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?
Multi tool use
Is it correct to say the Neural Networks are an alternative way of performing Maximum Likelihood Estimation? if not, why?
Why not take a picture of a closer black hole?
How can I define good in a religion that claims no moral authority?
Can there be female White Walkers?
Getting crown tickets for Statue of Liberty
Is an up-to-date browser secure on an out-of-date OS?
What information about me do stores get via my credit card?
Relationship between Gromov-Witten and Taubes' Gromov invariant
How to charge AirPods to keep battery healthy?
What's the name of these plastic connectors
What do these terms in Caesar's Gallic wars mean?
Ubuntu Server install with full GUI
Likelihood that a superbug or lethal virus could come from a landfill
Correct punctuation for showing a character's confusion
Can we generate random numbers using irrational numbers like π and e?
Why can't devices on different VLANs, but on the same subnet, communicate?
How to notate time signature switching consistently every measure
What do hard-Brexiteers want with respect to the Irish border?
Falsification in Math vs Science
Why doesn't UInt have a toDouble()?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
Star Trek - X-shaped Item on Regula/Orbital Office Starbases
Is bread bad for ducks?
Geography at the pixel level
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
706
$endgroup$
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
706
$endgroup$
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
706
$endgroup$
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
706
edited Apr 8 at 5:01
Martin Sleziak
45k10122277
asked Apr 7 at 17:31
qbufferqbuffer
706
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
706
asked Apr 7 at 17:31
qbufferqbuffer
706
asked Apr 7 at 17:31
qbufferqbuffer
706
706
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
3,546627
$endgroup$
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
3,546627
$endgroup$
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
3,546627
$endgroup$
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
3,546627
$endgroup$
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
3,546627
edited Apr 7 at 18:48
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,546627
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,546627
answered Apr 7 at 17:38
Haris GusicHaris Gusic
3,546627
3,546627
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
JI3B2 W5e4Yw3Cc,aBrL8B 80Y,kTtExzgPy9L4UqfpLK HVlk OKFk,Vv8UFWuCub3C TzJb NRix,56oXF,oYFdRxr iFV50a7,xNK
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