Mixing
Number of size 1 partitions of the empty set
$begingroup$
Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.
We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number
of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = bigcup X$. Then $X$ is called a partition of
$Y$ if either (i) $X = Y = varnothing$; or (ii) $X neq varnothing land varnothing notin X$."
As far as I can tell the only set $X$ with $bigcup X = varnothing$ and $|X| = 1$ is $X = {varnothing}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $varnothing$, and $S(0,1)$ should be $0$, not $1$.
Am I missing something here?
elementary-set-theory set-partition
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
$endgroup$
add a comment |
$begingroup$
Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.
We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number
of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = bigcup X$. Then $X$ is called a partition of
$Y$ if either (i) $X = Y = varnothing$; or (ii) $X neq varnothing land varnothing notin X$."
As far as I can tell the only set $X$ with $bigcup X = varnothing$ and $|X| = 1$ is $X = {varnothing}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $varnothing$, and $S(0,1)$ should be $0$, not $1$.
Am I missing something here?
elementary-set-theory set-partition
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
$endgroup$
$begingroup$
You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22
add a comment |
$begingroup$
Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.
We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number
of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = bigcup X$. Then $X$ is called a partition of
$Y$ if either (i) $X = Y = varnothing$; or (ii) $X neq varnothing land varnothing notin X$."
As far as I can tell the only set $X$ with $bigcup X = varnothing$ and $|X| = 1$ is $X = {varnothing}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $varnothing$, and $S(0,1)$ should be $0$, not $1$.
Am I missing something here?
elementary-set-theory set-partition
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
$endgroup$
Disclaimer: This is a homework problem, but I'm just asking for clarification, not a solution.
We're asked to prove $S(0,1) = 1$, where $S(n,k)$ is "the number
of different partitions of [a set of size $n$] into $k$ mutually disjoint subsets", where a partition is defined like so: "Let $X$ be a collection of pairwise disjoint sets and let $Y = bigcup X$. Then $X$ is called a partition of
$Y$ if either (i) $X = Y = varnothing$; or (ii) $X neq varnothing land varnothing notin X$."
As far as I can tell the only set $X$ with $bigcup X = varnothing$ and $|X| = 1$ is $X = {varnothing}$, but this does not satisfy either (i) or (ii) of the defintion of a partition. So it seems to me that there are no size 1 partitions of $varnothing$, and $S(0,1)$ should be $0$, not $1$.
Am I missing something here?
elementary-set-theory set-partition
elementary-set-theory set-partition
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
edited Oct 29 '17 at 6:40
Peter Taylor
8,99212342
8,99212342
asked Sep 2 '13 at 20:15
AlecAlec
5261417
asked Sep 2 '13 at 20:15
AlecAlec
5261417
asked Sep 2 '13 at 20:15
AlecAlec
5261417
5261417
$begingroup$
You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22
add a comment |
$begingroup$
You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22
$begingroup$
You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22
$begingroup$
You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You are right according to that definition. By the way, it can be simplified to "... is called a partition if $emptysetnotin X$."
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
$endgroup$
add a comment |
$begingroup$
Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).
Condition (ii) implies that all parts in a partition are non-empty.
This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
I think the author intended the "special case" clause (i) to be "$X={emptyset}$ and $Y=emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
I believe you're mistaken when you say that the set $X={varnothing}$ satisfies $bigcup X = varnothing$. It actually satisfies $bigcup X = {varnothing}$.
The only $X$ satisfying $bigcup X = varnothing$ is $X=varnothing$, the empty set itself.
Does that help?
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
$endgroup$
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
|
show 1 more comment
$begingroup$
Empty relation on empty set is a equivalence relation .And no.of equivalence relations is equal to no.of partitions .
There is only on equivalence relation on empty set than empty set have one partition also
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right according to that definition. By the way, it can be simplified to "... is called a partition if $emptysetnotin X$."
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
$endgroup$
add a comment |
$begingroup$
You are right according to that definition. By the way, it can be simplified to "... is called a partition if $emptysetnotin X$."
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
$endgroup$
add a comment |
$begingroup$
You are right according to that definition. By the way, it can be simplified to "... is called a partition if $emptysetnotin X$."
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
$endgroup$
You are right according to that definition. By the way, it can be simplified to "... is called a partition if $emptysetnotin X$."
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
answered Sep 2 '13 at 20:19
Hagen von EitzenHagen von Eitzen
280k23271503
280k23271503
add a comment |
add a comment |
$begingroup$
Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).
Condition (ii) implies that all parts in a partition are non-empty.
This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).
Condition (ii) implies that all parts in a partition are non-empty.
This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).
Condition (ii) implies that all parts in a partition are non-empty.
This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
Condition (i) describes the partition of the empty set into $0$ non-empty parts (the "non-empty" property being vacuously true).
Condition (ii) implies that all parts in a partition are non-empty.
