Set Theory $Msetminusemptyset$ possible? [closed]

-1

$begingroup$

M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?

elementary-set-theory

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edited Jan 10 at 13:49

Aniruddh Agarwal

1218

asked Jan 10 at 11:59

user627643

$endgroup$

closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39

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." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply $$M setminus emptyset =M$$
  $endgroup$
  – Crostul
  Jan 10 at 12:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer but why is it exactly the same ? i know it is a bit tautological
  $endgroup$
  – user627643
  Jan 10 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
  $endgroup$
  – Git Gud
  Jan 10 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $Mbackslash{emptyset}$?
  $endgroup$
  – J.G.
  Jan 10 at 12:20
  

  
  
  
  

add a comment | 

-1

$begingroup$

M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?

elementary-set-theory

share|cite|improve this question

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

asked Jan 10 at 11:59

user627643

$endgroup$

closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39

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." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply $$M setminus emptyset =M$$
  $endgroup$
  – Crostul
  Jan 10 at 12:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer but why is it exactly the same ? i know it is a bit tautological
  $endgroup$
  – user627643
  Jan 10 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
  $endgroup$
  – Git Gud
  Jan 10 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $Mbackslash{emptyset}$?
  $endgroup$
  – J.G.
  Jan 10 at 12:20
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?

elementary-set-theory

share|cite|improve this question

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

asked Jan 10 at 11:59

user627643

$endgroup$

M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?

elementary-set-theory

elementary-set-theory

share|cite|improve this question

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

asked Jan 10 at 11:59

user627643

share|cite|improve this question

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

asked Jan 10 at 11:59

user627643

share|cite|improve this question

share|cite|improve this question

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

edited Jan 10 at 13:49

Aniruddh Agarwal

1218

1218

asked Jan 10 at 11:59

user627643

asked Jan 10 at 11:59

user627643

asked Jan 10 at 11:59

user627643

closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39

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." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

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

closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39

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." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply $$M setminus emptyset =M$$
  $endgroup$
  – Crostul
  Jan 10 at 12:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer but why is it exactly the same ? i know it is a bit tautological
  $endgroup$
  – user627643
  Jan 10 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
  $endgroup$
  – Git Gud
  Jan 10 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $Mbackslash{emptyset}$?
  $endgroup$
  – J.G.
  Jan 10 at 12:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is simply $$M setminus emptyset =M$$
  $endgroup$
  – Crostul
  Jan 10 at 12:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for the answer but why is it exactly the same ? i know it is a bit tautological
  $endgroup$
  – user627643
  Jan 10 at 12:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
  $endgroup$
  – Git Gud
  Jan 10 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $Mbackslash{emptyset}$?
  $endgroup$
  – J.G.
  Jan 10 at 12:20
  

  
  
  
  

$begingroup$
It is simply $$M setminus emptyset =M$$
$endgroup$
– Crostul
Jan 10 at 12:01

$begingroup$
It is simply $$M setminus emptyset =M$$
$endgroup$
– Crostul
Jan 10 at 12:01

$begingroup$
thanks for the answer but why is it exactly the same ? i know it is a bit tautological
$endgroup$
– user627643
Jan 10 at 12:02

$begingroup$
thanks for the answer but why is it exactly the same ? i know it is a bit tautological
$endgroup$
– user627643
Jan 10 at 12:02

$begingroup$
Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
$endgroup$
– Git Gud
Jan 10 at 12:03

$begingroup$
Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
$endgroup$
– Git Gud
Jan 10 at 12:03

$begingroup$
Did you mean $Mbackslash{emptyset}$?
$endgroup$
– J.G.
Jan 10 at 12:20

$begingroup$
Did you mean $Mbackslash{emptyset}$?
$endgroup$
– J.G.
Jan 10 at 12:20

add a comment | 

3 Answers
3

active

oldest

votes

2

$begingroup$

We have $M setminus emptyset =M$.

Proof:

1. If $x in M setminus emptyset$, then $x in M$




2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

$endgroup$

add a comment | 

1

$begingroup$

We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.

Hope it is helpful

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

$endgroup$

add a comment | 

0

$begingroup$

$X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

$endgroup$

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

We have $M setminus emptyset =M$.

Proof:

1. If $x in M setminus emptyset$, then $x in M$




2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

$endgroup$

add a comment | 

2

$begingroup$

We have $M setminus emptyset =M$.

Proof:

1. If $x in M setminus emptyset$, then $x in M$




2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

$endgroup$

add a comment | 

2

2

2

$begingroup$

We have $M setminus emptyset =M$.

Proof:

1. If $x in M setminus emptyset$, then $x in M$




2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

$endgroup$

We have $M setminus emptyset =M$.

Proof:

1. If $x in M setminus emptyset$, then $x in M$




2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 12:13

FredFred

45.7k1848

answered Jan 10 at 12:13

FredFred

45.7k1848

answered Jan 10 at 12:13

FredFred

45.7k1848

45.7k1848

add a comment | 

add a comment | 

1

$begingroup$

We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.

Hope it is helpful

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

$endgroup$

add a comment | 

1

$begingroup$

We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.

Hope it is helpful

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

$endgroup$

add a comment | 

1

1

1

$begingroup$

We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.

Hope it is helpful

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

$endgroup$

We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.

Hope it is helpful

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 12:20

MartundMartund

1,623213

answered Jan 10 at 12:20

MartundMartund

1,623213

answered Jan 10 at 12:20

MartundMartund

1,623213

1,623213

add a comment | 

add a comment | 

0

$begingroup$

$X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

$endgroup$

add a comment | 

0

$begingroup$

$X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

$endgroup$

add a comment | 

0

0

0

$begingroup$

$X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

$endgroup$

$X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 12:16

BermudesBermudes

17713

answered Jan 10 at 12:16

BermudesBermudes

17713

answered Jan 10 at 12:16

BermudesBermudes

17713

17713

add a comment | 

add a comment | 

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