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$$R=\{(A,B)\mid A \subseteq B\}$$

$R$ is assumed to be a relation on a collection of sets

Since $A$ is a subset of itself, the relation is reflexive.

And if $A$ is a subset of $B$ which is, in turn, a subset of $C$, then $A$ must be a subset of $C$ making the relation transitive.

However, $A$ might not equal $B$ and in that case, the relation will not be symmetric nor anti-symmetric (since there might be some case where $A = B$ and $B$ is, therefore, a subset of $A$). I'm a little confused about this last bit. Is my thought process right or will the relation be symmetric too?

Edit: $R$ is assumed to be a relation on a collection of sets.

Bernard
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Musk
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    You're correct it's not symmetric. – CyclotomicField Oct 08 '20 at 18:01
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    This is underspecified. What is this relation over? It must be over something like $\mathcal{P}(X)$ (that is, $\mathcal{P}(X)$ is both the domain and codomain of $R$) with $X$ some arbitrary set or something like that. You needed to include that in your question definition. – JMoravitz Oct 08 '20 at 18:02
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    In the case that $X$ is nonempty and we are indeed over $\mathcal{P}(X)$, then yes, $R$ will be reflexive, transitive and anti-symmetric. In the case that we are not over $\mathcal{P}(X)$ but are instead over some subset of $\mathcal{P}(X)$, then it depends. If it was something like $\binom{X}{3}$ (the set of all three-element subsets of $X$) then it is reflexive, transitive, symmetric, and anti-symmetric. It is the trivial identity relation where something is only related to itself. – JMoravitz Oct 08 '20 at 18:06
  • @JMoravitz Thanks! I've updated the post. Could you explain how it would be both symmetric and anti-symmetric instead of simply neither?

    PS this is a problem from class so I just have the details the prof gave us.

     – Musk Oct 08 '20 at 18:13
  • https://math.stackexchange.com/questions/1475354/can-a-relation-be-both-symmetric-and-antisymmetric-or-neither/1475381#1475381 – JMoravitz Oct 08 '20 at 18:15
  • Perhaps you have misheard or misunderstood the definitions of symmetry and antisymmetry. You might have thought that anti-symmetry means that if $a\sim b$ then $b\nsim a$. That is not the case. Antisymmetry means that if $a\sim b$ and $b\sim a$ then $a=b$. Having an element related to itself is fine. What is not fine for antisymmetry is an element related to a different element while that different element is also related to the first. – JMoravitz Oct 08 '20 at 18:20
  • @JMoravitz so am I right in assuming that this relation is neither symmetric nor anti-symmetric seeing as it would have single and double-sided arrows (aka relations) if taken on an arbitrary collection of sets, where the conditions for reflexivity and transitivity are satisfied but there would be cases where neither clauses for symmetry nor anti-symmetry are satisfied? – Musk Oct 08 '20 at 18:26
  • No. All arrows if any exist would be single-sided for this problem. It could have many loops but those don't matter. Recall that $A\subseteq B$ and $B\subseteq A$ would imply that $A=B$. Your relation will always be antisymmetric. It is possible that it is also symmetric but only in the instance that there is never any $A\neq B$ such that $A\subseteq B$ in your collection of sets. – JMoravitz Oct 08 '20 at 18:27
  • @JMoravitz Right, I just checked out the definition again from https://en.wikipedia.org/wiki/Antisymmetric_relation and https://en.wikipedia.org/wiki/Symmetric_relation. And this is what I've understood. aRb such that a is a sibling of b is symmetric but not anti-symmetric. And cRd where c is equal to d is both symmetric and anti? – Musk Oct 08 '20 at 18:31

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As you noted, it is Reflexive because: $\forall A~ (A\subseteq A)$ by definition of subset.

Likewise it is Transitive because: $\forall A~\forall B~\forall C~(A\subseteq B\wedge B\subseteq C\to A\subseteq C)$ by definition of subset.

However, it is actually Antisymmetric because $\forall A~\forall B~((A\subseteq B\wedge B\subseteq A)\to A=B)$ by definition of subset.

Next, it is only not symmetric when: $\exists A~\exists B~(A\subseteq B\wedge B\not\subseteq A)$ . Therefore, this is dependent on what collection of sets the relationship is held over, as it will only be not-symmetric when at least one set is a proper subset of another.

Graham Kemp
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