Find a set $X$ such that $Xcap P(X)neqemptyset$

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• Find a set $X$ such that $Xcap P(X)neqemptyset$

My answer is: Take $X=left emptyset right$. Then, $left emptyset rightcap P(left emptyset right)=left emptyset rightneq emptyset$. So we are done.

Can you check my answer and proof-writing?

elementary-set-theory proof-writing

share|cite|improve this question

asked Jul 31 at 18:37

Kahler

1,206721

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
  – Michal Dvořák
  Jul 31 at 18:39
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
  – InterstellarProbe
  Jul 31 at 18:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any nonempty set $X$ will work.
  – pointguard0
  Jul 31 at 18:43
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
  – Kahler
  Jul 31 at 18:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @pointguard0 not every non empty set will work. For example, let $X=1$.
  – Anurag A
  Jul 31 at 18:45
  
  

  
  
  
  

 | 
show 8 more comments

up vote
2
down vote

favorite

• Find a set $X$ such that $Xcap P(X)neqemptyset$

My answer is: Take $X=left emptyset right$. Then, $left emptyset rightcap P(left emptyset right)=left emptyset rightneq emptyset$. So we are done.

Can you check my answer and proof-writing?

elementary-set-theory proof-writing

share|cite|improve this question

asked Jul 31 at 18:37

Kahler

1,206721

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
  – Michal Dvořák
  Jul 31 at 18:39
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
  – InterstellarProbe
  Jul 31 at 18:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any nonempty set $X$ will work.
  – pointguard0
  Jul 31 at 18:43
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
  – Kahler
  Jul 31 at 18:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @pointguard0 not every non empty set will work. For example, let $X=1$.
  – Anurag A
  Jul 31 at 18:45
  
  

  
  
  
  

 | 
show 8 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

• Find a set $X$ such that $Xcap P(X)neqemptyset$

My answer is: Take $X=left emptyset right$. Then, $left emptyset rightcap P(left emptyset right)=left emptyset rightneq emptyset$. So we are done.

Can you check my answer and proof-writing?

elementary-set-theory proof-writing

share|cite|improve this question

asked Jul 31 at 18:37

Kahler

1,206721

• Find a set $X$ such that $Xcap P(X)neqemptyset$

My answer is: Take $X=left emptyset right$. Then, $left emptyset rightcap P(left emptyset right)=left emptyset rightneq emptyset$. So we are done.

Can you check my answer and proof-writing?

elementary-set-theory proof-writing

share|cite|improve this question

asked Jul 31 at 18:37

Kahler

1,206721

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 18:37

Kahler

1,206721

asked Jul 31 at 18:37

Kahler

1,206721

asked Jul 31 at 18:37

Kahler

1,206721

1,206721

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
  – Michal Dvořák
  Jul 31 at 18:39
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
  – InterstellarProbe
  Jul 31 at 18:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any nonempty set $X$ will work.
  – pointguard0
  Jul 31 at 18:43
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
  – Kahler
  Jul 31 at 18:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @pointguard0 not every non empty set will work. For example, let $X=1$.
  – Anurag A
  Jul 31 at 18:45
  
  

  
  
  
  

 | 
show 8 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
  – Michal Dvořák
  Jul 31 at 18:39
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
  – InterstellarProbe
  Jul 31 at 18:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Any nonempty set $X$ will work.
  – pointguard0
  Jul 31 at 18:43
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
  – Kahler
  Jul 31 at 18:45
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @pointguard0 not every non empty set will work. For example, let $X=1$.
  – Anurag A
  Jul 31 at 18:45
  
  

  
  
  
  

2

2

But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
– Michal Dvořák
Jul 31 at 18:39

But isn't $P(emptyset)=emptyset$, then $emptyset$ is really something different than $emptyset$? Also, I think this has no solution, since If $X$ is any set with some elements $x$, then $P(X)$ is set of subsets with elements $x$.
– Michal Dvořák
Jul 31 at 18:39

