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Let A be A={1,2,3}, let K be the set of all symmetric and not reflexive relations of A. Is K $$ K=\{(\phi,\phi)\quad((\phi,\phi),(1,2),(2,1))\quad ((\phi,\phi),(1,3),(3,1))\quad ((\phi,\phi),(2,3),(3,2))\quad((\phi,\phi),(1,2),(2,1),(1,3),(3,1))\quad ((\phi,\phi),(1,2),(2,1),(2,3),(3,2))\quad ((\phi,\phi),(1,3),(3,1),(2,3),(3,2))\quad ((\phi,\phi),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2))\} $$

???

quasi
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Arthur Losnikov
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  • Is $K$ a group or just a set? And what is $\phi$? – 57Jimmy Feb 09 '18 at 15:39
  • @Arthur Losnikov: A relation on $A$ is a set (possibly empty) of ordered pairs of elements of $A$. But note: $\varnothing$ is not an element of $A$, so $(\varnothing,\varnothing)$ is not an ordered pair of elements of $A$, hence, it should not be in any of the relations. – quasi Feb 09 '18 at 15:42
  • ϕ is an empty set. K is a group of all possible symmetric and not reflexive relations of A. – Arthur Losnikov Feb 09 '18 at 15:43
  • The word "group" should not be used (since it has another meaning). Instead, just refer to $K$ as a set, or as a collection. – quasi Feb 09 '18 at 15:45
  • K is P(AxA) here. Is it a group or a set? – Arthur Losnikov Feb 09 '18 at 15:46
  • Isn't the empty set a subset of any set? – Arthur Losnikov Feb 09 '18 at 15:49
  • Sure, the empty set is a subset of any set, hence it's a subset of $A$. But the empty set is not an element of $A$, hence it shouldn't appear as a component of any ordered pair of elements of $A$. – quasi Feb 09 '18 at 16:13
  • @quasi so, the right set K would be as written without (∅,∅)? In all relations – Arthur Losnikov Feb 09 '18 at 16:21
  • Yes.${}{}{}{}{}$ – quasi Feb 09 '18 at 16:34
  • But your set $K$ is missing lots of relations. There are $49$ relations on $A$ which are symmetric but not transitive. – quasi Feb 09 '18 at 16:35
  • symmetric but not reflexive. but how come the least element of K is (ϕ,ϕ) if its not an element of K? – Arthur Losnikov Feb 09 '18 at 17:00
  • I misread the problem. I thought the requirement was "symmetric but not transitive". – quasi Feb 09 '18 at 17:31
  • The analysis for the requirement "symmetric but not reflexive" is easier. There $56$ such relations on $A$. – quasi Feb 09 '18 at 17:52
  • As to your question, "is it a set ot a group?", as I indicated, "set"or "collection" is more correct terminology. – quasi Feb 09 '18 at 17:59
  • As to your question "how come the least element of K is (ϕ,ϕ) if its not an element of K?", note that a relation on $A$ is a subset of $A\times A$. In other words, any relation $R$ on $A$ is a set of ordered pairs of elements of $A$. Note that the ordered pair $(\varnothing,\varnothing)$ is not a subset of $A\times A$. It's also not an element of $A \times A$. – quasi Feb 09 '18 at 18:01

2 Answers2

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If $R$ is a relation not reflexive, we cannot have $(1,1),(2,2),(3,3) \in R$ together but for example you may have $(1,1) \in R$ and $(2,2),(3,3) \notin R$. Or $(1,1),(2,2) \in R$ and $(3,3) \notin R$. So if you want to list them systematically, answer should also include the relations:

$$K=\big\{\{(1,1)\},\{(2,2)\},\{(3,3)\},\{(1,1),(2,2)\},\{(1,1),(3,3)\},\{(2,2),(3,3)\},\{(1,1),(1,2),(2,1)\},\{(1,1),(1,3),(3,1)\},\{(1,1),(2,3),(3,2)\},... \big\}$$

and none of the elements of $K$ should not include $(\emptyset,\emptyset)$ since $\emptyset$ is not an element but a representation of empty set.

ArsenBerk
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Let $K$ be the set of relations $R$ on $A$ such that $R$ is symmetric, but not reflexive.

Then $R \in K$ if and only if $R=S\cup T$, where $S$ is a proper subset of $$\{(1,1),(2,2),(3,3)\}$$ and $T$ is the union of zero or more of the three sets $$\{(1,2),(2,1)\},\{(2,3),(3,2)\},\{(3,1),(1,3)\}$$

It follows that there are $2^3-1=7$ choices for $S$, and $2^3=8$ choices for $T$, so there are $(7)(8)=56$ possibilities for $R$.

Thus, $K$ has $56$ elements.

quasi
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  • What is the least element of this K set? – Arthur Losnikov Feb 09 '18 at 18:12
  • The empty set is an element of $K$, so that qualifies as a kind of "least" element. But $K$ is not a set of numbers. It's a set of sets (since relations on $A$ are subsets of $A\times A$). – quasi Feb 09 '18 at 18:15
  • To form an element of $R$ of $K$, choose any proper subset $S$ of $${(1,1),(2,2),(3,3)}$$ and let $T$ be the union of zero or more of the three sets $${(1,2),(2,1)},{(2,3),(3,2)},{(3,1),(1,3)}$$and then let $R=S\cup T$. – quasi Feb 09 '18 at 18:22
  • Then why the empty set is an element of K if its not there? – Arthur Losnikov Feb 09 '18 at 18:22
  • The empty set is a subset of $A\times A$. As a relation, the empty set is symmetric, but not reflexive, so it qualifies as an element of $K$. – quasi Feb 09 '18 at 18:23
  • It seems you're often blurring the meanings of element, set, subset, ordered pair, etc. – quasi Feb 09 '18 at 18:27