Properties of "membership relation" in naive set theory

Question

I'm reading Devlin, Fundamentals of contemporary set theory, I am at page 13:

Definition Let $R$ denote any binary relation on a set $x$. We say:

$R$ is reflexive if $(\forall a\in x) (aRa)$
$R$ is symmetric if $(\forall a,b\in x) (aRb\rightarrow bRa)$
$R$ is antisymmetric if $(\forall a,b\in x) [(aRb \wedge a\neq b)\rightarrow (\neg bRa)]$
$R$ is connected if $(\forall a,b\in x)[(a\neq b)\rightarrow (aRb \vee bRa)]$
$R$ is transitive if $(\forall a,b,c\in x)[(aRb \wedge bRc)\rightarrow (aRc)]$.

Excercise: Which of the above properties does the membership relation, $\in$, on a set satisfy?

My answer is NONE. But I'm very confused by this question, since $\in$ relates objects of different nature, i.e. elements and sets of elements, so I actually don't know how to deal with this particular relation. Can you help me clarifying what is going on? Consider that this is the first chapter of the book, hence set theory is not developed in an axiomatic way at this stage, and I think the exercise should be done with no use of axiomatizations, only using intuition or evidence.

counterexamples:

reflexivity: an element is not a member of itself
symmetry: take $a$ an element, $b=\{a,c\}$
antisymmetry: take $a=\{b,c\}$ and $b=\{a,d\}$, with $c\neq d$
connection: $a$ and $b$ different elements
transitivity: take $a$ an element,$b=\{a,d\}$ and $c=\{b,e\}$

The elements of sets can be sets themselves. And in set theory, one usually doesn't consider anything except sets (and classes), no Urelemente (atoms) [although there are set theories that use them, usually one doesn't]. But anyway, for example all elements of a power set are sets, whether one uses Urelemente or not. However, the $\in$ relation has one of the listed properties. — Daniel Fischer, Dec 09 '13 at 14:53
@DanielFischer Maybe antisimmetry? But this leads to another question: is every set an element of itself? — Danae Kissinger, Dec 09 '13 at 15:05
How does the axiom of regularity forbid self containing sets? or Example of set which contains itself — Martin Sleziak, Dec 09 '13 at 15:10
To me it seems rather surprising that such question appears in the book before introducing axioms of ZFC. Axiom of regularity seems to be relevant for this question. But maybe the author does not want you to find out which of the properties are true for arbitrary set $x$, maybe he wants you just to try some special cases; for example $x=\emptyset$, $x={\emptyset}$, $x={{\emptyset}}$, $x={\emptyset,{\emptyset}}$ and similar simple sets. (Obtained from the empty set.) — Martin Sleziak, Dec 09 '13 at 15:14
I think that your counterexample for antisymmetry is not correct. Also $a\ne b$ does not imply automatically $a\notin b$, so you have to choose appropriate sets $a$, $b$ to get a counterexample. — Martin Sleziak, Dec 09 '13 at 16:20

Martin Sleziak · Accepted Answer · 2015-02-19T13:53:33.740

I have interpreted the exercise in the following way: We are working in ZFC. In particular, this means that any element of a set is also a set. I have looked on the relation $\in$ for some simple sets which can be obtained from the empty set $\emptyset$.

When I look at the table of contents of Devlin's Joy of Sets (Fundamentals of Contemporary Set Theory), I saw that the author deals with axiomatic set theory only in the second chapter of this book. So it is possible that I have somehow misinterpreted what he wanted in this exercise.

Let us start by two trivial observations. If $x=\emptyset$, then all properties are vacuously true. If the set $x$ has one element, then $R$ is connected and antisymmetric, since we never have $a\ne b$.

The relation $\in$ on any set $x$ must be antisymmetric, since $a\in b\in a$ contradicts Axiom of regularity.

If $x=\{\emptyset,\{\emptyset\}\}$ then the relation $\in$ is not symmetric, just take $a=\emptyset$ and $b=\{\emptyset\}$. Notice that relation $\in$ on this set is connected and it is also transitive.

Let $x=\{\emptyset,\{\{\emptyset\}\}\}$. If we denote $a=\emptyset$ and $b=\{\{\emptyset\}\}$, then neither $a\in b$, nor $b\in a$. So the relation $\in$ on this set is not connected. However, as it is the empty relation, it is transitive.

Let $x=\{a,b,c\}$ where $a=\emptyset$, $b=\{\emptyset\}$ and $c=\{\{\emptyset\}\}$. Then $a\in b$ and $b\in c$, but $a\notin c$, so the relation $\in$ on this set is not transitive. It is also not connected, since neither $a\in c$ nor $c\in a$.

Stu · Answer 2 · 2015-02-19T13:49:45.637

I have looked at the exercise in Keith Devlin's book. Although I do not have a copy to hand I think it was exactly as given above. In this case the author is referring to the idea of a membership relation in general, rather than any special relations subsisting between elements which may (or may not) qualify them as members of a particular set.

We may ask whether a given element, x, is "related" to an element, y, where y belongs to the complement of some set S. In this case, we may define x as an element of S iff x is not related to y; whereas x is (defined as) an element of the complement otherwise. In this way, set-membership is established as an equivalence relation between its elements. (ie the properties of reflexivity, symmetry and transitivity define the membership relation).

nb. I have presented the membership relation in this way in order to avoid the apparent circularity involved if we try to assert that set-membership is self-evidently an equivalence relation. There may be better arguments out there, but I don't know of one.

Properties of "membership relation" in naive set theory

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