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Why is a set not a partition of itself?

Set with relative complement forms partitioncompute the number of subset of 1,2,3,4Name for a collection of subsets of a set $E$ such that every element of $E$ is in some member?Does $Bbb Z_0$,$Bbb Z_1$, $Bbb Z_2 ,cdots$, $Bbb Z_m-1$ form a partition of $Bbb Z$?Partition generated $sigma$-algebraDefinition of Partition in Analysis vs TopologyProof there's a partition of set where for each $i$, $A_i cap B_i neq emptyset$Partition an infinite set into countable setsShowing a Set is a Partition of AnotherEquivalent definition of partition in set theory

4

$begingroup$

Take the set 1,2,3,4, why is 1,2,3,4 not a partition of this, which condition does it not meet?

By my understanding, a partition of a finite set $S$ is any set $ S_1,...S_n $ of n subsets of $S$, which satisfy,

1. 
   $S_i ne emptyset$ for all $1 leq i leq n$,


2. 
   $S_i cap S_j = emptyset$ for all $1 leq i,j leq n$, $i neq j$
   


3. $S_1 cup cdots cup S_n = S$

Seeing as $S$ is a subset of $S$, which part of the definition breaks down here?

Note I think $1,2,3,4$ is a partition of $1,2,3,4$... could this be explained to?

combinatorics elementary-set-theory

share|cite|improve this question

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A partition of $S$ is a set of subsets of $S$ (with these three conditions.) Every element of the partition is a subset of $S$. Since $1 in1,2,3,4$ and $1$ is not a subset of $S$, $1,2,3,4$ is not a partition of $S$.
  $endgroup$
  – Jair Taylor
  May 12 at 17:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  (Your conditions were incorrect - I have edited them.)
  $endgroup$
  – Jair Taylor
  May 12 at 17:57
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Take the set 1,2,3,4, why is 1,2,3,4 not a partition of this, which condition does it not meet?

By my understanding, a partition of a finite set $S$ is any set $ S_1,...S_n $ of n subsets of $S$, which satisfy,

1. 
   $S_i ne emptyset$ for all $1 leq i leq n$,


2. 
   $S_i cap S_j = emptyset$ for all $1 leq i,j leq n$, $i neq j$
   


3. $S_1 cup cdots cup S_n = S$

Seeing as $S$ is a subset of $S$, which part of the definition breaks down here?

Note I think $1,2,3,4$ is a partition of $1,2,3,4$... could this be explained to?

combinatorics elementary-set-theory

share|cite|improve this question

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A partition of $S$ is a set of subsets of $S$ (with these three conditions.) Every element of the partition is a subset of $S$. Since $1 in1,2,3,4$ and $1$ is not a subset of $S$, $1,2,3,4$ is not a partition of $S$.
  $endgroup$
  – Jair Taylor
  May 12 at 17:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  (Your conditions were incorrect - I have edited them.)
  $endgroup$
  – Jair Taylor
  May 12 at 17:57
  
  

  
  
  
  

add a comment | 

$begingroup$

Take the set 1,2,3,4, why is 1,2,3,4 not a partition of this, which condition does it not meet?

By my understanding, a partition of a finite set $S$ is any set $ S_1,...S_n $ of n subsets of $S$, which satisfy,

1. 
   $S_i ne emptyset$ for all $1 leq i leq n$,


2. 
   $S_i cap S_j = emptyset$ for all $1 leq i,j leq n$, $i neq j$
   


3. $S_1 cup cdots cup S_n = S$

Seeing as $S$ is a subset of $S$, which part of the definition breaks down here?

Note I think $1,2,3,4$ is a partition of $1,2,3,4$... could this be explained to?

combinatorics elementary-set-theory

share|cite|improve this question

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

$endgroup$

share|cite|improve this question

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

share|cite|improve this question

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

edited May 12 at 17:56

Jair Taylor

9,30932244

edited May 12 at 17:56

Jair Taylor

9,30932244

asked May 12 at 17:06

GooJGooJ

235

asked May 12 at 17:06

GooJGooJ

235

3 Answers
3

active

oldest

votes

12

$begingroup$

A partition of a set $A$ is a subset of its power set.

$1,2,3,4$ is, but $1,2,3,4$ is not, a partition of $1,2,3,4$.

$1,2,3,4$ is an element, not a subset of the power set of $1,2,3,4$.

share|cite|improve this answer

answered May 12 at 17:08

CY AriesCY Aries

19k11844

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Another instructive reply is that $1,2,3,4$ is a partition.
  $endgroup$
  – Lee Mosher
  May 12 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which part of the definition above breaks down here?*
  $endgroup$
  – GooJ
  May 12 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $S$ is a subset of $S$ and hence it is an element of the power set of $S$. As mentioned, a partition of $S$ should be a subset (not an element) of the power set of $S$.
  $endgroup$
  – CY Aries
  May 12 at 17:13
  

  
  
  
  

add a comment | 

7

$begingroup$

a partition of a finite set $S$ is any set $S_1, dots, S_n$ of $n$ subsets of $S$...

