Find a set such that $Ain B$ and $Asubseteq B$












0












$begingroup$



Find a set such that $Ain B$ and $Asubseteq B$.




I was thinking about the $mathcal P(A)$ to be the answer of this but I'm not sure about $Asubseteq B$..






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try $A=emptyset$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:12










  • $begingroup$
    Try ${A,{A}}$.
    $endgroup$
    – Michael Hoppe
    Jan 17 at 7:15










  • $begingroup$
    Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
    $endgroup$
    – manooooh
    Jan 17 at 7:18












  • $begingroup$
    @MichaelHoppe What do you mean by ${A, {A}}$?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:19






  • 1




    $begingroup$
    @manooooh no difference, and i will edit it
    $endgroup$
    – C. Cristi
    Jan 17 at 7:20
















0












$begingroup$



Find a set such that $Ain B$ and $Asubseteq B$.




I was thinking about the $mathcal P(A)$ to be the answer of this but I'm not sure about $Asubseteq B$..






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try $A=emptyset$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:12










  • $begingroup$
    Try ${A,{A}}$.
    $endgroup$
    – Michael Hoppe
    Jan 17 at 7:15










  • $begingroup$
    Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
    $endgroup$
    – manooooh
    Jan 17 at 7:18












  • $begingroup$
    @MichaelHoppe What do you mean by ${A, {A}}$?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:19






  • 1




    $begingroup$
    @manooooh no difference, and i will edit it
    $endgroup$
    – C. Cristi
    Jan 17 at 7:20














0












0








0





$begingroup$



Find a set such that $Ain B$ and $Asubseteq B$.




I was thinking about the $mathcal P(A)$ to be the answer of this but I'm not sure about $Asubseteq B$..






elementary-set-theory







share|cite|improve this question











$endgroup$





Find a set such that $Ain B$ and $Asubseteq B$.




I was thinking about the $mathcal P(A)$ to be the answer of this but I'm not sure about $Asubseteq B$..






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 10:54









Hanul Jeon

17.7k42881




17.7k42881










asked Jan 17 at 7:10









C. CristiC. Cristi

1,655318




1,655318












  • $begingroup$
    Try $A=emptyset$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:12










  • $begingroup$
    Try ${A,{A}}$.
    $endgroup$
    – Michael Hoppe
    Jan 17 at 7:15










  • $begingroup$
    Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
    $endgroup$
    – manooooh
    Jan 17 at 7:18












  • $begingroup$
    @MichaelHoppe What do you mean by ${A, {A}}$?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:19






  • 1




    $begingroup$
    @manooooh no difference, and i will edit it
    $endgroup$
    – C. Cristi
    Jan 17 at 7:20


















  • $begingroup$
    Try $A=emptyset$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 17 at 7:12










  • $begingroup$
    Try ${A,{A}}$.
    $endgroup$
    – Michael Hoppe
    Jan 17 at 7:15










  • $begingroup$
    Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
    $endgroup$
    – manooooh
    Jan 17 at 7:18












  • $begingroup$
    @MichaelHoppe What do you mean by ${A, {A}}$?
    $endgroup$
    – C. Cristi
    Jan 17 at 7:19






  • 1




    $begingroup$
    @manooooh no difference, and i will edit it
    $endgroup$
    – C. Cristi
    Jan 17 at 7:20
















$begingroup$
Try $A=emptyset$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:12




$begingroup$
Try $A=emptyset$.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 7:12












$begingroup$
Try ${A,{A}}$.
$endgroup$
– Michael Hoppe
Jan 17 at 7:15




$begingroup$
Try ${A,{A}}$.
$endgroup$
– Michael Hoppe
Jan 17 at 7:15












$begingroup$
Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
$endgroup$
– manooooh
Jan 17 at 7:18






$begingroup$
Could you please post the statement with a proper notation (the title says "$Asubset B$" but in the message it says "$Asubseteq B$"), and explain what is the difference between "Find a set" and "Give an example"?
$endgroup$
– manooooh
Jan 17 at 7:18














$begingroup$
@MichaelHoppe What do you mean by ${A, {A}}$?
$endgroup$
– C. Cristi
Jan 17 at 7:19




$begingroup$
@MichaelHoppe What do you mean by ${A, {A}}$?
$endgroup$
– C. Cristi
Jan 17 at 7:19




1




1




$begingroup$
@manooooh no difference, and i will edit it
$endgroup$
– C. Cristi
Jan 17 at 7:20




$begingroup$
@manooooh no difference, and i will edit it
$endgroup$
– C. Cristi
Jan 17 at 7:20










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $A={x}$ and $B={{x}, x}$



Then, $Ain B$ and $Asubseteq B$; since $forall yin A$, $y in B$



$P(A)= {x| xsubseteq A }$



Suppose $A={x}$, then $P(A)=B={{x}, {emptyset }, {x, emptyset }}$



As you can see, $Ain P(A)$, but $Anot subseteq P(A)$. Since, $xnotin P(A)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
    $endgroup$
    – C. Cristi
    Jan 17 at 14:17






  • 1




    $begingroup$
    @C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Jan 18 at 5:17



















0












$begingroup$

Every ordinal and its successor as defined by Von Neumann satisfy this property.



