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How is this relation reflexive?
Need help counting equivalence classes.Is this relation reflexive, symmetric and transitive?Proving an equivalence relation(specifically transitivity)Equivalence relation example. How is this even reflexive?Where is the transistivity in this equivalence relationIdentity relation vs Reflexive RelationHow is this an equivalence relation?truefalse claims in relations and equivalence relationsHow is this case a reflexive relation?Is this relation reflexive if it “chains” to itself?
$begingroup$
Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.
Prove that $mathcalR$ is an equivalence relation on $mathcalX$.
From my understanding, the definition of reflexive is:
$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
However, for this problem, you can have the relation with these two sets:
$1$ and $1,2$
Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.
discrete-mathematics relations equivalence-relations
asked 8 hours ago
qbufferqbuffer
625
$endgroup$
add a comment |
$begingroup$
Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.
Prove that $mathcalR$ is an equivalence relation on $mathcalX$.
From my understanding, the definition of reflexive is:
$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
However, for this problem, you can have the relation with these two sets:
$1$ and $1,2$
Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.
discrete-mathematics relations equivalence-relations
asked 8 hours ago
qbufferqbuffer
625
$endgroup$
4
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
5
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago
add a comment |
$begingroup$
Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.
Prove that $mathcalR$ is an equivalence relation on $mathcalX$.
From my understanding, the definition of reflexive is:
$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
However, for this problem, you can have the relation with these two sets:
$1$ and $1,2$
Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.
discrete-mathematics relations equivalence-relations
asked 8 hours ago
qbufferqbuffer
625
$endgroup$
Let $mathcalX$ be the set of all nonempty subsets of the set $1,2,3,...,10$. Define the relation $mathcalR$ on $mathcalX$ by: $forall A, B in mathcalX, A mathcalR B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $1,2,3 mathcalR 1,3,5,8$ because the smallest element of $1,2,3$ is $1$ which is also the smallest element of $1,3,5,8$.
Prove that $mathcalR$ is an equivalence relation on $mathcalX$.
From my understanding, the definition of reflexive is:
$$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
However, for this problem, you can have the relation with these two sets:
$1$ and $1,2$
Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.
discrete-mathematics relations equivalence-relations
discrete-mathematics relations equivalence-relations
asked 8 hours ago
qbufferqbuffer
625
asked 8 hours ago
qbufferqbuffer
625
asked 8 hours ago
qbufferqbuffer
625
asked 8 hours ago
qbufferqbuffer
625
asked 8 hours ago
qbufferqbuffer
625
625
4
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
5
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago
add a comment |
4
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
5
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago
4
4
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
5
5
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.
To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.
To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
$endgroup$
add a comment |
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.
To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
$endgroup$
add a comment |
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.
To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
$endgroup$
Why are you testing reflexivity by looking at two different elements of $mathcalX$? The definition of reflexivity says that a relation is reflexive iff each element of $mathcal X$ is in relation with itself.
To check whether $mathcal R$ is reflexive, just take one element of $mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x in mathcal X$. Of course, this is true because $min(x) = min(x)$ is always true, which is intuitive. In other words, $x mathcalR x$ for all $x in mathcal X$, which is exactly what you needed to prove that $mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $xneq y$) can be in the relation $mathcal R$. The fact that $1mathcal R 1,2$ does not contradict the fact that $1,2mathcal R 1,2$ as well.
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
edited 6 hours ago
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
answered 8 hours ago
Haris GusicHaris Gusic
3,331525
3,331525
add a comment |
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
$endgroup$
A binary relation $R$ over a set $mathcalX$ is reflexive if every element of $mathcalX$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcalR text is reflexive iff forall x in mathcalX, x mathcalR x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
answered 7 hours ago
s0ulr3aper07s0ulr3aper07
658112
658112
add a comment |
add a comment |
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4
$begingroup$
Reflexive means that every element is related to itself. Thus, for reflexivity you have to consider one set only. Ok, we have that $ 1 mathcal R 1,2 $ but we have also $ 1 mathcal R 1 $ and $ 1,2 mathcal R 1,2 $
$endgroup$
– Mauro ALLEGRANZA
8 hours ago
5
$begingroup$
Note: “reflexive” does not mean that if $x$ is related to $y$, then $x=y$. It means that if $x=y$, then $x$ is related to $y$.
$endgroup$
– Arturo Magidin
7 hours ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
7 hours ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
6 hours ago