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$rho$ antisym., irreflexive, $rho ^t$ its transitive closure. Prove or disprove: $rho ^t$ reflexive...
Let $rho$ be an antisymmetric, irreflexive relation, and $rho ^t$ its transitive closure.
Prove or disprove: $rho ^t$ is reflexive $Rightarrow$ $rho ^t$ is symmetric.
I tried to prove that $rho ^t$ is symmetric by proving that if $(x,y)in rho ^t$, than $(x,y)in rho ^t$, but whatever I do, I just can't prove that this is true.
Also, I tried to find an example for where the statement isn't true, but
whenever I try to make a relation that is antisymmetric and irreflexive and I find its transitive closure which is reflexive, I get that $rho ^t$ is symmetric which makes me think that the statement is correct.
Thanks in advance!
relations
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
add a comment |
Let $rho$ be an antisymmetric, irreflexive relation, and $rho ^t$ its transitive closure.
Prove or disprove: $rho ^t$ is reflexive $Rightarrow$ $rho ^t$ is symmetric.
I tried to prove that $rho ^t$ is symmetric by proving that if $(x,y)in rho ^t$, than $(x,y)in rho ^t$, but whatever I do, I just can't prove that this is true.
Also, I tried to find an example for where the statement isn't true, but
whenever I try to make a relation that is antisymmetric and irreflexive and I find its transitive closure which is reflexive, I get that $rho ^t$ is symmetric which makes me think that the statement is correct.
Thanks in advance!
relations
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
add a comment |
Let $rho$ be an antisymmetric, irreflexive relation, and $rho ^t$ its transitive closure.
Prove or disprove: $rho ^t$ is reflexive $Rightarrow$ $rho ^t$ is symmetric.
I tried to prove that $rho ^t$ is symmetric by proving that if $(x,y)in rho ^t$, than $(x,y)in rho ^t$, but whatever I do, I just can't prove that this is true.
Also, I tried to find an example for where the statement isn't true, but
whenever I try to make a relation that is antisymmetric and irreflexive and I find its transitive closure which is reflexive, I get that $rho ^t$ is symmetric which makes me think that the statement is correct.
Thanks in advance!
relations
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
Let $rho$ be an antisymmetric, irreflexive relation, and $rho ^t$ its transitive closure.
Prove or disprove: $rho ^t$ is reflexive $Rightarrow$ $rho ^t$ is symmetric.
I tried to prove that $rho ^t$ is symmetric by proving that if $(x,y)in rho ^t$, than $(x,y)in rho ^t$, but whatever I do, I just can't prove that this is true.
Also, I tried to find an example for where the statement isn't true, but
whenever I try to make a relation that is antisymmetric and irreflexive and I find its transitive closure which is reflexive, I get that $rho ^t$ is symmetric which makes me think that the statement is correct.
Thanks in advance!
relations
relations
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
edited Nov 22 '18 at 10:55
Asaf Karagila♦
302k32427757
302k32427757
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
asked Nov 21 '18 at 22:59
mathbbandstuffmathbbandstuff
278111
278111
add a comment |
add a comment |
1 Answer
1
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oldest
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The statement is false. Here is a counterexample : take two disjoint length-three cycles $x_1 rho x_2 rho x_3 rho x_1$ and $y_1 rho y_2 rho y_3 rho y_1$, and add just one more edge $x_1 rho y_1$.
The resulting $rho^t$ is reflexive (because every element is in some cycle) but not symmetric (look at what happens on $x_1$ and $y_1$).
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
The statement is false. Here is a counterexample : take two disjoint length-three cycles $x_1 rho x_2 rho x_3 rho x_1$ and $y_1 rho y_2 rho y_3 rho y_1$, and add just one more edge $x_1 rho y_1$.
The resulting $rho^t$ is reflexive (because every element is in some cycle) but not symmetric (look at what happens on $x_1$ and $y_1$).
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
add a comment |
The statement is false. Here is a counterexample : take two disjoint length-three cycles $x_1 rho x_2 rho x_3 rho x_1$ and $y_1 rho y_2 rho y_3 rho y_1$, and add just one more edge $x_1 rho y_1$.
The resulting $rho^t$ is reflexive (because every element is in some cycle) but not symmetric (look at what happens on $x_1$ and $y_1$).
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
add a comment |
The statement is false. Here is a counterexample : take two disjoint length-three cycles $x_1 rho x_2 rho x_3 rho x_1$ and $y_1 rho y_2 rho y_3 rho y_1$, and add just one more edge $x_1 rho y_1$.
The resulting $rho^t$ is reflexive (because every element is in some cycle) but not symmetric (look at what happens on $x_1$ and $y_1$).
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
The statement is false. Here is a counterexample : take two disjoint length-three cycles $x_1 rho x_2 rho x_3 rho x_1$ and $y_1 rho y_2 rho y_3 rho y_1$, and add just one more edge $x_1 rho y_1$.
The resulting $rho^t$ is reflexive (because every element is in some cycle) but not symmetric (look at what happens on $x_1$ and $y_1$).
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
answered Nov 22 '18 at 11:06
Ewan DelanoyEwan Delanoy
41.4k442104
41.4k442104
add a comment |
add a comment |
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