Show that $Xsubset Xcupleft{ Xright}$












3












$begingroup$



  • Show that If $X$ is an any set, then $Xsubset Xcupleft{ Xright}$


Proof. Let $tin X$. We must show $tin Xcupleft{ Xright}$, that is we need to show either $tin X$ or $tinleft{ Xright}$, so we know that $tin X$, hence we are done.



Can you check my proof?






proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 0:31






  • 1




    $begingroup$
    Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:31








  • 1




    $begingroup$
    @YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 0:35






  • 1




    $begingroup$
    @KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
    $endgroup$
    – Shubham Johri
    Dec 25 '18 at 0:41








  • 1




    $begingroup$
    Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:42
















3












$begingroup$



  • Show that If $X$ is an any set, then $Xsubset Xcupleft{ Xright}$


Proof. Let $tin X$. We must show $tin Xcupleft{ Xright}$, that is we need to show either $tin X$ or $tinleft{ Xright}$, so we know that $tin X$, hence we are done.



Can you check my proof?






proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 0:31






  • 1




    $begingroup$
    Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:31








  • 1




    $begingroup$
    @YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 0:35






  • 1




    $begingroup$
    @KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
    $endgroup$
    – Shubham Johri
    Dec 25 '18 at 0:41








  • 1




    $begingroup$
    Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:42














3












3








3





$begingroup$



  • Show that If $X$ is an any set, then $Xsubset Xcupleft{ Xright}$


Proof. Let $tin X$. We must show $tin Xcupleft{ Xright}$, that is we need to show either $tin X$ or $tinleft{ Xright}$, so we know that $tin X$, hence we are done.



Can you check my proof?






proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$





  • Show that If $X$ is an any set, then $Xsubset Xcupleft{ Xright}$


Proof. Let $tin X$. We must show $tin Xcupleft{ Xright}$, that is we need to show either $tin X$ or $tinleft{ Xright}$, so we know that $tin X$, hence we are done.



Can you check my proof?






proof-verification elementary-set-theory




proof-verification elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 0:29







PozcuKushimotoStreet

















asked Dec 25 '18 at 0:23









PozcuKushimotoStreetPozcuKushimotoStreet

1,371923




1,371923








  • 3




    $begingroup$
    Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 0:31






  • 1




    $begingroup$
    Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:31








  • 1




    $begingroup$
    @YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 0:35






  • 1




    $begingroup$
    @KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
    $endgroup$
    – Shubham Johri
    Dec 25 '18 at 0:41








  • 1




    $begingroup$
    Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:42














  • 3




    $begingroup$
    Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 0:31






  • 1




    $begingroup$
    Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:31








  • 1




    $begingroup$
    @YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 0:35






  • 1




    $begingroup$
    @KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
    $endgroup$
    – Shubham Johri
    Dec 25 '18 at 0:41








  • 1




    $begingroup$
    Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
    $endgroup$
    – YoungMath
    Dec 25 '18 at 0:42








3




3




$begingroup$
Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
$endgroup$
– DonAntonio
Dec 25 '18 at 0:31




$begingroup$
Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but".
$endgroup$
– DonAntonio
Dec 25 '18 at 0:31




1




1




$begingroup$
Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
$endgroup$
– YoungMath
Dec 25 '18 at 0:31






$begingroup$
Ehm. Well. You're perfectly right. But the question unsettles me a little. :D
$endgroup$
– YoungMath
Dec 25 '18 at 0:31






1




1




$begingroup$
@YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 0:35




$begingroup$
@YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $Xsubseteq Xcupleft{Xright}$, from this do we get any problem?
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 0:35




1




1




$begingroup$
@KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
$endgroup$
– Shubham Johri
Dec 25 '18 at 0:41






$begingroup$
@KathySong Yes, $subset$ makes a difference. You need to show $exists ain Xcup{X}$ such that $anotin X$. That $a$ is $X$
$endgroup$
– Shubham Johri
Dec 25 '18 at 0:41






1




1




$begingroup$
Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
$endgroup$
– YoungMath
Dec 25 '18 at 0:42




$begingroup$
Oh well, that's subtile, you're right. Then, you need to show that $X neq X cup {X}$. However, this is simple since $X notin X$ due to the axiom of regularity.
$endgroup$
– YoungMath
Dec 25 '18 at 0:42










2 Answers
2






active

oldest

votes


















2












$begingroup$

Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X subseteq X cup {X}$ is already discussed in full detail, I want to make a remark about the case $X neq X cup {X }$.



The Axiom of Regularity reads
$$ forall x left( x neq emptyset Rightarrow exists y in x: y cap x = emptyset right).$$



Let us prove, that $X notin X$ for all sets $X$.



Soo, let $X$ be any set. Due to the axiom of pairing, ${X}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X cap {X} = emptyset$ since $X$ is the only element in ${X}$. Hence, $X notin X$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer :)
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 12:51



















0












$begingroup$

Exercise.

Show for all sets A and B, that A $subseteq$ A $cup$ B.



With that, your problem is just a special case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 1:50












  • $begingroup$
    @KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
    $endgroup$
    – William Elliot
    Dec 25 '18 at 1:58






  • 1




    $begingroup$
    Please read the comments before posting an answer. Thank you! :)
    $endgroup$
    – YoungMath
    Dec 25 '18 at 2:57











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X subseteq X cup {X}$ is already discussed in full detail, I want to make a remark about the case $X neq X cup {X }$.