You answered your own question. Which element of $1,2,3,4$ is a subset of $1,2,3,4$?

share|cite|improve this answer

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With the von Neumann definition of naturals and assuming we started at $1$, it turns out all the elements would be subsets... (The joys of the non-abstractness of set theory.)
  $endgroup$
  – Derek Elkins
  May 12 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins Interesting..
  $endgroup$
  – 雨が好きな人
  May 12 at 23:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins: If you're going to play that game, you should not go around calling them $1, 2, 3, 4$ in contexts where they are intended to be sets. Someone else might be using a different construction of the naturals. Also, starting the von Neumann construction at 1 instead of 0 is in very poor taste. You want the von Neumann naturals to be the fixed points of the cardinality operator, but that construction leaves them offset by one.
  $endgroup$
  – Kevin
  May 13 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin You won't get an argument from me. I prefer foundations where this kind of thing can't happen in the first place. And I, of course, agree with starting from $0$; I'm a programmer. I did see someone recently define von Neumann naturals starting at $1$, and it was a bit nauseating.
  $endgroup$
  – Derek Elkins
  May 13 at 4:19
  

  
  
  
  

add a comment | 

0

$begingroup$

The following remarks do not aim at proving as a general law that no set is a partition of itself, but at showing via couterexamples that it is not the case that any set is a partition of itself,

(1) A partition of a set S is, by definition, a family of sets.

So, if a set S is not itself a family of sets, it cannot be a partition of any set; and consequently, it cannot be a partition of itself.

(2) Lets consider a given set S that is a family of sets, say

S = 1, 2 , 2,3

Now suppose that S is a partition of itself.

That would mean that : Intersection(S) is empty

( for by definition, the intersection of the elements of a partition is empty).

But that is not true, for here we have

Intersection (S) = 1,2 Inter 2,3 = 2

(3) If S = 1, 2 , 2,3 were a partition of itself, then Union (S) would be equal to S ( this is also a necessary condition to be a partition).

But Union (S) = 1, 2 Union 2,3 = 1,2,3

and this is not equal to S.

share|cite|improve this answer

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

$endgroup$

add a comment | 

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12

$begingroup$

A partition of a set $A$ is a subset of its power set.

$1,2,3,4$ is, but $1,2,3,4$ is not, a partition of $1,2,3,4$.

$1,2,3,4$ is an element, not a subset of the power set of $1,2,3,4$.

share|cite|improve this answer

answered May 12 at 17:08

CY AriesCY Aries

19k11844

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Another instructive reply is that $1,2,3,4$ is a partition.
  $endgroup$
  – Lee Mosher
  May 12 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which part of the definition above breaks down here?*
  $endgroup$
  – GooJ
  May 12 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $S$ is a subset of $S$ and hence it is an element of the power set of $S$. As mentioned, a partition of $S$ should be a subset (not an element) of the power set of $S$.
  $endgroup$
  – CY Aries
  May 12 at 17:13
  

  
  
  
  

add a comment | 

12

$begingroup$

A partition of a set $A$ is a subset of its power set.

$1,2,3,4$ is, but $1,2,3,4$ is not, a partition of $1,2,3,4$.

$1,2,3,4$ is an element, not a subset of the power set of $1,2,3,4$.

share|cite|improve this answer

answered May 12 at 17:08

CY AriesCY Aries

19k11844

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Another instructive reply is that $1,2,3,4$ is a partition.
  $endgroup$
  – Lee Mosher
  May 12 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which part of the definition above breaks down here?*
  $endgroup$
  – GooJ
  May 12 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $S$ is a subset of $S$ and hence it is an element of the power set of $S$. As mentioned, a partition of $S$ should be a subset (not an element) of the power set of $S$.
  $endgroup$
  – CY Aries
  May 12 at 17:13
  

  
  
  
  

add a comment | 

$begingroup$

A partition of a set $A$ is a subset of its power set.

$1,2,3,4$ is, but $1,2,3,4$ is not, a partition of $1,2,3,4$.

$1,2,3,4$ is an element, not a subset of the power set of $1,2,3,4$.

share|cite|improve this answer

answered May 12 at 17:08

CY AriesCY Aries

19k11844

$endgroup$

share|cite|improve this answer

answered May 12 at 17:08

CY AriesCY Aries

19k11844

answered May 12 at 17:08

CY AriesCY Aries

19k11844

answered May 12 at 17:08

CY AriesCY Aries

19k11844

7

$begingroup$

a partition of a finite set $S$ is any set $S_1, dots, S_n$ of $n$ subsets of $S$...

You answered your own question. Which element of $1,2,3,4$ is a subset of $1,2,3,4$?

share|cite|improve this answer

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With the von Neumann definition of naturals and assuming we started at $1$, it turns out all the elements would be subsets... (The joys of the non-abstractness of set theory.)
  $endgroup$
  – Derek Elkins
  May 12 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins Interesting..
  $endgroup$
  – 雨が好きな人
  May 12 at 23:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins: If you're going to play that game, you should not go around calling them $1, 2, 3, 4$ in contexts where they are intended to be sets. Someone else might be using a different construction of the naturals. Also, starting the von Neumann construction at 1 instead of 0 is in very poor taste. You want the von Neumann naturals to be the fixed points of the cardinality operator, but that construction leaves them offset by one.
  $endgroup$
  – Kevin
  May 13 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin You won't get an argument from me. I prefer foundations where this kind of thing can't happen in the first place. And I, of course, agree with starting from $0$; I'm a programmer. I did see someone recently define von Neumann naturals starting at $1$, and it was a bit nauseating.
  $endgroup$
  – Derek Elkins
  May 13 at 4:19
  

  
  
  
  

add a comment | 

7

$begingroup$

a partition of a finite set $S$ is any set $S_1, dots, S_n$ of $n$ subsets of $S$...

