Hall's Marriage Theorem for correspondence
$begingroup$
Let $A={A_1,....A_n}$ be a collection of subsets of a finite set $X$. A selection for $A$ is the image of an injective function $f:Ato X$ such that $ f(A_i)in A_i$ for every $A_iin A$.
Hall's marriage theorem shows that , $A$ has a selection if and only if for each subset $Ssubseteq A$,
$$|S|leq |cup_{iin S} A_i|.$$
I wonder if it is possible to generalize this result and obtain a similar condition for a choice correspondence $f:Arightrightarrows X $ that selects for each $A_i$ more than one elements, i.e. $f(A_i)subset A_i$ and $|f(A_i)|=2$ and all selected elements are distinct.
Any ideas or suggestions?
combinatorics discrete-mathematics graph-theory
asked Nov 30 '18 at 4:48
samsam
134
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add a comment |
$begingroup$
Let $A={A_1,....A_n}$ be a collection of subsets of a finite set $X$. A selection for $A$ is the image of an injective function $f:Ato X$ such that $ f(A_i)in A_i$ for every $A_iin A$.
Hall's marriage theorem shows that , $A$ has a selection if and only if for each subset $Ssubseteq A$,
$$|S|leq |cup_{iin S} A_i|.$$
I wonder if it is possible to generalize this result and obtain a similar condition for a choice correspondence $f:Arightrightarrows X $ that selects for each $A_i$ more than one elements, i.e. $f(A_i)subset A_i$ and $|f(A_i)|=2$ and all selected elements are distinct.
Any ideas or suggestions?
combinatorics discrete-mathematics graph-theory
asked Nov 30 '18 at 4:48
samsam
134
$endgroup$
add a comment |
$begingroup$
Let $A={A_1,....A_n}$ be a collection of subsets of a finite set $X$. A selection for $A$ is the image of an injective function $f:Ato X$ such that $ f(A_i)in A_i$ for every $A_iin A$.
Hall's marriage theorem shows that , $A$ has a selection if and only if for each subset $Ssubseteq A$,
$$|S|leq |cup_{iin S} A_i|.$$
I wonder if it is possible to generalize this result and obtain a similar condition for a choice correspondence $f:Arightrightarrows X $ that selects for each $A_i$ more than one elements, i.e. $f(A_i)subset A_i$ and $|f(A_i)|=2$ and all selected elements are distinct.
Any ideas or suggestions?
combinatorics discrete-mathematics graph-theory
asked Nov 30 '18 at 4:48
samsam
134
$endgroup$
Let $A={A_1,....A_n}$ be a collection of subsets of a finite set $X$. A selection for $A$ is the image of an injective function $f:Ato X$ such that $ f(A_i)in A_i$ for every $A_iin A$.
Hall's marriage theorem shows that , $A$ has a selection if and only if for each subset $Ssubseteq A$,
$$|S|leq |cup_{iin S} A_i|.$$
I wonder if it is possible to generalize this result and obtain a similar condition for a choice correspondence $f:Arightrightarrows X $ that selects for each $A_i$ more than one elements, i.e. $f(A_i)subset A_i$ and $|f(A_i)|=2$ and all selected elements are distinct.
Any ideas or suggestions?
combinatorics discrete-mathematics graph-theory
combinatorics discrete-mathematics graph-theory
asked Nov 30 '18 at 4:48
samsam
134
asked Nov 30 '18 at 4:48
samsam
134
asked Nov 30 '18 at 4:48
samsam
134
asked Nov 30 '18 at 4:48
samsam
134
asked Nov 30 '18 at 4:48
samsam
134
134
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, this is a straightforward extension. Instead of the sequence
$$A_1,A_2,ldots,A_n$$
consider the sequence
$$A_1,A_1,A_2,A_2,ldots,A_n,A_n.$$
The Hall condition for this sequence amounts to
$$left|bigcup_{iin S}A_iright|ge2|S|.$$
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes, this is a straightforward extension. Instead of the sequence
$$A_1,A_2,ldots,A_n$$
consider the sequence
$$A_1,A_1,A_2,A_2,ldots,A_n,A_n.$$
The Hall condition for this sequence amounts to
$$left|bigcup_{iin S}A_iright|ge2|S|.$$
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
add a comment |
$begingroup$
Yes, this is a straightforward extension. Instead of the sequence
$$A_1,A_2,ldots,A_n$$
consider the sequence
$$A_1,A_1,A_2,A_2,ldots,A_n,A_n.$$
The Hall condition for this sequence amounts to
$$left|bigcup_{iin S}A_iright|ge2|S|.$$
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
add a comment |
$begingroup$
Yes, this is a straightforward extension. Instead of the sequence
$$A_1,A_2,ldots,A_n$$
consider the sequence
$$A_1,A_1,A_2,A_2,ldots,A_n,A_n.$$
The Hall condition for this sequence amounts to
$$left|bigcup_{iin S}A_iright|ge2|S|.$$
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
Yes, this is a straightforward extension. Instead of the sequence
$$A_1,A_2,ldots,A_n$$
consider the sequence
$$A_1,A_1,A_2,A_2,ldots,A_n,A_n.$$
The Hall condition for this sequence amounts to
$$left|bigcup_{iin S}A_iright|ge2|S|.$$
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Nov 30 '18 at 5:12
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
add a comment |
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
Great. Thanks for your answer!
$endgroup$
– sam
Nov 30 '18 at 5:27
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
$begingroup$
@sam This doesn't work very well with your (incorrect) definition of a choice function. You should define a choice function as an injection $f:{1,2,dots,n}to X$ instead of an injection $f:{A_1,A_2,dots,A_n}to X$ for just this reason, that you don't want to require the sets $A_i$ to be distinct.
$endgroup$
– bof
Nov 30 '18 at 12:51
add a comment |
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