Which version of the Pigeonhole principle is correct? One is far stronger than the otherIs Pigeonhole Principle the negation of Dedekind-infinite?About the Pigeonhole principleProof of the pigeonhole principle by contradictionHow to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.Examples of the Pigeonhole PrincipleThis proof of the Pigeonhole principle has a real (and not integer) number equal to a variable that needs to be an integerA generalization the pigeonhole principle?Pigeonhole principle to find the exact quantity.Pigeonhole principle for finding consecutive days in which exactly $33$ tasks are performed
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Which version of the Pigeonhole principle is correct? One is far stronger than the other
Is Pigeonhole Principle the negation of Dedekind-infinite?About the Pigeonhole principleProof of the pigeonhole principle by contradictionHow to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.Examples of the Pigeonhole PrincipleThis proof of the Pigeonhole principle has a real (and not integer) number equal to a variable that needs to be an integerA generalization the pigeonhole principle?Pigeonhole principle to find the exact quantity.Pigeonhole principle for finding consecutive days in which exactly $33$ tasks are performed
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
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One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
$endgroup$
add a comment
|
$begingroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
$endgroup$
17
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
Sep 15 at 22:59
add a comment
|
$begingroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
$endgroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
combinatorics discrete-mathematics pigeonhole-principle
asked Sep 15 at 14:28
James RonaldJames Ronald
1,5633 silver badges10 bronze badges
1,5633 silver badges10 bronze badges
17
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
Sep 15 at 22:59
add a comment
|
17
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
Sep 15 at 22:59
17
17
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
Sep 15 at 22:59
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
Sep 15 at 22:59
add a comment
|
4 Answers
4
active
oldest
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$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true for this pair of infinite sets, this is pretty much useless (thanks, tomasz:) because it can be false for other pairs of infinite sets.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
$endgroup$
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
|
show 4 more comments
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Besides the two good answers already present, I would like to mention that the second version is also a trivial corollary of the first: indeed, if you take $m$ disjoint subsets $A_1,ldots,A_m$ of $A$, each with less than $lceil n/mrceil$ elements, then the union $bigcup_j=1^m A_j$ has at most $mlfloor n/mrfloor<n$ elements; so by the "simple" Pigeonhole Principle there cannot be an injection $Atobigcup_j=1^m A_j$.
So, for finite sets, both versions are easily seen to be equivalent. As mentioned, the first one also applies to infinite sets.
$endgroup$
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$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
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1
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This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
1
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@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
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– Matthew Daly
Sep 15 at 14:38
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That makes sense, I was too narrow in my thinking: thanks!
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– James Ronald
Sep 15 at 14:52
add a comment
|
$begingroup$
Which version is more useful depends on the situation. There are many situations where all that's needed is one hole with more than one pigeon, and besides that, the number of pigeons in each hole is not important.
As an additional note, for infinite sets, the nature of the principles changes. The definition of cardinality is equivalence classes defined in terms of injectivity, so saying that if |A| > |B| then there's injective function is pretty much just a restatement of the definition. And for infinite sets, if |A| > |B|, then, to the extent that |A|/|B| is defined, it is equal to |A|. For instance, if you assign to each integer a set of real numbers, and each real number is in at least one such set, then at least one set has the same cardinality as the set of real numbers.
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4 Answers
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4 Answers
4
active
oldest
votes
active
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active
oldest
votes
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true for this pair of infinite sets, this is pretty much useless (thanks, tomasz:) because it can be false for other pairs of infinite sets.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
$endgroup$
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
|
show 4 more comments
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true for this pair of infinite sets, this is pretty much useless (thanks, tomasz:) because it can be false for other pairs of infinite sets.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
$endgroup$
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
|
show 4 more comments
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true for this pair of infinite sets, this is pretty much useless (thanks, tomasz:) because it can be false for other pairs of infinite sets.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
$endgroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true for this pair of infinite sets, this is pretty much useless (thanks, tomasz:) because it can be false for other pairs of infinite sets.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
edited Sep 17 at 15:54
answered Sep 15 at 14:35
Eric TowersEric Towers
41.8k2 gold badges30 silver badges83 bronze badges
41.8k2 gold badges30 silver badges83 bronze badges
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
|
show 4 more comments
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
3
3
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
Sep 15 at 14:51
18
18
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
I feel like even in the infinite case, "in any map from $mathbb R$ to $mathbb Z$, there is an integer with at least $|mathbb R|$ preimages" is still strictly more useful than "there is no injection from $mathbb R$ to $mathbb Z$".
$endgroup$
– Misha Lavrov
Sep 16 at 1:56
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
$begingroup$
Why do you think it is "useless information" that continuum many reals are sent to at least one integer? I think it can be quite useful. More importantly, the analogue is not even true for infinite sets. For example, there is a function $fcolon aleph_omegato mathbf N$, such that for every $n$, $lvert f^-1[n]times mathbf Nrvert<aleph_omega$ (it is true if the larger cardinal is regular, or at least has cofinality larger than the smaller cardinal --- this is the case for finite cardinals, of course).
$endgroup$
– tomasz
Sep 16 at 12:14
3
3
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
$begingroup$
I've actually used that "useless" strengthening for infinite sets quite a few times.
$endgroup$
– Richard Rast
Sep 16 at 12:18
2
2
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
$begingroup$
@EricTowers: It is not useless when it is actually true.
$endgroup$
– tomasz
Sep 16 at 13:15
|
show 4 more comments
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Besides the two good answers already present, I would like to mention that the second version is also a trivial corollary of the first: indeed, if you take $m$ disjoint subsets $A_1,ldots,A_m$ of $A$, each with less than $lceil n/mrceil$ elements, then the union $bigcup_j=1^m A_j$ has at most $mlfloor n/mrfloor<n$ elements; so by the "simple" Pigeonhole Principle there cannot be an injection $Atobigcup_j=1^m A_j$.