The Axiom of Regularity reads
$$ forall x left( x neq emptyset Rightarrow exists y in x: y cap x = emptyset right).$$



Let us prove, that $X notin X$ for all sets $X$.



Soo, let $X$ be any set. Due to the axiom of pairing, ${X}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X cap {X} = emptyset$ since $X$ is the only element in ${X}$. Hence, $X notin X$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer :)
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 12:51
















2












$begingroup$

Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X subseteq X cup {X}$ is already discussed in full detail, I want to make a remark about the case $X neq X cup {X }$.



The Axiom of Regularity reads
$$ forall x left( x neq emptyset Rightarrow exists y in x: y cap x = emptyset right).$$



Let us prove, that $X notin X$ for all sets $X$.



Soo, let $X$ be any set. Due to the axiom of pairing, ${X}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X cap {X} = emptyset$ since $X$ is the only element in ${X}$. Hence, $X notin X$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer :)
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 12:51














2












2








2





$begingroup$

Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X subseteq X cup {X}$ is already discussed in full detail, I want to make a remark about the case $X neq X cup {X }$.



The Axiom of Regularity reads
$$ forall x left( x neq emptyset Rightarrow exists y in x: y cap x = emptyset right).$$



Let us prove, that $X notin X$ for all sets $X$.



Soo, let $X$ be any set. Due to the axiom of pairing, ${X}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X cap {X} = emptyset$ since $X$ is the only element in ${X}$. Hence, $X notin X$.






share|cite|improve this answer









$endgroup$



Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X subseteq X cup {X}$ is already discussed in full detail, I want to make a remark about the case $X neq X cup {X }$.



The Axiom of Regularity reads
$$ forall x left( x neq emptyset Rightarrow exists y in x: y cap x = emptyset right).$$



Let us prove, that $X notin X$ for all sets $X$.



Soo, let $X$ be any set. Due to the axiom of pairing, ${X}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X cap {X} = emptyset$ since $X$ is the only element in ${X}$. Hence, $X notin X$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 2:56









YoungMathYoungMath

197111




197111












  • $begingroup$
    Thank you for answer :)
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 12:51


















  • $begingroup$
    Thank you for answer :)
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 12:51
















$begingroup$
Thank you for answer :)
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 12:51




$begingroup$
Thank you for answer :)
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 12:51











0












$begingroup$

Exercise.

Show for all sets A and B, that A $subseteq$ A $cup$ B.



With that, your problem is just a special case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 1:50












  • $begingroup$
    @KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
    $endgroup$
    – William Elliot
    Dec 25 '18 at 1:58






  • 1




    $begingroup$
    Please read the comments before posting an answer. Thank you! :)
    $endgroup$
    – YoungMath
    Dec 25 '18 at 2:57
















0












$begingroup$

Exercise.

Show for all sets A and B, that A $subseteq$ A $cup$ B.



With that, your problem is just a special case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 1:50












  • $begingroup$
    @KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
    $endgroup$
    – William Elliot
    Dec 25 '18 at 1:58






  • 1




    $begingroup$
    Please read the comments before posting an answer. Thank you! :)
    $endgroup$
    – YoungMath
    Dec 25 '18 at 2:57














0












0








0





$begingroup$

Exercise.

Show for all sets A and B, that A $subseteq$ A $cup$ B.



With that, your problem is just a special case.






share|cite|improve this answer









$endgroup$



Exercise.

Show for all sets A and B, that A $subseteq$ A $cup$ B.



With that, your problem is just a special case.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 1:43









William ElliotWilliam Elliot

8,5922720




8,5922720












  • $begingroup$
    Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 1:50












  • $begingroup$
    @KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
    $endgroup$
    – William Elliot
    Dec 25 '18 at 1:58






  • 1




    $begingroup$
    Please read the comments before posting an answer. Thank you! :)
    $endgroup$
    – YoungMath
    Dec 25 '18 at 2:57


















  • $begingroup$
    Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
    $endgroup$
    – PozcuKushimotoStreet
    Dec 25 '18 at 1:50












  • $begingroup$
    @KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
    $endgroup$
    – William Elliot
    Dec 25 '18 at 1:58






  • 1




    $begingroup$
    Please read the comments before posting an answer. Thank you! :)
    $endgroup$
    – YoungMath
    Dec 25 '18 at 2:57
















$begingroup$
Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 1:50






$begingroup$
Thanks for exercise. I would like to ask a question that how can I show $Xneq Xcupleft{Xright}$ Can you help?
$endgroup$
– PozcuKushimotoStreet
Dec 25 '18 at 1:50














$begingroup$
@KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
$endgroup$
– William Elliot
Dec 25 '18 at 1:58




$begingroup$
@KathySong. Equality would lead to X in X in contradiction to axiom of regularity.
$endgroup$
– William Elliot
Dec 25 '18 at 1:58




1




1




$begingroup$
Please read the comments before posting an answer. Thank you! :)
$endgroup$
– YoungMath
Dec 25 '18 at 2:57




$begingroup$
Please read the comments before posting an answer. Thank you! :)
$endgroup$
– YoungMath
Dec 25 '18 at 2:57


















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