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What is the size of a set of sets of the empty set , , ?

Set Operations with Empty setsOn the size of a non-empty family of non-empty sets such that every set in the family has a proper subset also in the familyIs $varnothing $ an empty set?Is the empty set a member of any collection of sets?Is the empty set countable?What is an Empty set?Subsets of the empty setWhat's the difference between a null set and an empty set?Do all non-empty sets of natural numbers contain a smallest number?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

$begingroup$

What is the size of a set of sets of the empty set , , ?

I am not sure if the empty set, in this case, can be considered as a set and make the size 3 or if it is 2 or 0.

thanks

elementary-set-theory

share|cite|improve this question

edited 7 hours ago

Andrés E. Caicedo

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TakobellTakobell

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  one $)$ is meant to be $}$ I believe. There are three elements in the set.
  $endgroup$
  – Alvin Lepik
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not an empty set.
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As you are probably aware, $$ represents the empty set. $$ represents the set containing the empty set, and so on. Your set has in it the empty set, the set containing the empty set, and the set containing the set that contains the empty set. How many distinct elements do you have?
  $endgroup$
  – Cameron Williams
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The empty set is a set. It just has nothing in it. This set has three things in it. On thing is the empty set, the second thing is a set containing the empty set. The third thing is a set containing a set containing the empty set. That's three things. The things may be nothing more than an empty bag, a bag with an empty bag in it, and a bag containing a bag with an empty bag; so the may have nothing of substance, but they are still things.
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  

add a comment
 | 

3

$begingroup$

What is the size of a set of sets of the empty set , , ?

I am not sure if the empty set, in this case, can be considered as a set and make the size 3 or if it is 2 or 0.

thanks

elementary-set-theory

share|cite|improve this question

edited 7 hours ago

Andrés E. Caicedo

67.1k88 gold badges170170 silver badges263263 bronze badges

asked 9 hours ago

TakobellTakobell

1633 bronze badges

New contributor

Takobell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  one $)$ is meant to be $}$ I believe. There are three elements in the set.
  $endgroup$
  – Alvin Lepik
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not an empty set.
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As you are probably aware, $$ represents the empty set. $$ represents the set containing the empty set, and so on. Your set has in it the empty set, the set containing the empty set, and the set containing the set that contains the empty set. How many distinct elements do you have?
  $endgroup$
  – Cameron Williams
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The empty set is a set. It just has nothing in it. This set has three things in it. On thing is the empty set, the second thing is a set containing the empty set. The third thing is a set containing a set containing the empty set. That's three things. The things may be nothing more than an empty bag, a bag with an empty bag in it, and a bag containing a bag with an empty bag; so the may have nothing of substance, but they are still things.
  $endgroup$
  – fleablood
  7 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

What is the size of a set of sets of the empty set , , ?

I am not sure if the empty set, in this case, can be considered as a set and make the size 3 or if it is 2 or 0.

thanks

elementary-set-theory

share|cite|improve this question

edited 7 hours ago

Andrés E. Caicedo

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asked 9 hours ago

TakobellTakobell

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$endgroup$

share|cite|improve this question

edited 7 hours ago

Andrés E. Caicedo

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asked 9 hours ago

TakobellTakobell

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share|cite|improve this question

edited 7 hours ago

Andrés E. Caicedo

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asked 9 hours ago

TakobellTakobell

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edited 7 hours ago

Andrés E. Caicedo

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edited 7 hours ago

Andrés E. Caicedo

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TakobellTakobell

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asked 9 hours ago

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3 Answers
3

active

oldest

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5

$begingroup$

Your set can be written as $a,b,c$ where $a$ happens to be the emptyset, $b$ happens to be the set containing the empty set, and $c$ happens to be the set containing the set containing the emptyset.

