example.

Consider the set: $ A = \{1, 2,3,4,5 \} $

True or false?

  1. $1 \in A$

  2. $\{1\} \in A$

  3. $1 \subset A$

  4. $\{1\} \subset A$

Number one is obviously true, because the element $1$ is a member of $A$.

Number two, as far as I understand, should be false because it makes no sense. Membership is between a set and an element, not between two sets. Correct?

Number three: I'm not sure what the answer is. Can the concept of inclusion be applied between an element and a set or only between sets?

Number four is obviously true, because the singleton set $\{1\}$ is a subset of $A$.

  • Are my answers correct?
  • Can the concept of membership be applied between two sets? Or is it only always between an element and a set?
  • Can the concept of inclusion be applied between an element and a set? Or is it only always between two sets?
  • Consider $B = \{1, 2, 3\}, C = \{\{1\}, 2, 3\}$.
    • Is $B=C$ true? (Can a singleton set be considered as an element?)
    • Is $1 \in C$ true?
    • Is $\{1\} \in C$ true?
    • Is $1 \subset C$ true?
    • Is $\{1\} \subset C$ true?

You're correct that 1 and 4 are true.

On the other hand, if we're doing Zermelo-Fraenkel set theory, then the truthvalues of 2 and 3 cannot be determined from the information given. For example, 2 says that $\{1\} \in A$, or in other words that $\{1\} = 1$ or $\{1\} = 2$ or $\{1\} = 3$ or $\{1\} = 4$ or $\{1\} = 5$. Are any of those statements true? Well they could be. In Zermelo's approach to defining the natural numbers, $\{1\} = 2$ is true, indeed this is the definition of $2$. However, in von Neumann's approach, the set $\{1\}$ is not a natural number and thus not an element of $A$.

Similarly, the truthvalue of 3 cannot be determined. It all comes down to how you define the notion "natural number."

On the other hand, other approaches to the foundations would deem 2 and 3 "ill-formed" and therefore nonsensical. They're not even false; they're just nonsense.

To answer your questions: Membership can be a relation between two sets (at least in ZFC). For example, $\mathbb{R} \in \{\mathbb{R}\}$ is true. Furthermore, there is no distinction between "elements" and "sets" in ZFC; everything is a set, and "element" is a relationship: we can say the set $\mathbb{R}$ is an element of $\{\mathbb{R}\}$, but there's no point in saying that something "is" an element, period.

This was quite a rushed answer because I need to get back to my assignments, so please comment with any questions you may have.

  • $\begingroup$ Thanks for the reply. As far as I understand, basically, infinitely nesting an element into a set is equal to the element itself, correct? $\{\{\{...\{1\}...\}\}\} = 1$ --- Also, I'm having a lot of trouble understanding how $\{1\} = 2$. Can you elaborate on that, please? $\endgroup$ – Vittorio Romeo Oct 8 '13 at 12:34
  • $\begingroup$ @user18921 IIRC $0:=\emptyset$ and the successor of a natural number $S(n+1):n\cup\{n\}$, so $1=\emptyset\cup\{\emptyset\}=\{\emptyset\}$ and $2=1\cup\{1\}=\{\emptyset,\{\emptyset\}\}\neq\{1\}$. $\endgroup$ – Michael Hoppe Oct 8 '13 at 12:40
  • $\begingroup$ @VittorioRomeo, in early mathematics, no real distinction was made between a set $X$ and the set $\{X\},$ and no real distinction was made between $\in$ and $\subset$. There came a point when they realized that there's an important distinction. Suppose $X = \{1,2,3\}.$ How many elements does $X$ have? How many elements does $\{X\}$ have? Thus $X \neq \{X\}$. $\endgroup$ – goblin Oct 8 '13 at 12:47

Sets can be members of other sets. Element is just a term for a member of a set, it is not synonymous with "not a set". For example, given a set $A$, the power set of $A$ denoted by $\mathcal P(A)$ is a set whose elements are sets.

Numbers themselves can be represented as sets, and in modern set theory they are, and in fact everything is a set in the context of modern set theory. However if you are doing naive set theory then it is possible that you are assuming that numbers (say up to $\Bbb R$ or so) are not actually sets, but rather some atomic entity.

