Mixing
What is the definition of a relation?
$begingroup$
Let $S$ be a set. An equivalence relation on $S$ is a subset $Rsubseteq Stimes S$ satisfying that $(s,s)in R$ for all $sin S$, $(s,t)in R implies (t,s)in R$, $(s,t)in R; land; (t,u)in Rimplies (s,u)in R$.
My question is: What is the definition of a relation?
Is a relation just any subset of $Stimes S$
discrete-mathematics elementary-set-theory definition relations equivalence-relations
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
$endgroup$
add a comment |
$begingroup$
Let $S$ be a set. An equivalence relation on $S$ is a subset $Rsubseteq Stimes S$ satisfying that $(s,s)in R$ for all $sin S$, $(s,t)in R implies (t,s)in R$, $(s,t)in R; land; (t,u)in Rimplies (s,u)in R$.
My question is: What is the definition of a relation?
Is a relation just any subset of $Stimes S$
discrete-mathematics elementary-set-theory definition relations equivalence-relations
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
$endgroup$
5
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39
add a comment |
$begingroup$
Let $S$ be a set. An equivalence relation on $S$ is a subset $Rsubseteq Stimes S$ satisfying that $(s,s)in R$ for all $sin S$, $(s,t)in R implies (t,s)in R$, $(s,t)in R; land; (t,u)in Rimplies (s,u)in R$.
My question is: What is the definition of a relation?
Is a relation just any subset of $Stimes S$
discrete-mathematics elementary-set-theory definition relations equivalence-relations
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
$endgroup$
Let $S$ be a set. An equivalence relation on $S$ is a subset $Rsubseteq Stimes S$ satisfying that $(s,s)in R$ for all $sin S$, $(s,t)in R implies (t,s)in R$, $(s,t)in R; land; (t,u)in Rimplies (s,u)in R$.
My question is: What is the definition of a relation?
Is a relation just any subset of $Stimes S$
discrete-mathematics elementary-set-theory definition relations equivalence-relations
discrete-mathematics elementary-set-theory definition relations equivalence-relations
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
asked Jan 25 at 15:33
John DoeJohn Doe
30921346
30921346
5
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39
add a comment |
5
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39
5
5
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, a relation on $S$ is just a subset of $Stimes S$, and any subset of $Stimes S$ is a relation on $S$.
It comes with a notational quirk, though. If $R$ is such a relation, we may write $sRt$ for $(s,t)in R$. You commonly see this in expressions like $2leq3$ (no mathematician in their right mind would seriously prefer to write $(2,3)in {}leq$ unless they, like I am here, are making some very specific point).
answered Jan 25 at 15:39
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
More generally a relation can defined as a binary predicate, that's a statement with two free variables $mathrm R_{x,y} $.
For example, $xin y $ is a relation respect to the free variables $x $ and $y $.
Now, given a relation $mathrm R_{x,y} $, its graph is the class
$$R={(x,y):mathrm R_{x,y}} $$
Moreover, given a class (or a set) $X $, the graph of $mathrm R_{x,y} $ restricted to $S$ is
$$R={(x,y)in Stimes S:mathrm R_{x,y}} $$
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Yes, a relation on $S$ is just a subset of $Stimes S$, and any subset of $Stimes S$ is a relation on $S$.
It comes with a notational quirk, though. If $R$ is such a relation, we may write $sRt$ for $(s,t)in R$. You commonly see this in expressions like $2leq3$ (no mathematician in their right mind would seriously prefer to write $(2,3)in {}leq$ unless they, like I am here, are making some very specific point).
answered Jan 25 at 15:39
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
Yes, a relation on $S$ is just a subset of $Stimes S$, and any subset of $Stimes S$ is a relation on $S$.
It comes with a notational quirk, though. If $R$ is such a relation, we may write $sRt$ for $(s,t)in R$. You commonly see this in expressions like $2leq3$ (no mathematician in their right mind would seriously prefer to write $(2,3)in {}leq$ unless they, like I am here, are making some very specific point).
answered Jan 25 at 15:39
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
Yes, a relation on $S$ is just a subset of $Stimes S$, and any subset of $Stimes S$ is a relation on $S$.
