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6

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Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.

elementary-set-theory

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asked yesterday

lthompsonlthompson

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put on hold as off-topic by user21820, Lord_Farin, TheSimpliFire, Saad, Alex Provost 5 hours ago

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If this question can be reworded to fit the rules in the help center, please edit the question.

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6

1

$begingroup$

Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.

elementary-set-theory

share|cite|improve this question

asked yesterday

lthompsonlthompson

16110

$endgroup$

put on hold as off-topic by user21820, Lord_Farin, TheSimpliFire, Saad, Alex Provost 5 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Lord_Farin, TheSimpliFire, Saad, Alex Provost

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

Is there any set whose every subset is equal to the set itself? It seems like this isn't possible, but maybe something similar is possible.

elementary-set-theory

share|cite|improve this question

asked yesterday

lthompsonlthompson

16110

$endgroup$

share|cite|improve this question

asked yesterday

lthompsonlthompson

16110

share|cite|improve this question

asked yesterday

lthompsonlthompson

16110

asked yesterday

lthompsonlthompson

16110

asked yesterday

lthompsonlthompson

16110

3 Answers
3

active

oldest

votes

21

$begingroup$

In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.

If you wish to do so otherwise, you’d violate the Axiom of Extensionality.

share|cite|improve this answer

answered yesterday

user458276user458276

926314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easier argument than going to the ZF axioms: the empty set is a subset of any set.
  $endgroup$
  – Daniel Robert-Nicoud
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To make @DanielRobert-Nicoud's point more explicit: since ∅ is a subset of any set S, the only set which could have all its subsets equal to S itself is ∅. Then, you can look at ∅ and observe by inspection that in fact all of its subsets are also ∅.
  $endgroup$
  – amalloy
  6 hours ago
  

  
  
  
  

add a comment | 

18

$begingroup$

The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.

share|cite|improve this answer

answered yesterday

Ross MillikanRoss Millikan

298k24200374

$endgroup$

add a comment | 

4

$begingroup$

I want to present a slightly more formal argument to what the others posted - they're right, under the typical axioms of ZFC, the only satisfying set is the empty set. But I want to validate why.

---

Let $S$ be some set, with subsets $A,B$.

We want all of the subsets of $S$ to be equal to $S$. Then $A=B=S$ if we apply these restrictions to $S$. As a result, $S$ has only one subset - itself. Why? Consider otherwise. If it has two distinct subsets $A,B$ that meet the condition, then $A=S$ and $B=S$, but since $A,B$ are distinct, $Aneq B$. But since $A=S=B$ implies $A=B$, we have a contradiction.

Let $n$ be the cardinality of $S$. Assume $S$ is finite. We then know that $scr{P}$$(S)$, its power set, has cardinality $2^n$ as a result. Obviously, then, the power set has greater cardinality than the set itself (this also holds for infinite sets and is known as Cantor's theorem).

So, $2^n > n$. We want the number of subsets to be $1$. Thus, take $n=0$. Then $2^0 = 1 > 0$. Since $n$ is the cardinality of $S$, it follows that $S$ has cardinality $0$; i.e. $S$ has no elements, and $S = emptyset$.

(This can of course be verified to meet the condition as the only subset of the empty set is itself.)

We can also make the consideration $S$ is not finite since I assume as much earlier. It is easy to show that no such $S$ meets the conditions: since it's infinite, it definitely has more than three elements; thus, we can fix a particular triple $x,y,zin S$ and have subsets ${x}neq {y,z}$ (and both are obviously not $S$ as well).

share|cite|improve this answer

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

$endgroup$

• 
  
  
  

  
  15
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but this is needlessly convoluted. Let $A$ be a set. Then $emptyset subseteq A$. If $A$ has op's property, then $emptyset = A$.
  $endgroup$
  – Martin Bidlingmaier
  15 hours ago
  

  
  
  
  

add a comment | 

21

$begingroup$

In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.