If this is indeed the case, then writing $1\subseteq A$ is indeed meaningless, because $1$ is not a set; however in the former case it is meaningful as you interpret $1$ as a particular set (e.g. $\{\varnothing\}$ is a common way, as $\varnothing$ is often represents $0$). If indeed $1$ is a set in your context, you need to know what is that set and then you can decide whether or not $1\subseteq A$ is true or false.


You may do it mechanically. If you remove $\{$ and $\}$ of $A$ you'll see a comma-separated list $$1,2,3,4,5$$ of the elements of $A$. So clearly (1) is true, since “1” is in that list and (2) is false since “$\{1\}$” does not appear in that list.

For $C$ the list of its element is $$\{1\},2,3$$ hence $\{1\}$ is an element of $C$.

Mechanically you'll get a subset by removing first the braces, then choose some elements from the element-list and put braces round those elements again. So (3) is false and (4) is right.

Obexample: for $M=\{\{0\},0\}$ it's true that $0\in M$, $\{0\}\in M$, $\{0\}\subset M$ and $\{\{0\}\}\subset M$ Do you see why?

  • $\begingroup$ I think it's a bit misleading to say that (2) is false since "$\{1\}$" does not appear in the list. Although that particular notation does not appear there, one still has to check whether "$\{1\}$" is equal to 1, 2, 3, 4, or 5 as user18921 pointed out. $\endgroup$ – Trevor Wilson Oct 8 '13 at 16:34
  • $\begingroup$ So in which case is $\{a\}=a$? $\endgroup$ – Michael Hoppe Nov 2 '16 at 21:28
  • $\begingroup$ That is never the case, by the axiom of regularity. I meant that you should check that $\{a\} \ne b$ for every $b$ in the list. For example, if $A = \{0,1,2,3,4\}$ then technically speaking we have $\{0\} \in A$ because $\{0\} = 1$, although this would probably only be used in a set theory class and not in OP's discrete math class. More generally, my point is that it's not quite enough to check whether a given notation appears in the list, because the same object might be denoted in different ways. (I'm not sure why I made such a nitpicky comment, or who user18921 is.) $\endgroup$ – Trevor Wilson Nov 3 '16 at 18:44

Some simple examples to clarify the distinction between membership and inclusion.

Suppose that Peter belongs to a football club, let's call this club the "Lions". Let's admit that a club can be defined as a set of persons.

Peter belongs / is a member of the set "Lions".

In that club, there are 3 teams, Team A, Team B, Team C.

Neither Team A, nor Team B nor Team C is a member of the set Lions: all the members of the club are persons, but a team is not a person, it is a set of persons.

True, Team A is a member of some set, namely, the set of the Lions's Teams. Team A belongs to the set T = { Team A, Team B, Team C}. Although there are 11 players in each team, the set T has only 3 elements.

The correct relation between each Team, say Team A, and the set Lions is the inclusion relation. Team A is not a member of the set Lions, it is a subset of the set Lions, it is included in that set.

True, all the elements of team A are alson elements of the set Lions ( this is the definition of " inclusion") but the set Team A is not itself an element of the set Lions.

Suppose the club Lions belongs to a football league. The football league is a set of cluds. The League is a set whose elements are clubs, not of persons, nor of teams.

Peter belongs to the set Lions, the set Lions belongs to the set League, but Peter does not belong to the set League: the set "League" is a set of clubs, not a set of persons.

Thus, the membership relation is not transitive!

One cannot say : a belongs to the set X, the set X is an element of Z, therefore necessarily a belongs to Z.

The inclusion relation is transitive.

Suppose the set { Peter; John, David} is a subset of Team A. Since Team A is, in turn, a subset of the set " Lions", it follows that the set { Peter, John, David} is ( by transitivity) a subset of the set Lions.

Of which set is the set Lions a subset? Not of the set League. That would mean ( according the definition of inclusion) that all the elements of Lions are also elements of League; in other words , that all the members of Lions are football clubs ( since League is a set of Clubs)! No ... The set Lions is a subset of the union of all the clubs ( that is the set of all persons that belong to one football club or another). Let's call this set P ( for players, all football players in the country, at least all those that belong to a football club). All the members of the set Lions are also members of the set P ( they are all players, football players): so the set Lions is included in P.

Remak.- Saying that the membership relation is not transitive does not mean that it is intransitive. There are ( so to say) special cases in which an element of an element of S is also an element of S.

Example. The number 0 = { } is an element of the number 1 = {0} . The number 1 is an element of the number 2={0,1}. And the number 0 is also an element of the number 2.


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