You should not have any trouble seeing that $a,b,c$ is of size three so long as they are all distinct. The only possible source of confusion in this is in recognizing that each of $emptyset, emptyset$ and $emptyset$ are not only valid possible elements of a set, but are distinct and different than one another. Indeed, they are all valid possible elements of sets and are all different than one another. Remember that the "depth" of each is relevant and is part of what makes these distinct.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TarekHich yes, well, I did include pointing out that $aneq b$, $aneq c$ and $bneq c$ that $a,b,c$ are distinct, and any set $a,b,c$ where $a,b,c$ are distinct is of size three.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the event that $x$ and $y$ are distinct and neither are equal to the emptyset, then $emptyset,x,y$ is still of size three. The only reason why $emptyset,a,b$ was of size two in your example in the comment was because $emptyset$ and $a$ were in reality the same element.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  TarekHich: "but a,b can be writen as ∅, a,b and its size is 2" and JMoravitz: "yes, well". Me: Um, no..... $a,b$ can not be written as $emptyset, a, b$. The empty set is a thing and it is not in $a,b$ and it is in $emptyset, a,b$. Maybe you meant say we can write $a,b$ as $ , a, b$ where "$ $" is nothing. But $emptyset = $ is not nothing. It is something. "$ $" isn't anything at all. $emptyset$ doesn't HAVE anything at all but it IS something.
  $endgroup$
  – fleablood
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood perhaps you missed where I said $a$ is the empty set. Are you implying x,y can't be written in the case that x=y or is somehow different than x? Remember we are dealing with sets and not multisets
  $endgroup$
  – JMoravitz
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

As you are only interested in the number of elements in the set and not in the elements themself, you basically just have to count the commas, so your set has three elements, so three is its size.

share|cite|improve this answer

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have mixed feelings about this answer. On the one hand it directly gets that a set has a list of things-- that the content of the things doesn't matter. But on the other hand it begs and avoids the questions. Also consider $apple, orange, banana,banana, peach, cherry$. That has has five commas but only four items. And $a,a$ has a comma indicating two items but it actually has only one.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

3. The empty set is indeed an element of this set.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

17099 bronze badges

$endgroup$

add a comment
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5

$begingroup$

Your set can be written as $a,b,c$ where $a$ happens to be the emptyset, $b$ happens to be the set containing the empty set, and $c$ happens to be the set containing the set containing the emptyset.

You should not have any trouble seeing that $a,b,c$ is of size three so long as they are all distinct. The only possible source of confusion in this is in recognizing that each of $emptyset, emptyset$ and $emptyset$ are not only valid possible elements of a set, but are distinct and different than one another. Indeed, they are all valid possible elements of sets and are all different than one another. Remember that the "depth" of each is relevant and is part of what makes these distinct.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TarekHich yes, well, I did include pointing out that $aneq b$, $aneq c$ and $bneq c$ that $a,b,c$ are distinct, and any set $a,b,c$ where $a,b,c$ are distinct is of size three.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the event that $x$ and $y$ are distinct and neither are equal to the emptyset, then $emptyset,x,y$ is still of size three. The only reason why $emptyset,a,b$ was of size two in your example in the comment was because $emptyset$ and $a$ were in reality the same element.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  TarekHich: "but a,b can be writen as ∅, a,b and its size is 2" and JMoravitz: "yes, well". Me: Um, no..... $a,b$ can not be written as $emptyset, a, b$. The empty set is a thing and it is not in $a,b$ and it is in $emptyset, a,b$. Maybe you meant say we can write $a,b$ as $ , a, b$ where "$ $" is nothing. But $emptyset = $ is not nothing. It is something. "$ $" isn't anything at all. $emptyset$ doesn't HAVE anything at all but it IS something.
  $endgroup$
  – fleablood
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood perhaps you missed where I said $a$ is the empty set. Are you implying x,y can't be written in the case that x=y or is somehow different than x? Remember we are dealing with sets and not multisets
  $endgroup$
  – JMoravitz
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

5

$begingroup$

Your set can be written as $a,b,c$ where $a$ happens to be the emptyset, $b$ happens to be the set containing the empty set, and $c$ happens to be the set containing the set containing the emptyset.

