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How is the relation “the smallest element is the same” reflexive?

Need help counting equivalence classes.Finding the smallest relation that is reflexive, transitive, and symmetricSmallest relation for reflexive, symmetry and transitivityEquivalence relation example. How is this even reflexive?Is antisymmetric the same as reflexive?Finding the smallest equivalence relation containing a specific list of ordered pairsHow is this an equivalence relation?truefalse claims in relations and equivalence relationsWhat is the least and greatest element in symmetric but not reflexive relation over ${1,2,3}$?How is this case a reflexive relation?

8

$begingroup$

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

share|cite|improve this question

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
  $endgroup$
  – qbuffer
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @qbuffer Have a look at the updated version of my answer.
  $endgroup$
  – Haris Gusic
  10 hours ago
  

  
  
  
  

add a comment | 

8

$begingroup$

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

share|cite|improve this question

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So it must be reflexive because both $A$ and $B$ belong to the same set $mathcal{X}$?
  $endgroup$
  – qbuffer
  11 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @qbuffer Have a look at the updated version of my answer.
  $endgroup$
  – Haris Gusic
  10 hours 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

share|cite|improve this question

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 hours ago

qbufferqbuffer

625

$endgroup$

share|cite|improve this question

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 hours ago

qbufferqbuffer

625

share|cite|improve this question

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 hours ago

qbufferqbuffer

625

edited 28 mins ago

Martin Sleziak

45k10122277

edited 28 mins ago

Martin Sleziak

45k10122277

asked 11 hours ago

qbufferqbuffer

625

asked 11 hours ago

qbufferqbuffer

625

2 Answers
2

active

oldest

votes

8

$begingroup$

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.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

$endgroup$

add a comment | 

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8

$begingroup$

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.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

$endgroup$

add a comment | 

8

$begingroup$

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.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

$endgroup$

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.

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

answered 11 hours ago

Haris GusicHaris Gusic

3,331525

4

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

$endgroup$

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.

share|cite|improve this answer

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

$endgroup$

share|cite|improve this answer

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112

answered 11 hours ago

s0ulr3aper07s0ulr3aper07

658112