How is the relation “the smallest element is the same” reflexive? The 2019 Stack Overflow Developer Survey Results Are InNeed 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?
Reference request: Oldest number theory books with (unsolved) exercises?
One word riddle: Vowel in the middle
What does Linus Torvalds mean when he says that Git "never ever" tracks a file?
Why was M87 targetted for the Event Horizon Telescope instead of Sagittarius A*?
Why did Acorn's A3000 have red function keys?
Are spiders unable to hurt humans, especially very small spiders?
Did 3000BC Egyptians use meteoric iron weapons?
Is an up-to-date browser secure on an out-of-date OS?
Why hard-Brexiteers don't insist on a hard border to prevent illegal immigration after Brexit?
What do the Banks children have against barley water?
If a Druid sees an animal’s corpse, can they Wild Shape into that animal?
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Return to UK after having been refused entry years ago
Can a flute soloist sit?
What is the meaning of the verb "bear" in this context?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
Is "plugging out" electronic devices an American expression?
Are there incongruent pythagorean triangles with the same perimeter and same area?
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
Why is the maximum length of OpenWrt’s root password 8 characters?
What is the most effective way of iterating a std::vector and why?
Have you ever entered Singapore using a different passport or name?
What is the accessibility of a package's `Private` context variables?
Why not take a picture of a closer black hole?
How is the relation “the smallest element is the same” reflexive?
The 2019 Stack Overflow Developer Survey Results Are InNeed 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?
$begingroup$
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 elementary-set-theory relations equivalence-relations
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
726
$endgroup$
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 elementary-set-theory relations equivalence-relations
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
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
Apr 7 at 17:34
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
Apr 7 at 17:44
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49
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 elementary-set-theory relations equivalence-relations
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
726
$endgroup$
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 elementary-set-theory relations equivalence-relations
discrete-mathematics elementary-set-theory relations equivalence-relations
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
726
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
726
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
726
asked Apr 7 at 17:31
qbufferqbuffer
726
asked Apr 7 at 17:31
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
Apr 7 at 17:34
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
Apr 7 at 17:44
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49
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
Apr 7 at 17:34
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
Apr 7 at 17:44
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49
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
Apr 7 at 17:34
$begingroup$
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
Apr 7 at 17:34
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
Apr 7 at 17:44
$begingroup$
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
Apr 7 at 17:44
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49
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 Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
$endgroup$
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 Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3178532%2fhow-is-the-relation-the-smallest-element-is-the-same-reflexive%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
$endgroup$
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 Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
$endgroup$
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 Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
$endgroup$
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 Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
edited Apr 7 at 18:48
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,531627
3,531627
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 Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
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 Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
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 Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
$endgroup$
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 Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
answered Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
answered Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
answered Apr 7 at 17:44
s0ulr3aper07s0ulr3aper07
683112
683112
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3178532%2fhow-is-the-relation-the-smallest-element-is-the-same-reflexive%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
Apr 7 at 17:34
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
Apr 7 at 17:44
$begingroup$
So it must be reflexive because both $A$ and $B$ belong to the same set $mathcalX$?
$endgroup$
– qbuffer
Apr 7 at 18:00
$begingroup$
@qbuffer Have a look at the updated version of my answer.
$endgroup$
– Haris Gusic
Apr 7 at 18:49