Prove $Acap A = A$. The 2019 Stack Overflow Developer Survey Results Are InProve $(A triangle B) cap (Btriangle C) cap (Ctriangle A) = emptyset$Prove that $A subseteq B$ if and only if $A cap overlineB=emptyset.$Prove that $Acap(Bbigtriangleup C)subseteq (A cap B) bigtriangleup C$Prove that $(A cap B') cap (B cap A') = A cap A' cap B cap B'$If $Asubseteq C$ then $ABcap C=A(Bcap C)$?Prove there are no sets $A,B,C$ such that $ A cap B neq emptyset$, $ A cap C = emptyset$, $ (A cap B) - C = emptyset$prove $A cap (B - A) = emptyset$Prove the set identity $(A cap B) setminus (Bcap C) = A cap (B setminus C)$How to prove that $A - (A cap B) = A - B$Prove by contradiction $[(Acap B)−(Bcap C)]−(Acap C)'=phi$.
What is the meaning of Triage in Cybersec world?
Is a "Democratic" Oligarchy-Style System Possible?
Apparent duplicates between Haynes service instructions and MOT
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Protecting Dualbooting Windows from dangerous code (like rm -rf)
Interpreting the 2019 New York Reproductive Health Act?
Why isn't airport relocation done gradually?
Can you compress metal and what would be the consequences?
Why is the maximum length of OpenWrt’s root password 8 characters?
When should I buy a clipper card after flying to OAK?
Why do UK politicians seemingly ignore opinion polls on Brexit?
Can a flute soloist sit?
FPGA - DIY Programming
Feature engineering suggestion required
Why not take a picture of a closer black hole?
Aging parents with no investments
The difference between dialogue marks
Right tool to dig six foot holes?
How to manage monthly salary
What do the Banks children have against barley water?
Should I use my personal e-mail address, or my workplace one, when registering to external websites for work purposes?
What do hard-Brexiteers want with respect to the Irish border?
Have you ever entered Singapore using a different passport or name?
Are there any other methods to apply to solving simultaneous equations?
Prove $Acap A = A$.
The 2019 Stack Overflow Developer Survey Results Are InProve $(A triangle B) cap (Btriangle C) cap (Ctriangle A) = emptyset$Prove that $A subseteq B$ if and only if $A cap overlineB=emptyset.$Prove that $Acap(Bbigtriangleup C)subseteq (A cap B) bigtriangleup C$Prove that $(A cap B') cap (B cap A') = A cap A' cap B cap B'$If $Asubseteq C$ then $ABcap C=A(Bcap C)$?Prove there are no sets $A,B,C$ such that $ A cap B neq emptyset$, $ A cap C = emptyset$, $ (A cap B) - C = emptyset$prove $A cap (B - A) = emptyset$Prove the set identity $(A cap B) setminus (Bcap C) = A cap (B setminus C)$How to prove that $A - (A cap B) = A - B$Prove by contradiction $[(Acap B)−(Bcap C)]−(Acap C)'=phi$.
$begingroup$
How can $Acap A = A$ be proved? Can it only be proved by the method of contradiction since I see it to be an obvious set their fact that can be observed?
elementary-set-theory
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How can $Acap A = A$ be proved? Can it only be proved by the method of contradiction since I see it to be an obvious set their fact that can be observed?
elementary-set-theory
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08
add a comment |
$begingroup$
How can $Acap A = A$ be proved? Can it only be proved by the method of contradiction since I see it to be an obvious set their fact that can be observed?
elementary-set-theory
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How can $Acap A = A$ be proved? Can it only be proved by the method of contradiction since I see it to be an obvious set their fact that can be observed?
elementary-set-theory
elementary-set-theory
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
edited Apr 5 at 14:50
YuiTo Cheng
2,3694937
2,3694937
asked Apr 5 at 13:58
Aditya Aditya
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 5 at 13:58
Aditya Aditya
61
asked Apr 5 at 13:58
Aditya Aditya
61
61
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08
add a comment |
1
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08
1
1
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can just look at what intersection, i.e. $cap$, produces for a set. Let $X,Y$ be sets, then
$$Xcap Y=zmid zin Xtext and zin Y$$
Thus
$$Acap A=xmid xin Atext and xin A=xmid xin A=A$$
answered Apr 5 at 14:01
blubblub
3,324929
$endgroup$
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
add a comment |
$begingroup$
Remember that $$boxedM= N iff Msubseteq N;; wedge ;; Nsubseteq M$$
Take any $xin Acap A$, then ($xin A$ and $xin A$) so $xin A$ so $boxedAcap Asubseteq A$
And vice versa, say $xin A $, then $xin Acap A$ so $boxedAsubseteq Acap A$
So $Acap A =A$.
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
By the double inclusion theory, if A ⊂ B and B ⊂ A, then A = B.
A ⊂ A∩A
- Let x ∈ A.