5

5

@MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
– InterstellarProbe
Jul 31 at 18:42

@MichalDvořák, no $P(emptyset )$ is the power set of the set containing the emptyset, which is a one-element set. so, the power set would be $emptyset, emptyset $, which is two elements.
– InterstellarProbe
Jul 31 at 18:42

2

2

Any nonempty set $X$ will work.
– pointguard0
Jul 31 at 18:43

Any nonempty set $X$ will work.
– pointguard0
Jul 31 at 18:43

2

2

@MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
– Kahler
Jul 31 at 18:45

@MichalDvořák $P(left emptyset right)=left emptyset ,left emptysetright right$?
– Kahler
Jul 31 at 18:45

3

3

@pointguard0 not every non empty set will work. For example, let $X=1$.
– Anurag A
Jul 31 at 18:45

@pointguard0 not every non empty set will work. For example, let $X=1$.
– Anurag A
Jul 31 at 18:45

 | 
show 8 more comments

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Any non-empty transitive set would work (if $X$ is transitive then $X subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = varnothing$ would work, which is your example.

Another collection of solutions is the van Neumann universes with non-zero finite index.

Also, more generally, the transitive closure of any non-empty set would work.

---

$X cap P(X) ne varnothing$ is equivalent to an element of $X$ being a subset of $X$.

If we start with any set $X$, we could pick any subset $Y subseteq X$, and then $Y$ would also be a subset of $Z := X cup Y$, but now $Y in Z$ also, so $Z$ would work.

In particular, taking $Y = X$ gives us $Z := X cup X$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.

share|cite|improve this answer

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
  – Henning Makholm
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's taking $Y = varnothing$ lol, but good point.
  – Kenny Lau
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In particular the power set of any set would work...
  – Kenny Lau
  Jul 31 at 19:33
  

  
  
  
  

add a comment | 

up vote
0
down vote

Only one thing you can be sure of when constructing a power set for some set $X$ that $emptyset in P(X)$ and $Xin P(X)$.

So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $Xin X$ is restricted by axiom of foundation. So anything in the form $emptyset,a,b,c,1,2,3,...$ would be a solution.

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Any non-empty transitive set would work (if $X$ is transitive then $X subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = varnothing$ would work, which is your example.

Another collection of solutions is the van Neumann universes with non-zero finite index.

Also, more generally, the transitive closure of any non-empty set would work.

---

$X cap P(X) ne varnothing$ is equivalent to an element of $X$ being a subset of $X$.

If we start with any set $X$, we could pick any subset $Y subseteq X$, and then $Y$ would also be a subset of $Z := X cup Y$, but now $Y in Z$ also, so $Z$ would work.

In particular, taking $Y = X$ gives us $Z := X cup X$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.

share|cite|improve this answer

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
  – Henning Makholm
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's taking $Y = varnothing$ lol, but good point.
  – Kenny Lau
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In particular the power set of any set would work...
  – Kenny Lau
  Jul 31 at 19:33
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Any non-empty transitive set would work (if $X$ is transitive then $X subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = varnothing$ would work, which is your example.

Another collection of solutions is the van Neumann universes with non-zero finite index.

Also, more generally, the transitive closure of any non-empty set would work.

---

$X cap P(X) ne varnothing$ is equivalent to an element of $X$ being a subset of $X$.

If we start with any set $X$, we could pick any subset $Y subseteq X$, and then $Y$ would also be a subset of $Z := X cup Y$, but now $Y in Z$ also, so $Z$ would work.

In particular, taking $Y = X$ gives us $Z := X cup X$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.

share|cite|improve this answer

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
  – Henning Makholm
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's taking $Y = varnothing$ lol, but good point.
  – Kenny Lau
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In particular the power set of any set would work...
  – Kenny Lau
  Jul 31 at 19:33
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Any non-empty transitive set would work (if $X$ is transitive then $X subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = varnothing$ would work, which is your example.

Another collection of solutions is the van Neumann universes with non-zero finite index.

