Mixing
The partial order induced by a preorder, and the functions respecting this induced p.o.
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Let $X$ be a non-empty set and let $preceq$ be a preorder on $X$.
Is there a standard name for the induced equivalence relation '$equiv$' on $X$ that satisfies: for every $x, y in X$, $x equiv y$ iff $x preceq y$ and $x succeq y$? Is there a standard name for the collection of equivalence classes associated with '$equiv$'?
Let $Y$ be a non-empty set. Is there a standard name for those functions $f:Xrightarrow Y$ that respect the induced equivalence relation described above, i.e. those functions satisfying: whenever $x, y in X$ are such that $x preceq y$ and $x succeq y$, $f(x) = f(y)$?
terminology order-theory
asked Jul 17 at 16:35
Evan Aad
5,36511748
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up vote
0
down vote
favorite
Let $X$ be a non-empty set and let $preceq$ be a preorder on $X$.
Is there a standard name for the induced equivalence relation '$equiv$' on $X$ that satisfies: for every $x, y in X$, $x equiv y$ iff $x preceq y$ and $x succeq y$? Is there a standard name for the collection of equivalence classes associated with '$equiv$'?
Let $Y$ be a non-empty set. Is there a standard name for those functions $f:Xrightarrow Y$ that respect the induced equivalence relation described above, i.e. those functions satisfying: whenever $x, y in X$ are such that $x preceq y$ and $x succeq y$, $f(x) = f(y)$?
terminology order-theory
asked Jul 17 at 16:35
Evan Aad
5,36511748
I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a non-empty set and let $preceq$ be a preorder on $X$.
Is there a standard name for the induced equivalence relation '$equiv$' on $X$ that satisfies: for every $x, y in X$, $x equiv y$ iff $x preceq y$ and $x succeq y$? Is there a standard name for the collection of equivalence classes associated with '$equiv$'?
Let $Y$ be a non-empty set. Is there a standard name for those functions $f:Xrightarrow Y$ that respect the induced equivalence relation described above, i.e. those functions satisfying: whenever $x, y in X$ are such that $x preceq y$ and $x succeq y$, $f(x) = f(y)$?
terminology order-theory
asked Jul 17 at 16:35
Evan Aad
5,36511748
Let $X$ be a non-empty set and let $preceq$ be a preorder on $X$.
Is there a standard name for the induced equivalence relation '$equiv$' on $X$ that satisfies: for every $x, y in X$, $x equiv y$ iff $x preceq y$ and $x succeq y$? Is there a standard name for the collection of equivalence classes associated with '$equiv$'?
Let $Y$ be a non-empty set. Is there a standard name for those functions $f:Xrightarrow Y$ that respect the induced equivalence relation described above, i.e. those functions satisfying: whenever $x, y in X$ are such that $x preceq y$ and $x succeq y$, $f(x) = f(y)$?
terminology order-theory
asked Jul 17 at 16:35
Evan Aad
5,36511748
asked Jul 17 at 16:35
Evan Aad
5,36511748
asked Jul 17 at 16:35
Evan Aad
5,36511748
asked Jul 17 at 16:35
Evan Aad
5,36511748
5,36511748
I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59
add a comment |Â
I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59
I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59
add a comment |Â
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I think what you mean in 2. is the functions whose kernel is the equivalence relation you defined in 1. (Of course this doesn't answer anything.)
– amrsa
Jul 17 at 16:41
@amrsa: I'm not sure kernel is an adequate terminology here: firstly because $f$ may not be a homomorphism (no structure is specified on $Y$), and secondly because it can happen for some function $f$ and for some $x, y in X$ that $f(x) = f(y)$, but either $x notpreceq y$ or $x notsucceq y$ (or both).
– Evan Aad
Jul 17 at 16:47
I see. I didn't anticipate your second objection. About the first, if $Y$ has no structure, it's the same as being an anti-chain, and then it would only be a homomorphism if $x preceq y$ implied $f(x)=f(y)$, so again, your objection makes sense.
– amrsa
Jul 17 at 18:59