If you wish to do so otherwise, you’d violate the Axiom of Extensionality.

share|cite|improve this answer

answered yesterday

user458276user458276

926314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easier argument than going to the ZF axioms: the empty set is a subset of any set.
  $endgroup$
  – Daniel Robert-Nicoud
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To make @DanielRobert-Nicoud's point more explicit: since ∅ is a subset of any set S, the only set which could have all its subsets equal to S itself is ∅. Then, you can look at ∅ and observe by inspection that in fact all of its subsets are also ∅.
  $endgroup$
  – amalloy
  6 hours ago
  

  
  
  
  

add a comment | 

21

$begingroup$

In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.

If you wish to do so otherwise, you’d violate the Axiom of Extensionality.

share|cite|improve this answer

answered yesterday

user458276user458276

926314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Easier argument than going to the ZF axioms: the empty set is a subset of any set.
  $endgroup$
  – Daniel Robert-Nicoud
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To make @DanielRobert-Nicoud's point more explicit: since ∅ is a subset of any set S, the only set which could have all its subsets equal to S itself is ∅. Then, you can look at ∅ and observe by inspection that in fact all of its subsets are also ∅.
  $endgroup$
  – amalloy
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In standard foundations (by which I mean ZF, or ZFC) the empty set works:
If $Ssubset emptyset$, then $S = emptyset$.

If you wish to do so otherwise, you’d violate the Axiom of Extensionality.

share|cite|improve this answer

answered yesterday

user458276user458276

926314

$endgroup$

share|cite|improve this answer

answered yesterday

user458276user458276

926314

answered yesterday

user458276user458276

926314

answered yesterday

user458276user458276

926314

18

$begingroup$

The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.

share|cite|improve this answer

answered yesterday

Ross MillikanRoss Millikan

298k24200374

$endgroup$

add a comment | 

18

$begingroup$

The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.

share|cite|improve this answer

answered yesterday

Ross MillikanRoss Millikan

298k24200374

$endgroup$

add a comment | 

$begingroup$

The empty set has only itself as a subset. This is the only example because every set has the empty set as a subset.

share|cite|improve this answer

answered yesterday

Ross MillikanRoss Millikan

298k24200374

$endgroup$

share|cite|improve this answer

answered yesterday

Ross MillikanRoss Millikan

298k24200374

answered yesterday

Ross MillikanRoss Millikan

298k24200374

answered yesterday

Ross MillikanRoss Millikan

298k24200374

4

$begingroup$

I want to present a slightly more formal argument to what the others posted - they're right, under the typical axioms of ZFC, the only satisfying set is the empty set. But I want to validate why.

---

Let $S$ be some set, with subsets $A,B$.

We want all of the subsets of $S$ to be equal to $S$. Then $A=B=S$ if we apply these restrictions to $S$. As a result, $S$ has only one subset - itself. Why? Consider otherwise. If it has two distinct subsets $A,B$ that meet the condition, then $A=S$ and $B=S$, but since $A,B$ are distinct, $Aneq B$. But since $A=S=B$ implies $A=B$, we have a contradiction.

Let $n$ be the cardinality of $S$. Assume $S$ is finite. We then know that $scr{P}$$(S)$, its power set, has cardinality $2^n$ as a result. Obviously, then, the power set has greater cardinality than the set itself (this also holds for infinite sets and is known as Cantor's theorem).

So, $2^n > n$. We want the number of subsets to be $1$. Thus, take $n=0$. Then $2^0 = 1 > 0$. Since $n$ is the cardinality of $S$, it follows that $S$ has cardinality $0$; i.e. $S$ has no elements, and $S = emptyset$.

(This can of course be verified to meet the condition as the only subset of the empty set is itself.)