This is consistent with the definition of the Stirling Numbers of the second kind. The number $S(0,1)=0$ since there are no partitions of the empty set into exactly 1 non-empty part.
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 2 '13 at 20:58
Rebecca J. StonesRebecca J. Stones
21k22781
21k22781
add a comment |
add a comment |
$begingroup$
I think the author intended the "special case" clause (i) to be "$X={emptyset}$ and $Y=emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
I think the author intended the "special case" clause (i) to be "$X={emptyset}$ and $Y=emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
I think the author intended the "special case" clause (i) to be "$X={emptyset}$ and $Y=emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
I think the author intended the "special case" clause (i) to be "$X={emptyset}$ and $Y=emptyset$." That would make $S(0,1)=1$ and it would not be subject to Hagen von Eitzen's simplification of what was actually written.
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
answered Sep 2 '13 at 21:06
Andreas BlassAndreas Blass
49.8k451108
49.8k451108
add a comment |
add a comment |
$begingroup$
I believe you're mistaken when you say that the set $X={varnothing}$ satisfies $bigcup X = varnothing$. It actually satisfies $bigcup X = {varnothing}$.
The only $X$ satisfying $bigcup X = varnothing$ is $X=varnothing$, the empty set itself.
Does that help?
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
$endgroup$
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
|
show 1 more comment
$begingroup$
I believe you're mistaken when you say that the set $X={varnothing}$ satisfies $bigcup X = varnothing$. It actually satisfies $bigcup X = {varnothing}$.
The only $X$ satisfying $bigcup X = varnothing$ is $X=varnothing$, the empty set itself.
Does that help?
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
$endgroup$
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
|
show 1 more comment
$begingroup$
I believe you're mistaken when you say that the set $X={varnothing}$ satisfies $bigcup X = varnothing$. It actually satisfies $bigcup X = {varnothing}$.
The only $X$ satisfying $bigcup X = varnothing$ is $X=varnothing$, the empty set itself.
Does that help?
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
$endgroup$
I believe you're mistaken when you say that the set $X={varnothing}$ satisfies $bigcup X = varnothing$. It actually satisfies $bigcup X = {varnothing}$.
The only $X$ satisfying $bigcup X = varnothing$ is $X=varnothing$, the empty set itself.
Does that help?
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
answered Sep 2 '13 at 20:20
G Tony JacobsG Tony Jacobs
25.8k43585
25.8k43585
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
|
show 1 more comment
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
2
2
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
$bigcup X={,zmid exists yin Xcolon zin y,}$, so $bigcup{emptyset}=emptyset$.
$endgroup$
– Hagen von Eitzen
Sep 2 '13 at 20:21
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
$begingroup$
It looks like you're saying that the union over a non-empty set is empty. That seems wrong to me. What am I missing?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 20:23
3
3
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
$begingroup$
You’re not paying careful enough attention to the definition of $bigcup X$. In order for $x$ to be an element of $bigcup{varnothing}$, there must be an element $u$ of ${varnothing}$ such that $xin u$. But the only element of ${varnothing}$ is $varnothing$, and no $x$ is in $varnothing$, so $bigcup{varnothing}$ must be empty.
$endgroup$
– Brian M. Scott
Sep 2 '13 at 20:26
2
2
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
It may be helpful to think of it in the following way (even though it is more complicated than necessary.) For any two sets $A$ and $B$ we have $bigcup{A,B} = A cup B$. Now let $A = B = emptyset$.
$endgroup$
– Trevor Wilson
Sep 2 '13 at 21:10
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
$begingroup$
I think I understand. Does that mean it's correct to say that, for any set $X$, we have $bigcup {X} = X$?
$endgroup$
– G Tony Jacobs
Sep 2 '13 at 21:14
|
show 1 more comment
$begingroup$
Empty relation on empty set is a equivalence relation .And no.of equivalence relations is equal to no.of partitions .
There is only on equivalence relation on empty set than empty set have one partition also
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
$endgroup$
add a comment |
$begingroup$
Empty relation on empty set is a equivalence relation .And no.of equivalence relations is equal to no.of partitions .
There is only on equivalence relation on empty set than empty set have one partition also
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
$endgroup$
add a comment |
$begingroup$
Empty relation on empty set is a equivalence relation .And no.of equivalence relations is equal to no.of partitions .
There is only on equivalence relation on empty set than empty set have one partition also
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
$endgroup$
Empty relation on empty set is a equivalence relation .And no.of equivalence relations is equal to no.of partitions .
There is only on equivalence relation on empty set than empty set have one partition also
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
answered Jan 14 at 4:25
Jagdish palariyaJagdish palariya
1
1
add a comment |
add a comment |
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You are correct. There are no size-1 partitions of the empty set.
$endgroup$
– MJD
Sep 2 '13 at 20:22