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Having an immense amount of trouble trying to figure this problem out, and the more I think and ask about it the more confused I seem to get. I think I have finally figured it out so can someone who truly knows the correct answer justify this?

Problem:Let $A=\{a,b,c\}$,Let $\mathcal{P}(A)=\{S:S\subseteq A\}$

Q1.Is the relation "is a subset of" a relation on $A$?

Q2.Is the relation "is a subset of" a relation on $\mathcal{P}(A)$?

Q1.Attempt:

I would say the relation "is a subset of" is NOT a relation on $A$

Since $A \times A=\{(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b)(c,c)\}$

Knowing $R \subseteq A \times A$

It is clear to me $a,b,c$ are elements, not sets, so an element in the relation on $A$ takes the form of $(a,a)$ for $a \in A$.So a "subset relation" cannot be defined on $A$.

Since $a \subseteq a$ is false for all $a \in A$ because $a$ is an element not a set

Q2.However a subset relation can be defined on $\mathcal{P}(A)$

Since elements of a relation on $\mathcal{P}(A)$ are actually ordered pairs of sets, now it is possible to define a "subset relation" on $\mathcal{P}(A)$.

For example $(\{a\},\{a,b,c\})\in \mathcal{P}(A) \times \mathcal{P}(A)$

and since for a relation on $\mathcal{P}(A)$, $R \subseteq \mathcal{P}(A) \times \mathcal{P}(A)$ and $\{a\} \subseteq \{a,b,c\}$

Now it is possible to define the relation "is a subset" of on $\mathcal{P}(A)$

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  • $\begingroup$ If you want to be pedantic... it is common to view everything in mathematics as a set, even elements of sets. The natural number zero for example could be viewed as $\emptyset$ while the natural number $2$ is viewed as $\{\{\emptyset\},\emptyset\}$ etc... Now... which answer your teacher or book is looking for for Q1 will vary depending on how pedantic they are being, but I would argue in favor of saying that it is a relation on $A$, though some might disagree. Read about Zermelo-Fraenkel set theory. $\endgroup$ – JMoravitz Jul 22 '19 at 19:05
  • $\begingroup$ As for Q2, it absolutely is a relation on $\mathcal{P}(A)$ and no one would argue about that. $\endgroup$ – JMoravitz Jul 22 '19 at 19:07
  • $\begingroup$ @JMoravitz is my reasoning correct as to why it is a relation?Since an element of $\mathcal{P}(A) \times \mathcal{P}(A)$ is the cartesian product of two sets then a relation can be defined since there exists pairs of two subsets? $\endgroup$ – user686544 Jul 22 '19 at 19:10
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    $\begingroup$ Yes, though be prepared to defend that statement and even to get the question temporarily marked as being incorrect and needing to approach the teacher to get the grade adjusted. $\endgroup$ – JMoravitz Jul 22 '19 at 19:19
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    $\begingroup$ That being said, even if you were to argue against the elements themselves being interpreted as being sets, that doesn't prevent the relation from being a relation... it would just be an empty relation since under that interpretation nothing in $A$ would be a subset of anything else in $A$. All that is required for $\subset$ to be a relation on $A$ is for you to be able to either say "true" or "false" when asked "what is the truth value of the expression '$a\subset b$'?" If you happen to say false to everything, that doesn't stop it from being a relation. $\endgroup$ – JMoravitz Jul 22 '19 at 19:29
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To summarize my comments above, all that a set needs to be to be called a relation is to be a subset of a cartesian product of sets. Nothing more, nothing less. With regards to Q1, after some rewording the question asks if the set of ordered pairs $\{(x,y)\in A\times A~:~x\subseteq y\}$ is a relation.

Regardless of your point of view or how pedantic you want to be about whether the elements of $A$ are able to themselves be sets or not, as soon as you have a decided upon interpretation of how the elements of $A$ behave and are defined you should be able to agree that for each $x,y\in A$ the statement $x\subseteq y$ has a truth value as being clearly true or clearly false with no ambiguity. We can say that $5\not\subseteq 3$ for instance, whether that is because you treat $5$ and $3$ as sets and you see that $5$ is not a subset of $3$, or whether you choose to treat them as atomic elements who are not sets and you say that $5\not\subseteq 3$ for the reason that these are not sets.

@MatthewLeingang is correct when he poitns out that $(\{a\},\{a\})\not\in A\times A$, however he is incorrect in assuming that this has any relevance to the problem at hand whatsoever. We never cared about whether $(\{a\},\{a\})$ was an element of our relation or not (and it not being in our relation wouldn't have mattered anyways), we only cared about seeing which elements (if any) of $A\times A$ are included in our relation.

The only reason why the above would not have been a relation is in the event that the condition for an ordered pair to be in the set is ill-defined or undefined. The set $\{(x,y)\in A\times A~:~x\heartsuit y\}$ is such an example since we don't yet know the meaning of the symbol $\heartsuit$.

Otherwise, just because a relation is empty does not stop it from being a relation. Again, even if you disagree on whether or not $a,b,c$ should be considered pure "elements" and not sets or not, that does not stop the described relation from being well defined.


That all being said, depending on which style of set theory you are studying, the most common foundations to set theory that people study in the modern age is Zermelo-Fraenkel set theory. In an attempt to make things as clean and unambiguous as possible, we restrict our attention to pure sets and every mathematical object, number, set, function, operation, statement, etc... are all also considered sets. (There are things that ZF can't handle, such as proper classes. Read about other set theories in the wiki link above such as ZFG)

Under this interpretation, even if we didn't intend it, $a,b,c$ are all actually sets too! And as such, not only is the relation defined, just like before, it is even nonempty as it is guaranteed to contain at least the pairs $(a,a),(b,b)$ and $(c,c)$ and depending on the values of $a,b,c$ possibly more.

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  • $\begingroup$ Thank you for the explanation $\endgroup$ – user686544 Jul 23 '19 at 0:40
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With respect to JMoravitz's point of view, I think you understand the problem as intended by the instructor/problem author, and should not extend to the case that elements of $A$ might contain themselves as subsets.

I'm inferring the point of the problem is to understand the difference between something being an element of a set versus a subset. The notation $A = \{a,b,c\}$ implies (not logically, but idiomatically) that $A$ is a set containing three elements which are not sets. A representative interpretation might be three integers, or three points in the plane.

Assuming this, then you can say $\{a\} \subset \{a\}$ but $(\{a\},\{a\}) \notin A \times A$ (since $\{a\} \notin A$), so is-a-subset-of is not a relation on $A$.

I'm sure your instructor would be happy to clarify the problem or discuss these counterintuitive edge cases.

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