Set Theory $Msetminusemptyset$ possible? [closed]
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M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?
elementary-set-theory
edited Jan 10 at 13:49
Aniruddh Agarwal
1218
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closed as off-topic by mrtaurho, Adrian Keister, Holo, Namaste, Lord_Farin Jan 10 at 18:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?
elementary-set-theory
edited Jan 10 at 13:49
Aniruddh Agarwal
1218
$endgroup$
closed as off-topic by mrtaurho, Adrian Keister, Holo, Namaste, 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, Namaste, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
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It is simply $$M setminus emptyset =M$$
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– Crostul
Jan 10 at 12:01
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thanks for the answer but why is it exactly the same ? i know it is a bit tautological
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– user627643
Jan 10 at 12:02
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Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
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– Git Gud
Jan 10 at 12:03
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Did you mean $Mbackslash{emptyset}$?
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– J.G.
Jan 10 at 12:20
add a comment |
$begingroup$
M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?
elementary-set-theory
edited Jan 10 at 13:49
Aniruddh Agarwal
1218
$endgroup$
M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?
elementary-set-theory
elementary-set-theory
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
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
closed as off-topic by mrtaurho, Adrian Keister, Holo, Namaste, 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, Namaste, 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, Namaste, 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, Namaste, 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
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We have $M setminus emptyset =M$.
Proof:
If $x in M setminus emptyset$, then $x in M$
If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$
answered Jan 10 at 12:13
FredFred
48.6k11849
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add a comment |
$begingroup$
We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.
Hope it is helpful
answered Jan 10 at 12:20
MartundMartund
1,945213
$endgroup$
add a comment |
$begingroup$
$X setminus Y$ 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 $M setminus emptyset = M$.
answered Jan 10 at 12:16
BermudesBermudes
309114
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $M setminus emptyset =M$.
Proof:
If $x in M setminus emptyset$, then $x in M$
If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$
answered Jan 10 at 12:13
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have $M setminus emptyset =M$.
Proof:
If $x in M setminus emptyset$, then $x in M$
If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$
answered Jan 10 at 12:13
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have $M setminus emptyset =M$.
Proof:
If $x in M setminus emptyset$, then $x in M$
If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$
answered Jan 10 at 12:13
FredFred
48.6k11849
$endgroup$
We have $M setminus emptyset =M$.
Proof:
If $x in M setminus emptyset$, then $x in M$
If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$
answered Jan 10 at 12:13
FredFred
48.6k11849
answered Jan 10 at 12:13
FredFred
48.6k11849
answered Jan 10 at 12:13
FredFred
48.6k11849
answered Jan 10 at 12:13
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
$begingroup$
We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.
Hope it is helpful
answered Jan 10 at 12:20
MartundMartund
1,945213
$endgroup$
add a comment |
$begingroup$
We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.
Hope it is helpful
answered Jan 10 at 12:20
MartundMartund
1,945213
$endgroup$
add a comment |
$begingroup$
We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.
Hope it is helpful
answered Jan 10 at 12:20
MartundMartund
1,945213
$endgroup$
We know that,
$$AB=Acap B^c$$
$$Longrightarrow Msetminusemptyset=Mcap U=M$$
where $U$ is the universal set.
Hope it is helpful
answered Jan 10 at 12:20
MartundMartund
1,945213
answered Jan 10 at 12:20
MartundMartund
1,945213
answered Jan 10 at 12:20
MartundMartund
1,945213
answered Jan 10 at 12:20
MartundMartund
1,945213
1,945213
add a comment |
add a comment |
$begingroup$
$X setminus Y$ 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 $M setminus emptyset = M$.
answered Jan 10 at 12:16
BermudesBermudes
309114
$endgroup$
add a comment |
$begingroup$
$X setminus Y$ 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 $M setminus emptyset = M$.
answered Jan 10 at 12:16
BermudesBermudes
309114
$endgroup$
add a comment |
$begingroup$
$X setminus Y$ 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 $M setminus emptyset = M$.
answered Jan 10 at 12:16
BermudesBermudes
309114
$endgroup$
$X setminus Y$ 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 $M setminus emptyset = M$.
answered Jan 10 at 12:16
BermudesBermudes
309114
answered Jan 10 at 12:16
BermudesBermudes
309114
answered Jan 10 at 12:16
BermudesBermudes
309114
answered Jan 10 at 12:16
BermudesBermudes
309114
309114
add a comment |
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