You answered your own question. Which element of $1,2,3,4$ is a subset of $1,2,3,4$?

share|cite|improve this answer

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With the von Neumann definition of naturals and assuming we started at $1$, it turns out all the elements would be subsets... (The joys of the non-abstractness of set theory.)
  $endgroup$
  – Derek Elkins
  May 12 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins Interesting..
  $endgroup$
  – 雨が好きな人
  May 12 at 23:32
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @DerekElkins: If you're going to play that game, you should not go around calling them $1, 2, 3, 4$ in contexts where they are intended to be sets. Someone else might be using a different construction of the naturals. Also, starting the von Neumann construction at 1 instead of 0 is in very poor taste. You want the von Neumann naturals to be the fixed points of the cardinality operator, but that construction leaves them offset by one.
  $endgroup$
  – Kevin
  May 13 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin You won't get an argument from me. I prefer foundations where this kind of thing can't happen in the first place. And I, of course, agree with starting from $0$; I'm a programmer. I did see someone recently define von Neumann naturals starting at $1$, and it was a bit nauseating.
  $endgroup$
  – Derek Elkins
  May 13 at 4:19
  

  
  
  
  

add a comment | 

$begingroup$

a partition of a finite set $S$ is any set $S_1, dots, S_n$ of $n$ subsets of $S$...

You answered your own question. Which element of $1,2,3,4$ is a subset of $1,2,3,4$?

share|cite|improve this answer

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

$endgroup$

share|cite|improve this answer

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

answered May 12 at 17:13

雨が好きな人雨が好きな人

2,127417

0

$begingroup$

The following remarks do not aim at proving as a general law that no set is a partition of itself, but at showing via couterexamples that it is not the case that any set is a partition of itself,

(1) A partition of a set S is, by definition, a family of sets.

So, if a set S is not itself a family of sets, it cannot be a partition of any set; and consequently, it cannot be a partition of itself.

(2) Lets consider a given set S that is a family of sets, say

S = 1, 2 , 2,3

Now suppose that S is a partition of itself.

That would mean that : Intersection(S) is empty

( for by definition, the intersection of the elements of a partition is empty).

But that is not true, for here we have

Intersection (S) = 1,2 Inter 2,3 = 2

(3) If S = 1, 2 , 2,3 were a partition of itself, then Union (S) would be equal to S ( this is also a necessary condition to be a partition).

But Union (S) = 1, 2 Union 2,3 = 1,2,3

and this is not equal to S.

share|cite|improve this answer

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

$endgroup$

add a comment | 

0

$begingroup$

The following remarks do not aim at proving as a general law that no set is a partition of itself, but at showing via couterexamples that it is not the case that any set is a partition of itself,

(1) A partition of a set S is, by definition, a family of sets.

So, if a set S is not itself a family of sets, it cannot be a partition of any set; and consequently, it cannot be a partition of itself.

(2) Lets consider a given set S that is a family of sets, say

S = 1, 2 , 2,3

Now suppose that S is a partition of itself.

That would mean that : Intersection(S) is empty

( for by definition, the intersection of the elements of a partition is empty).

But that is not true, for here we have

Intersection (S) = 1,2 Inter 2,3 = 2

(3) If S = 1, 2 , 2,3 were a partition of itself, then Union (S) would be equal to S ( this is also a necessary condition to be a partition).

But Union (S) = 1, 2 Union 2,3 = 1,2,3

and this is not equal to S.

share|cite|improve this answer

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

$endgroup$

add a comment | 

$begingroup$

The following remarks do not aim at proving as a general law that no set is a partition of itself, but at showing via couterexamples that it is not the case that any set is a partition of itself,

(1) A partition of a set S is, by definition, a family of sets.

So, if a set S is not itself a family of sets, it cannot be a partition of any set; and consequently, it cannot be a partition of itself.

(2) Lets consider a given set S that is a family of sets, say

S = 1, 2 , 2,3

Now suppose that S is a partition of itself.

That would mean that : Intersection(S) is empty

( for by definition, the intersection of the elements of a partition is empty).

But that is not true, for here we have

Intersection (S) = 1,2 Inter 2,3 = 2

(3) If S = 1, 2 , 2,3 were a partition of itself, then Union (S) would be equal to S ( this is also a necessary condition to be a partition).

But Union (S) = 1, 2 Union 2,3 = 1,2,3

and this is not equal to S.

share|cite|improve this answer

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

$endgroup$

share|cite|improve this answer

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118

answered May 13 at 20:52

Eleonore Saint JamesEleonore Saint James

892118