So, for finite sets, both versions are easily seen to be equivalent. As mentioned, the first one also applies to infinite sets.
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Besides the two good answers already present, I would like to mention that the second version is also a trivial corollary of the first: indeed, if you take $m$ disjoint subsets $A_1,ldots,A_m$ of $A$, each with less than $lceil n/mrceil$ elements, then the union $bigcup_j=1^m A_j$ has at most $mlfloor n/mrfloor<n$ elements; so by the "simple" Pigeonhole Principle there cannot be an injection $Atobigcup_j=1^m A_j$.
So, for finite sets, both versions are easily seen to be equivalent. As mentioned, the first one also applies to infinite sets.
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add a comment
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$begingroup$
Besides the two good answers already present, I would like to mention that the second version is also a trivial corollary of the first: indeed, if you take $m$ disjoint subsets $A_1,ldots,A_m$ of $A$, each with less than $lceil n/mrceil$ elements, then the union $bigcup_j=1^m A_j$ has at most $mlfloor n/mrfloor<n$ elements; so by the "simple" Pigeonhole Principle there cannot be an injection $Atobigcup_j=1^m A_j$.
So, for finite sets, both versions are easily seen to be equivalent. As mentioned, the first one also applies to infinite sets.
$endgroup$
Besides the two good answers already present, I would like to mention that the second version is also a trivial corollary of the first: indeed, if you take $m$ disjoint subsets $A_1,ldots,A_m$ of $A$, each with less than $lceil n/mrceil$ elements, then the union $bigcup_j=1^m A_j$ has at most $mlfloor n/mrfloor<n$ elements; so by the "simple" Pigeonhole Principle there cannot be an injection $Atobigcup_j=1^m A_j$.
So, for finite sets, both versions are easily seen to be equivalent. As mentioned, the first one also applies to infinite sets.
answered Sep 16 at 2:25
Martin ArgeramiMartin Argerami
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This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
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1
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This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
1
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@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
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– Matthew Daly
Sep 15 at 14:38
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That makes sense, I was too narrow in my thinking: thanks!
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– James Ronald
Sep 15 at 14:52
add a comment
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$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
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1
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This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
1
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@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
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– Matthew Daly
Sep 15 at 14:38
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That makes sense, I was too narrow in my thinking: thanks!
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– James Ronald
Sep 15 at 14:52
add a comment
|
$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
$endgroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
edited Sep 15 at 14:40
answered Sep 15 at 14:33
Matthew DalyMatthew Daly
10.7k4 gold badges18 silver badges40 bronze badges
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1
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This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
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– Matthew Daly
Sep 15 at 14:38
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
Sep 15 at 14:52
add a comment
|
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
Sep 15 at 14:38
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
Sep 15 at 14:52
1
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
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– Ethan Bolker
Sep 15 at 14:37
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
Sep 15 at 14:37
1
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
Sep 15 at 14:38
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
Sep 15 at 14:38
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
Sep 15 at 14:52
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
Sep 15 at 14:52
add a comment
|
$begingroup$
Which version is more useful depends on the situation. There are many situations where all that's needed is one hole with more than one pigeon, and besides that, the number of pigeons in each hole is not important.
As an additional note, for infinite sets, the nature of the principles changes. The definition of cardinality is equivalence classes defined in terms of injectivity, so saying that if |A| > |B| then there's injective function is pretty much just a restatement of the definition. And for infinite sets, if |A| > |B|, then, to the extent that |A|/|B| is defined, it is equal to |A|. For instance, if you assign to each integer a set of real numbers, and each real number is in at least one such set, then at least one set has the same cardinality as the set of real numbers.
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add a comment
|
$begingroup$
Which version is more useful depends on the situation. There are many situations where all that's needed is one hole with more than one pigeon, and besides that, the number of pigeons in each hole is not important.
As an additional note, for infinite sets, the nature of the principles changes. The definition of cardinality is equivalence classes defined in terms of injectivity, so saying that if |A| > |B| then there's injective function is pretty much just a restatement of the definition. And for infinite sets, if |A| > |B|, then, to the extent that |A|/|B| is defined, it is equal to |A|. For instance, if you assign to each integer a set of real numbers, and each real number is in at least one such set, then at least one set has the same cardinality as the set of real numbers.
$endgroup$
add a comment
|
$begingroup$
Which version is more useful depends on the situation. There are many situations where all that's needed is one hole with more than one pigeon, and besides that, the number of pigeons in each hole is not important.
As an additional note, for infinite sets, the nature of the principles changes. The definition of cardinality is equivalence classes defined in terms of injectivity, so saying that if |A| > |B| then there's injective function is pretty much just a restatement of the definition. And for infinite sets, if |A| > |B|, then, to the extent that |A|/|B| is defined, it is equal to |A|. For instance, if you assign to each integer a set of real numbers, and each real number is in at least one such set, then at least one set has the same cardinality as the set of real numbers.
$endgroup$
Which version is more useful depends on the situation. There are many situations where all that's needed is one hole with more than one pigeon, and besides that, the number of pigeons in each hole is not important.
As an additional note, for infinite sets, the nature of the principles changes. The definition of cardinality is equivalence classes defined in terms of injectivity, so saying that if |A| > |B| then there's injective function is pretty much just a restatement of the definition. And for infinite sets, if |A| > |B|, then, to the extent that |A|/|B| is defined, it is equal to |A|. For instance, if you assign to each integer a set of real numbers, and each real number is in at least one such set, then at least one set has the same cardinality as the set of real numbers.
answered Sep 16 at 15:01
AcccumulationAcccumulation
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It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
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– YiFan
Sep 15 at 22:59