How is the relation “the smallest element is the same” reflexive?
$begingroup$
Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.
Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.
From my understanding, the definition of reflexive is:
$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 elementary-set-theory relations equivalence-relations
edited yesterday
Martin Sleziak
45k10122277
asked 2 days ago
qbufferqbuffer
726
$endgroup$
add a comment |
$begingroup$
Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.
Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.
From my understanding, the definition of reflexive is:
$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 elementary-set-theory relations equivalence-relations
edited yesterday
Martin Sleziak
45k10122277
asked 2 days ago
qbufferqbuffer
726
$endgroup$
5
$begingroup$
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
2 days ago
7
$begingroup$
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
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago
add a comment |
$begingroup$
Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.
Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.
From my understanding, the definition of reflexive is:
$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 elementary-set-theory relations equivalence-relations
edited yesterday
Martin Sleziak
45k10122277
asked 2 days ago
qbufferqbuffer
726
$endgroup$
Let $mathcal{X}$ be the set of all nonempty subsets of the set ${1,2,3,...,10}$. Define the relation $mathcal{R}$ on $mathcal{X}$ by: $forall A, B in mathcal{X}, A mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, ${1,2,3} mathcal{R} {1,3,5,8}$ because the smallest element of ${1,2,3}$ is $1$ which is also the smallest element of ${1,3,5,8}$.
Prove that $mathcal{R}$ is an equivalence relation on $mathcal{X}$.
From my understanding, the definition of reflexive is:
$$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 elementary-set-theory relations equivalence-relations
discrete-mathematics elementary-set-theory relations equivalence-relations
edited yesterday
Martin Sleziak
45k10122277
asked 2 days ago
qbufferqbuffer
726
edited yesterday
Martin Sleziak
45k10122277
asked 2 days ago
qbufferqbuffer
726
edited yesterday
Martin Sleziak
45k10122277
edited yesterday
Martin Sleziak
45k10122277
edited yesterday
Martin Sleziak
45k10122277
45k10122277
asked 2 days ago
qbufferqbuffer
726
asked 2 days ago
qbufferqbuffer
726
asked 2 days ago
qbufferqbuffer
726
726
5
$begingroup$
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
2 days ago
7
$begingroup$
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
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago
add a comment |
5
$begingroup$
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
2 days ago
7
$begingroup$
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
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago
5
5
$begingroup$
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
2 days 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
2 days ago
7
7
$begingroup$
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
2 days 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
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? 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, $x mathcal{R} x$ 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 ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.
answered 2 days ago
Haris GusicHaris Gusic
3,516627
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 2 days ago
s0ulr3aper07s0ulr3aper07
683112
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? 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, $x mathcal{R} x$ 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 ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.
answered 2 days ago
Haris GusicHaris Gusic
3,516627
$endgroup$
add a comment |
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? 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, $x mathcal{R} x$ 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 ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.
answered 2 days ago
Haris GusicHaris Gusic
3,516627
$endgroup$
add a comment |
$begingroup$
Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? 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, $x mathcal{R} x$ 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 ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.
answered 2 days ago
Haris GusicHaris Gusic
3,516627
$endgroup$
Why are you testing reflexivity by looking at two different elements of $mathcal{X}$? 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, $x mathcal{R} x$ 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 ${1}mathcal R {1,2}$ does not contradict the fact that ${1,2}mathcal R {1,2}$ as well.
answered 2 days ago
Haris GusicHaris Gusic
3,516627
edited 2 days ago
answered 2 days ago
Haris GusicHaris Gusic
3,516627
answered 2 days ago
Haris GusicHaris Gusic
3,516627
answered 2 days ago
Haris GusicHaris Gusic
3,516627
3,516627
add a comment |
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 2 days ago
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 2 days ago
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
add a comment |
$begingroup$
A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 2 days ago
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
A binary relation $R$ over a set $mathcal{X}$ is reflexive if every element of $mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$mathcal{R} text{ is reflexive iff } forall x in mathcal{X}, x mathcal{R} x$$
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 2 days ago
s0ulr3aper07s0ulr3aper07
683112
answered 2 days ago
s0ulr3aper07s0ulr3aper07
683112
answered 2 days ago
s0ulr3aper07s0ulr3aper07
683112
answered 2 days ago
s0ulr3aper07s0ulr3aper07
683112
683112
add a comment |
add a comment |
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$begingroup$
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
2 days ago
7
$begingroup$
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
2 days ago
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
$endgroup$
– qbuffer
2 days ago
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
2 days ago