- If x∈A we have, in particular, that x is an element of set A. Then x∈A∩A, because x∈A and x∈A by the intersection definition. So x ∈ A∩A;
- Therefore, A ⊂ A∩A;
A∩A ⊂ A
Let X ∈ A∩A.
Then x∈A and x∈A, by intersection definition. So, in particular, we got x∈A.
Therefore, A∩A ⊂ A.
Thus, by double inclusion, A∩A = A
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
$endgroup$
add a comment |
$begingroup$
The set A$cap$B is definied as x: x$in$A and x$in$B
The set A$cap$A is, by consequence, x: x$in$A and x$in$A
The set A is obviously x: x$in$A
Now, given any element G$in$A, G$in$A$
cap$A, as the definition suggests.
Also, for any element G$in$A$cap$A, G$in$A has to be satisfied.
Thus, every element of set A is also in the set A$cap$A and vice versa
That should be enough to declare the sets equal.
answered Apr 5 at 14:11
amanaman
33111
$endgroup$
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
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
);
);
Aditya is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3175967%2fprove-a-cap-a-a%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can just look at what intersection, i.e. $cap$, produces for a set. Let $X,Y$ be sets, then
$$Xcap Y=zmid zin Xtext and zin Y$$
Thus
$$Acap A=xmid xin Atext and xin A=xmid xin A=A$$
answered Apr 5 at 14:01
blubblub
3,324929
$endgroup$
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
add a comment |
$begingroup$
You can just look at what intersection, i.e. $cap$, produces for a set. Let $X,Y$ be sets, then
$$Xcap Y=zmid zin Xtext and zin Y$$
Thus
$$Acap A=xmid xin Atext and xin A=xmid xin A=A$$
answered Apr 5 at 14:01
blubblub
3,324929
$endgroup$
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
add a comment |
$begingroup$
You can just look at what intersection, i.e. $cap$, produces for a set. Let $X,Y$ be sets, then
$$Xcap Y=zmid zin Xtext and zin Y$$
Thus
$$Acap A=xmid xin Atext and xin A=xmid xin A=A$$
answered Apr 5 at 14:01
blubblub
3,324929
$endgroup$
You can just look at what intersection, i.e. $cap$, produces for a set. Let $X,Y$ be sets, then
$$Xcap Y=zmid zin Xtext and zin Y$$
Thus
$$Acap A=xmid xin Atext and xin A=xmid xin A=A$$
answered Apr 5 at 14:01
blubblub
3,324929
answered Apr 5 at 14:01
blubblub
3,324929
answered Apr 5 at 14:01
blubblub
3,324929
answered Apr 5 at 14:01
blubblub
3,324929
3,324929
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
add a comment |
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
Sin (Q)/Cos (Q)
$endgroup$
– Aditya
Apr 5 at 16:51
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
$begingroup$
@Aditya ... What?
$endgroup$
– blub
Apr 5 at 17:20
add a comment |
$begingroup$
Remember that $$boxedM= N iff Msubseteq N;; wedge ;; Nsubseteq M$$
Take any $xin Acap A$, then ($xin A$ and $xin A$) so $xin A$ so $boxedAcap Asubseteq A$
And vice versa, say $xin A $, then $xin Acap A$ so $boxedAsubseteq Acap A$
So $Acap A =A$.
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Remember that $$boxedM= N iff Msubseteq N;; wedge ;; Nsubseteq M$$
Take any $xin Acap A$, then ($xin A$ and $xin A$) so $xin A$ so $boxedAcap Asubseteq A$
And vice versa, say $xin A $, then $xin Acap A$ so $boxedAsubseteq Acap A$
So $Acap A =A$.
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
add a comment |
$begingroup$
Remember that $$boxedM= N iff Msubseteq N;; wedge ;; Nsubseteq M$$
Take any $xin Acap A$, then ($xin A$ and $xin A$) so $xin A$ so $boxedAcap Asubseteq A$
And vice versa, say $xin A $, then $xin Acap A$ so $boxedAsubseteq Acap A$
So $Acap A =A$.
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
$endgroup$
Remember that $$boxedM= N iff Msubseteq N;; wedge ;; Nsubseteq M$$
Take any $xin Acap A$, then ($xin A$ and $xin A$) so $xin A$ so $boxedAcap Asubseteq A$
And vice versa, say $xin A $, then $xin Acap A$ so $boxedAsubseteq Acap A$
So $Acap A =A$.
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
edited Apr 5 at 14:17
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
answered Apr 5 at 14:04
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
add a comment |
add a comment |
$begingroup$
By the double inclusion theory, if A ⊂ B and B ⊂ A, then A = B.
A ⊂ A∩A
- Let x ∈ A.
- If x∈A we have, in particular, that x is an element of set A. Then x∈A∩A, because x∈A and x∈A by the intersection definition. So x ∈ A∩A;
- Therefore, A ⊂ A∩A;
A∩A ⊂ A
Let X ∈ A∩A.
Then x∈A and x∈A, by intersection definition. So, in particular, we got x∈A.
Therefore, A∩A ⊂ A.