You should not have any trouble seeing that $a,b,c$ is of size three so long as they are all distinct. The only possible source of confusion in this is in recognizing that each of $emptyset, emptyset$ and $emptyset$ are not only valid possible elements of a set, but are distinct and different than one another. Indeed, they are all valid possible elements of sets and are all different than one another. Remember that the "depth" of each is relevant and is part of what makes these distinct.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TarekHich yes, well, I did include pointing out that $aneq b$, $aneq c$ and $bneq c$ that $a,b,c$ are distinct, and any set $a,b,c$ where $a,b,c$ are distinct is of size three.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the event that $x$ and $y$ are distinct and neither are equal to the emptyset, then $emptyset,x,y$ is still of size three. The only reason why $emptyset,a,b$ was of size two in your example in the comment was because $emptyset$ and $a$ were in reality the same element.
  $endgroup$
  – JMoravitz
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  TarekHich: "but a,b can be writen as ∅, a,b and its size is 2" and JMoravitz: "yes, well". Me: Um, no..... $a,b$ can not be written as $emptyset, a, b$. The empty set is a thing and it is not in $a,b$ and it is in $emptyset, a,b$. Maybe you meant say we can write $a,b$ as $ , a, b$ where "$ $" is nothing. But $emptyset = $ is not nothing. It is something. "$ $" isn't anything at all. $emptyset$ doesn't HAVE anything at all but it IS something.
  $endgroup$
  – fleablood
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood perhaps you missed where I said $a$ is the empty set. Are you implying x,y can't be written in the case that x=y or is somehow different than x? Remember we are dealing with sets and not multisets
  $endgroup$
  – JMoravitz
  7 hours ago
  
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Your set can be written as $a,b,c$ where $a$ happens to be the emptyset, $b$ happens to be the set containing the empty set, and $c$ happens to be the set containing the set containing the emptyset.

You should not have any trouble seeing that $a,b,c$ is of size three so long as they are all distinct. The only possible source of confusion in this is in recognizing that each of $emptyset, emptyset$ and $emptyset$ are not only valid possible elements of a set, but are distinct and different than one another. Indeed, they are all valid possible elements of sets and are all different than one another. Remember that the "depth" of each is relevant and is part of what makes these distinct.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

answered 8 hours ago

JMoravitzJMoravitz

54.4k44 gold badges4343 silver badges9393 bronze badges

2

$begingroup$

As you are only interested in the number of elements in the set and not in the elements themself, you basically just have to count the commas, so your set has three elements, so three is its size.

share|cite|improve this answer

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have mixed feelings about this answer. On the one hand it directly gets that a set has a list of things-- that the content of the things doesn't matter. But on the other hand it begs and avoids the questions. Also consider $apple, orange, banana,banana, peach, cherry$. That has has five commas but only four items. And $a,a$ has a comma indicating two items but it actually has only one.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

As you are only interested in the number of elements in the set and not in the elements themself, you basically just have to count the commas, so your set has three elements, so three is its size.

share|cite|improve this answer

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have mixed feelings about this answer. On the one hand it directly gets that a set has a list of things-- that the content of the things doesn't matter. But on the other hand it begs and avoids the questions. Also consider $apple, orange, banana,banana, peach, cherry$. That has has five commas but only four items. And $a,a$ has a comma indicating two items but it actually has only one.
  $endgroup$
  – fleablood
  3 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

As you are only interested in the number of elements in the set and not in the elements themself, you basically just have to count the commas, so your set has three elements, so three is its size.

share|cite|improve this answer

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

answered 8 hours ago

MatthiasMatthias

8844 bronze badges

2

$begingroup$

3. The empty set is indeed an element of this set.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

17099 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

3. The empty set is indeed an element of this set.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

17099 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

3. The empty set is indeed an element of this set.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

17099 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

7903766279037662

17099 bronze badges

answered 8 hours ago

7903766279037662

17099 bronze badges

answered 8 hours ago

7903766279037662

17099 bronze badges