It comes with a notational quirk, though. If $R$ is such a relation, we may write $sRt$ for $(s,t)in R$. You commonly see this in expressions like $2leq3$ (no mathematician in their right mind would seriously prefer to write $(2,3)in {}leq$ unless they, like I am here, are making some very specific point).
answered Jan 25 at 15:39
ArthurArthur
119k7118202
$endgroup$
Yes, a relation on $S$ is just a subset of $Stimes S$, and any subset of $Stimes S$ is a relation on $S$.
It comes with a notational quirk, though. If $R$ is such a relation, we may write $sRt$ for $(s,t)in R$. You commonly see this in expressions like $2leq3$ (no mathematician in their right mind would seriously prefer to write $(2,3)in {}leq$ unless they, like I am here, are making some very specific point).
answered Jan 25 at 15:39
ArthurArthur
119k7118202
edited Jan 25 at 15:53
answered Jan 25 at 15:39
ArthurArthur
119k7118202
answered Jan 25 at 15:39
ArthurArthur
119k7118202
answered Jan 25 at 15:39
ArthurArthur
119k7118202
119k7118202
add a comment |
add a comment |
$begingroup$
More generally a relation can defined as a binary predicate, that's a statement with two free variables $mathrm R_{x,y} $.
For example, $xin y $ is a relation respect to the free variables $x $ and $y $.
Now, given a relation $mathrm R_{x,y} $, its graph is the class
$$R={(x,y):mathrm R_{x,y}} $$
Moreover, given a class (or a set) $X $, the graph of $mathrm R_{x,y} $ restricted to $S$ is
$$R={(x,y)in Stimes S:mathrm R_{x,y}} $$
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
add a comment |
$begingroup$
More generally a relation can defined as a binary predicate, that's a statement with two free variables $mathrm R_{x,y} $.
For example, $xin y $ is a relation respect to the free variables $x $ and $y $.
Now, given a relation $mathrm R_{x,y} $, its graph is the class
$$R={(x,y):mathrm R_{x,y}} $$
Moreover, given a class (or a set) $X $, the graph of $mathrm R_{x,y} $ restricted to $S$ is
$$R={(x,y)in Stimes S:mathrm R_{x,y}} $$
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
add a comment |
$begingroup$
More generally a relation can defined as a binary predicate, that's a statement with two free variables $mathrm R_{x,y} $.
For example, $xin y $ is a relation respect to the free variables $x $ and $y $.
Now, given a relation $mathrm R_{x,y} $, its graph is the class
$$R={(x,y):mathrm R_{x,y}} $$
Moreover, given a class (or a set) $X $, the graph of $mathrm R_{x,y} $ restricted to $S$ is
$$R={(x,y)in Stimes S:mathrm R_{x,y}} $$
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
$endgroup$
More generally a relation can defined as a binary predicate, that's a statement with two free variables $mathrm R_{x,y} $.
For example, $xin y $ is a relation respect to the free variables $x $ and $y $.
Now, given a relation $mathrm R_{x,y} $, its graph is the class
$$R={(x,y):mathrm R_{x,y}} $$
Moreover, given a class (or a set) $X $, the graph of $mathrm R_{x,y} $ restricted to $S$ is
$$R={(x,y)in Stimes S:mathrm R_{x,y}} $$
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
edited Jan 25 at 16:38
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
answered Jan 25 at 16:29
Fabio LucchiniFabio Lucchini
9,34111426
9,34111426
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
add a comment |
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
$begingroup$
$in$ is a bit special, technically. At least if you build your math on standard (ZF), it's not a relation.
$endgroup$
– Arthur
Jan 25 at 16:43
add a comment |
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5
$begingroup$
Yes, that's the definition.
$endgroup$
– José Carlos Santos
Jan 25 at 15:38
$begingroup$
@JoséCarlosSantos: Ok, please write this as an answer so that I can accept. (Nevermind, I just got an answer)
$endgroup$
– John Doe
Jan 25 at 15:39