Thus, by double inclusion, A∩A = A
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
$endgroup$
add a comment |
$begingroup$
By the double inclusion theory, if A ⊂ B and B ⊂ A, then A = B.
A ⊂ A∩A
- Let x ∈ A.
- If x∈A we have, in particular, that x is an element of set A. Then x∈A∩A, because x∈A and x∈A by the intersection definition. So x ∈ A∩A;
- Therefore, A ⊂ A∩A;
A∩A ⊂ A
Let X ∈ A∩A.
Then x∈A and x∈A, by intersection definition. So, in particular, we got x∈A.
Therefore, A∩A ⊂ A.
Thus, by double inclusion, A∩A = A
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
$endgroup$
add a comment |
$begingroup$
By the double inclusion theory, if A ⊂ B and B ⊂ A, then A = B.
A ⊂ A∩A
- Let x ∈ A.
- If x∈A we have, in particular, that x is an element of set A. Then x∈A∩A, because x∈A and x∈A by the intersection definition. So x ∈ A∩A;
- Therefore, A ⊂ A∩A;
A∩A ⊂ A
Let X ∈ A∩A.
Then x∈A and x∈A, by intersection definition. So, in particular, we got x∈A.
Therefore, A∩A ⊂ A.
Thus, by double inclusion, A∩A = A
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
$endgroup$
By the double inclusion theory, if A ⊂ B and B ⊂ A, then A = B.
A ⊂ A∩A
- Let x ∈ A.
- If x∈A we have, in particular, that x is an element of set A. Then x∈A∩A, because x∈A and x∈A by the intersection definition. So x ∈ A∩A;
- Therefore, A ⊂ A∩A;
A∩A ⊂ A
Let X ∈ A∩A.
Then x∈A and x∈A, by intersection definition. So, in particular, we got x∈A.
Therefore, A∩A ⊂ A.
Thus, by double inclusion, A∩A = A
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
answered Apr 6 at 14:06
Daniel Sehn ColaoDaniel Sehn Colao
314
314
add a comment |
add a comment |
$begingroup$
The set A$cap$B is definied as x: x$in$A and x$in$B
The set A$cap$A is, by consequence, x: x$in$A and x$in$A
The set A is obviously x: x$in$A
Now, given any element G$in$A, G$in$A$
cap$A, as the definition suggests.
Also, for any element G$in$A$cap$A, G$in$A has to be satisfied.
Thus, every element of set A is also in the set A$cap$A and vice versa
That should be enough to declare the sets equal.
answered Apr 5 at 14:11
amanaman
33111
$endgroup$
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
add a comment |
$begingroup$
The set A$cap$B is definied as x: x$in$A and x$in$B
The set A$cap$A is, by consequence, x: x$in$A and x$in$A
The set A is obviously x: x$in$A
Now, given any element G$in$A, G$in$A$
cap$A, as the definition suggests.
Also, for any element G$in$A$cap$A, G$in$A has to be satisfied.
Thus, every element of set A is also in the set A$cap$A and vice versa
That should be enough to declare the sets equal.
answered Apr 5 at 14:11
amanaman
33111
$endgroup$
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
add a comment |
$begingroup$
The set A$cap$B is definied as x: x$in$A and x$in$B
The set A$cap$A is, by consequence, x: x$in$A and x$in$A
The set A is obviously x: x$in$A
Now, given any element G$in$A, G$in$A$
cap$A, as the definition suggests.
Also, for any element G$in$A$cap$A, G$in$A has to be satisfied.
Thus, every element of set A is also in the set A$cap$A and vice versa
That should be enough to declare the sets equal.
answered Apr 5 at 14:11
amanaman
33111
$endgroup$
The set A$cap$B is definied as x: x$in$A and x$in$B
The set A$cap$A is, by consequence, x: x$in$A and x$in$A
The set A is obviously x: x$in$A
Now, given any element G$in$A, G$in$A$
cap$A, as the definition suggests.
Also, for any element G$in$A$cap$A, G$in$A has to be satisfied.
Thus, every element of set A is also in the set A$cap$A and vice versa
That should be enough to declare the sets equal.
answered Apr 5 at 14:11
amanaman
33111
answered Apr 5 at 14:11
amanaman
33111
answered Apr 5 at 14:11
amanaman
33111
answered Apr 5 at 14:11
amanaman
33111
33111
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
add a comment |
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
$begingroup$
Why the down-votes? Can you tell me how to improve my answer?
$endgroup$
– aman
Apr 5 at 14:31
add a comment |
Aditya is a new contributor. Be nice, and check out our Code of Conduct.
Aditya is a new contributor. Be nice, and check out our Code of Conduct.
Aditya is a new contributor. Be nice, and check out our Code of Conduct.
Aditya is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3175967%2fprove-a-cap-a-a%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
1
$begingroup$
Apply Extensionality Axiom : $forall x (x in Q leftrightarrow x in R) to Q=R$
$endgroup$
– Mauro ALLEGRANZA
Apr 5 at 14:08