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$A$ is a nonempty set. The empty relation $\emptyset$ is a relation on $A$. Is the relation $R=\emptyset$ invertible in $(R(A),\circ)$?

I think no because there is no $S\in R(A)$ such that $\emptyset\circ S = I$, the identity relation.

If $A$ is an empty set??

Thank you.

plop
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Annai
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    Please, use MathJax to type the math in your questions. – jjagmath Aug 17 '22 at 00:17
  • I guess a more threshold problem is how you define the identity relation on the empty set. – hardmath Aug 17 '22 at 00:40
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    Is it true that for each $(x,y)\in R$, $(y,x)\in R$? – John Douma Aug 17 '22 at 00:42
  • That's right. $\emptyset\circ S=\emptyset\neq I$, when $A\neq\emptyset$. On the other hand, for $A=\emptyset$, then $I=\emptyset$. In this case $\emptyset=\emptyset^T=\emptyset^{-1}$. – plop Aug 17 '22 at 00:43
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    It depends on what you mean by “invertible,” which depends (hopefully) on some motivating problem context. Can you add any such context to your question? – user3716267 Aug 17 '22 at 02:13
  • @user3716267 Inverse relation – plop Aug 17 '22 at 02:19
  • @user85667 That’s probably not what’s meant by “invertible” here. Every relation has an inverse, but not every relation is invertible. – BrianO Aug 17 '22 at 06:22
  • @BrianO Every relation has a transpose, not an inverse. – plop Aug 17 '22 at 11:40
  • @user85667 Call it what you will. “Inverse” as the name for the reverse relation is quite old & well-attested but, agreed, it’s misleading in that it isn’t an inverse in the algebraic sense. If $R$ is a function then nobody calls $R^{-1}$ its transpose. – BrianO Aug 18 '22 at 04:24
  • @BrianO You: "That's probably not what's meant by invertible here". Second line of the question: "I think no because there is no S∈R(A) such that ∅∘S=I, the identity relation." Just don't waste my time. – plop Aug 18 '22 at 13:15
  • @user85667 It’ll be a pleasure to never interact with you again. Good luck with that winning personality. – BrianO Aug 18 '22 at 19:57
  • @BrianO LOL! Yet, you @ me again. – plop Aug 18 '22 at 20:01

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