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

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2

$begingroup$

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 Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  one $)$ is meant to be $}$ I believe. There are three elements in the set.
  $endgroup$
  – Alvin Lepik
  Sep 20 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not an empty set.
  $endgroup$
  – Lord Shark the Unknown
  Sep 20 at 14:22
  

  
  
  
  


• 
  
  
  

  
  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
  Sep 20 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be easier to read if you use ,and ; to add a little space, as $;,,,,;$
  $endgroup$
  – DanielWainfleet
  Sep 21 at 2:24
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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 Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  one $)$ is meant to be $}$ I believe. There are three elements in the set.
  $endgroup$
  – Alvin Lepik
  Sep 20 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not an empty set.
  $endgroup$
  – Lord Shark the Unknown
  Sep 20 at 14:22
  

  
  
  
  


• 
  
  
  

  
  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
  Sep 20 at 14:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would be easier to read if you use ,and ; to add a little space, as $;,,,,;$
  $endgroup$
  – DanielWainfleet
  Sep 21 at 2:24
  

  
  
  
  

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 Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

$endgroup$

share|cite|improve this question

edited Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

share|cite|improve this question

edited Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

edited Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

edited Sep 20 at 15:19

Andrés E. Caicedo

68.1k88 gold badges171171 silver badges268268 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

asked Sep 20 at 14:15

TakobellTakobell

8166 bronze badges

3 Answers
3

active

oldest

votes

8

$begingroup$

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 Sep 20 at 15:25

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  Sep 20 at 15:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:24
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:59
  
  

  
  
  
  

 | 
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 Sep 20 at 14:19

MatthiasMatthias

14277 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
  Sep 20 at 19:33
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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

share|cite|improve this answer

answered Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges

$endgroup$

add a comment
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8

$begingroup$

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 Sep 20 at 15:25

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  Sep 20 at 15:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:24
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:59
  
  

  
  
  
  

 | 
show 4 more comments

8

$begingroup$

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 Sep 20 at 15:25

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but a,b can be writen as ∅, a,b and its size is 2
  $endgroup$
  – Takobell
  Sep 20 at 15:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:24
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Sep 20 at 15:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Sep 20 at 15:59
  
  

  
  
  
  

 | 
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 Sep 20 at 15:25

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 20 at 15:25

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 bronze badges

answered Sep 20 at 14:30

JMoravitzJMoravitz

57.6k44 gold badges4646 silver badges9494 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 Sep 20 at 14:19

MatthiasMatthias

14277 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
  Sep 20 at 19:33
  

  
  
  
  

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 Sep 20 at 14:19

MatthiasMatthias

14277 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
  Sep 20 at 19:33
  

  
  
  
  

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 Sep 20 at 14:19

MatthiasMatthias

14277 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 20 at 14:19

MatthiasMatthias

14277 bronze badges

answered Sep 20 at 14:19

MatthiasMatthias

14277 bronze badges

answered Sep 20 at 14:19

MatthiasMatthias

14277 bronze badges

2

$begingroup$

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

share|cite|improve this answer

answered Sep 20 at 14:19

7903766279037662

1,1691212 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 Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

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

share|cite|improve this answer

answered Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges

answered Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges

answered Sep 20 at 14:19

7903766279037662

1,1691212 bronze badges