Also, more generally, the transitive closure of any non-empty set would work.

---

$X cap P(X) ne varnothing$ is equivalent to an element of $X$ being a subset of $X$.

If we start with any set $X$, we could pick any subset $Y subseteq X$, and then $Y$ would also be a subset of $Z := X cup Y$, but now $Y in Z$ also, so $Z$ would work.

In particular, taking $Y = X$ gives us $Z := X cup X$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.

share|cite|improve this answer

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

Any non-empty transitive set would work (if $X$ is transitive then $X subseteq P(X)$), and in particular any non-zero ordinal would work, and in particular any non-zero natural number would work, and in particular $1 = varnothing$ would work, which is your example.

Another collection of solutions is the van Neumann universes with non-zero finite index.

Also, more generally, the transitive closure of any non-empty set would work.

---

$X cap P(X) ne varnothing$ is equivalent to an element of $X$ being a subset of $X$.

If we start with any set $X$, we could pick any subset $Y subseteq X$, and then $Y$ would also be a subset of $Z := X cup Y$, but now $Y in Z$ also, so $Z$ would work.

In particular, taking $Y = X$ gives us $Z := X cup X$, i.e. the successor of any set works, and in particular the successor of the empty set works, which is also again your example.

share|cite|improve this answer

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 19:08

edited Jul 31 at 19:08

edited Jul 31 at 19:08

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

answered Jul 31 at 19:03

Kenny Lau

17.7k2156

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
  – Henning Makholm
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's taking $Y = varnothing$ lol, but good point.
  – Kenny Lau
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In particular the power set of any set would work...
  – Kenny Lau
  Jul 31 at 19:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
  – Henning Makholm
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That's taking $Y = varnothing$ lol, but good point.
  – Kenny Lau
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In particular the power set of any set would work...
  – Kenny Lau
  Jul 31 at 19:33
  

  
  
  
  

Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
– Henning Makholm
Jul 31 at 19:32

Even more generally (and more obviously) than the first two points, any set that contains $varnothing$ would work.
– Henning Makholm
Jul 31 at 19:32

That's taking $Y = varnothing$ lol, but good point.
– Kenny Lau
Jul 31 at 19:32

That's taking $Y = varnothing$ lol, but good point.
– Kenny Lau
Jul 31 at 19:32

In particular the power set of any set would work...
– Kenny Lau
Jul 31 at 19:33

In particular the power set of any set would work...
– Kenny Lau
Jul 31 at 19:33

add a comment | 

up vote
0
down vote

Only one thing you can be sure of when constructing a power set for some set $X$ that $emptyset in P(X)$ and $Xin P(X)$.

So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $Xin X$ is restricted by axiom of foundation. So anything in the form $emptyset,a,b,c,1,2,3,...$ would be a solution.

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

add a comment | 

up vote
0
down vote

Only one thing you can be sure of when constructing a power set for some set $X$ that $emptyset in P(X)$ and $Xin P(X)$.

So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $Xin X$ is restricted by axiom of foundation. So anything in the form $emptyset,a,b,c,1,2,3,...$ would be a solution.

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

add a comment | 

up vote
0
down vote

up vote
0
down vote

Only one thing you can be sure of when constructing a power set for some set $X$ that $emptyset in P(X)$ and $Xin P(X)$.

So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $Xin X$ is restricted by axiom of foundation. So anything in the form $emptyset,a,b,c,1,2,3,...$ would be a solution.

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

Only one thing you can be sure of when constructing a power set for some set $X$ that $emptyset in P(X)$ and $Xin P(X)$.

So, for our solution we want something to be both in $P(X)$ and $X$ itself. But $Xin X$ is restricted by axiom of foundation. So anything in the form $emptyset,a,b,c,1,2,3,...$ would be a solution.

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 18:53

Michal Dvořák

48912

answered Jul 31 at 18:53

Michal Dvořák

48912

answered Jul 31 at 18:53

Michal Dvořák

48912

48912

add a comment | 

add a comment | 

 

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