See Von Neumann definition of ordinals






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
    $endgroup$
    – Mees de Vries
    Jan 17 at 9:56












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let $A={x}$ and $B={{x}, x}$



Then, $Ain B$ and $Asubseteq B$; since $forall yin A$, $y in B$



$P(A)= {x| xsubseteq A }$



Suppose $A={x}$, then $P(A)=B={{x}, {emptyset }, {x, emptyset }}$



As you can see, $Ain P(A)$, but $Anot subseteq P(A)$. Since, $xnotin P(A)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
    $endgroup$
    – C. Cristi
    Jan 17 at 14:17






  • 1




    $begingroup$
    @C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Jan 18 at 5:17
















2












$begingroup$

Let $A={x}$ and $B={{x}, x}$



Then, $Ain B$ and $Asubseteq B$; since $forall yin A$, $y in B$



$P(A)= {x| xsubseteq A }$



Suppose $A={x}$, then $P(A)=B={{x}, {emptyset }, {x, emptyset }}$



As you can see, $Ain P(A)$, but $Anot subseteq P(A)$. Since, $xnotin P(A)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
    $endgroup$
    – C. Cristi
    Jan 17 at 14:17






  • 1




    $begingroup$
    @C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Jan 18 at 5:17














2












2








2





$begingroup$

Let $A={x}$ and $B={{x}, x}$



Then, $Ain B$ and $Asubseteq B$; since $forall yin A$, $y in B$



$P(A)= {x| xsubseteq A }$



Suppose $A={x}$, then $P(A)=B={{x}, {emptyset }, {x, emptyset }}$



As you can see, $Ain P(A)$, but $Anot subseteq P(A)$. Since, $xnotin P(A)$






share|cite|improve this answer











$endgroup$



Let $A={x}$ and $B={{x}, x}$



Then, $Ain B$ and $Asubseteq B$; since $forall yin A$, $y in B$



$P(A)= {x| xsubseteq A }$



Suppose $A={x}$, then $P(A)=B={{x}, {emptyset }, {x, emptyset }}$



As you can see, $Ain P(A)$, but $Anot subseteq P(A)$. Since, $xnotin P(A)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 9:26

























answered Jan 17 at 9:20









Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

537217




537217












  • $begingroup$
    So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
    $endgroup$
    – C. Cristi
    Jan 17 at 14:17






  • 1




    $begingroup$
    @C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Jan 18 at 5:17


















  • $begingroup$
    So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
    $endgroup$
    – C. Cristi
    Jan 17 at 14:17






  • 1




    $begingroup$
    @C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    Jan 18 at 5:17
















$begingroup$
So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
$endgroup$
– C. Cristi
Jan 17 at 14:17




$begingroup$
So inclusion $subseteq$ means that the set $A$ is in that set and $in$ means that all elements of $A$ are in that set, right?
$endgroup$
– C. Cristi
Jan 17 at 14:17




1




1




$begingroup$
@C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 18 at 5:17




$begingroup$
@C.Cristi $Asubseteq B$ means $forall x in A(xin B)$.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 18 at 5:17











0












$begingroup$

Every ordinal and its successor as defined by Von Neumann satisfy this property.



See Von Neumann definition of ordinals






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
    $endgroup$
    – Mees de Vries
    Jan 17 at 9:56
















0












$begingroup$

Every ordinal and its successor as defined by Von Neumann satisfy this property.



See Von Neumann definition of ordinals






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
    $endgroup$
    – Mees de Vries
    Jan 17 at 9:56














0












0








0





$begingroup$

Every ordinal and its successor as defined by Von Neumann satisfy this property.



See Von Neumann definition of ordinals






share|cite|improve this answer









$endgroup$



Every ordinal and its successor as defined by Von Neumann satisfy this property.



See Von Neumann definition of ordinals







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 7:22









mathcounterexamples.netmathcounterexamples.net

26.9k22158




26.9k22158












  • $begingroup$
    In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
    $endgroup$
    – Mees de Vries
    Jan 17 at 9:56


















  • $begingroup$
    In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
    $endgroup$
    – Mees de Vries
    Jan 17 at 9:56
















$begingroup$
In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
$endgroup$
– Mees de Vries
Jan 17 at 9:56




$begingroup$
In fact, if you extend the domain of the successor (class) function, every set and its successor satisfy this property. :-)
$endgroup$
– Mees de Vries
Jan 17 at 9:56


















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