We can also make the consideration $S$ is not finite since I assume as much earlier. It is easy to show that no such $S$ meets the conditions: since it's infinite, it definitely has more than three elements; thus, we can fix a particular triple $x,y,zin S$ and have subsets ${x}neq {y,z}$ (and both are obviously not $S$ as well).

share|cite|improve this answer

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

$endgroup$

• 
  
  
  

  
  15
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but this is needlessly convoluted. Let $A$ be a set. Then $emptyset subseteq A$. If $A$ has op's property, then $emptyset = A$.
  $endgroup$
  – Martin Bidlingmaier
  15 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

I want to present a slightly more formal argument to what the others posted - they're right, under the typical axioms of ZFC, the only satisfying set is the empty set. But I want to validate why.

---

Let $S$ be some set, with subsets $A,B$.

We want all of the subsets of $S$ to be equal to $S$. Then $A=B=S$ if we apply these restrictions to $S$. As a result, $S$ has only one subset - itself. Why? Consider otherwise. If it has two distinct subsets $A,B$ that meet the condition, then $A=S$ and $B=S$, but since $A,B$ are distinct, $Aneq B$. But since $A=S=B$ implies $A=B$, we have a contradiction.

Let $n$ be the cardinality of $S$. Assume $S$ is finite. We then know that $scr{P}$$(S)$, its power set, has cardinality $2^n$ as a result. Obviously, then, the power set has greater cardinality than the set itself (this also holds for infinite sets and is known as Cantor's theorem).

So, $2^n > n$. We want the number of subsets to be $1$. Thus, take $n=0$. Then $2^0 = 1 > 0$. Since $n$ is the cardinality of $S$, it follows that $S$ has cardinality $0$; i.e. $S$ has no elements, and $S = emptyset$.

(This can of course be verified to meet the condition as the only subset of the empty set is itself.)

We can also make the consideration $S$ is not finite since I assume as much earlier. It is easy to show that no such $S$ meets the conditions: since it's infinite, it definitely has more than three elements; thus, we can fix a particular triple $x,y,zin S$ and have subsets ${x}neq {y,z}$ (and both are obviously not $S$ as well).

share|cite|improve this answer

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

$endgroup$

• 
  
  
  

  
  15
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, but this is needlessly convoluted. Let $A$ be a set. Then $emptyset subseteq A$. If $A$ has op's property, then $emptyset = A$.
  $endgroup$
  – Martin Bidlingmaier
  15 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to present a slightly more formal argument to what the others posted - they're right, under the typical axioms of ZFC, the only satisfying set is the empty set. But I want to validate why.

---

Let $S$ be some set, with subsets $A,B$.

We want all of the subsets of $S$ to be equal to $S$. Then $A=B=S$ if we apply these restrictions to $S$. As a result, $S$ has only one subset - itself. Why? Consider otherwise. If it has two distinct subsets $A,B$ that meet the condition, then $A=S$ and $B=S$, but since $A,B$ are distinct, $Aneq B$. But since $A=S=B$ implies $A=B$, we have a contradiction.

Let $n$ be the cardinality of $S$. Assume $S$ is finite. We then know that $scr{P}$$(S)$, its power set, has cardinality $2^n$ as a result. Obviously, then, the power set has greater cardinality than the set itself (this also holds for infinite sets and is known as Cantor's theorem).

So, $2^n > n$. We want the number of subsets to be $1$. Thus, take $n=0$. Then $2^0 = 1 > 0$. Since $n$ is the cardinality of $S$, it follows that $S$ has cardinality $0$; i.e. $S$ has no elements, and $S = emptyset$.

(This can of course be verified to meet the condition as the only subset of the empty set is itself.)

We can also make the consideration $S$ is not finite since I assume as much earlier. It is easy to show that no such $S$ meets the conditions: since it's infinite, it definitely has more than three elements; thus, we can fix a particular triple $x,y,zin S$ and have subsets ${x}neq {y,z}$ (and both are obviously not $S$ as well).

share|cite|improve this answer

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

$endgroup$

share|cite|improve this answer

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

answered yesterday

Eevee TrainerEevee Trainer

7,38821338

answered yesterday

Eevee TrainerEevee Trainer

7,38821338