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

Do one quarter of Swedes named 'Ali' have a criminal record?

How to keep track of pairs of points

How can you weaponize a thermos?

How to indicate "photograph by"

How often are there lunar eclipses on Jupiter

Miniseries in post-rapture US with good/evil conflict

Is there a sonic boom when flying nearby?

Would the Disguise Self spell benefit Stealth checks made to hide?

Why is SpaceX not also working on a smaller version of Starship?

Why wasn't Vaders armour made of Beskar?

How can I offer my prayers for an atheist?

What is a recently obsolete computer storage device that would be significantly difficult to extract data from?

Can a Rogue exploit a tiny familiar for automatic Sneak Attack in melee?

Do I even like doing research?

SOQL subquery access variables

Match blood types in C

Why is it runway "1" instead of "01" in America?

Convention for inverted signal

Best fighting style for a pacifist

What are the units of the product of two signals?

Bought a book that is in the public domain ... but the T&A of company says I can't redistribute it

Contacted by head of school regarding an issue - should I be worried?

Suppose I capture encrypted data that I want to decrypt. Could I use a server farm to decrypt?

What are the benefits of multi-file programming?



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
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;








20















$begingroup$


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?










share|cite|improve this question









$endgroup$










  • 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

















20















$begingroup$


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?










share|cite|improve this question









$endgroup$










  • 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













20













20









20


3



$begingroup$


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?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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












  • 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










4 Answers
4






active

oldest

votes


















30

















$begingroup$

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.






share|cite|improve this answer












$endgroup$










  • 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


















11

















$begingroup$

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.






share|cite|improve this answer










$endgroup$






















    5

















    $begingroup$

    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.






    share|cite|improve this answer












    $endgroup$










    • 1




      $begingroup$
      This argument is correct for finite sets. See @EricTowers answer.
      $endgroup$
      – 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

















    $begingroup$

    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.






    share|cite|improve this answer










    $endgroup$
















      Your Answer








      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );














      draft saved

      draft discarded
















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3357414%2fwhich-version-of-the-pigeonhole-principle-is-correct-one-is-far-stronger-than-t%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown


























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      30

















      $begingroup$

      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.






      share|cite|improve this answer












      $endgroup$










      • 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















      30

















      $begingroup$

      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.






      share|cite|improve this answer












      $endgroup$










      • 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













      30















      30











      30







      $begingroup$

      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.






      share|cite|improve this answer












      $endgroup$



      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.







      share|cite|improve this answer















      share|cite|improve this answer




      share|cite|improve this answer








      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












      • 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













      11

















      $begingroup$

      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.






      share|cite|improve this answer










      $endgroup$



















        11

















        $begingroup$

        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.






        share|cite|improve this answer










        $endgroup$

















          11















          11











          11







          $begingroup$

          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.






          share|cite|improve this answer










          $endgroup$



          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.







          share|cite|improve this answer













          share|cite|improve this answer




          share|cite|improve this answer










          answered Sep 16 at 2:25









          Martin ArgeramiMartin Argerami

          137k12 gold badges86 silver badges192 bronze badges




          137k12 gold badges86 silver badges192 bronze badges
























              5

















              $begingroup$

              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.






              share|cite|improve this answer












              $endgroup$










              • 1




                $begingroup$
                This argument is correct for finite sets. See @EricTowers answer.
                $endgroup$
                – 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















              5

















              $begingroup$

              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.






              share|cite|improve this answer












              $endgroup$










              • 1




                $begingroup$
                This argument is correct for finite sets. See @EricTowers answer.
                $endgroup$
                – 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













              5















              5











              5







              $begingroup$

              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.






              share|cite|improve this answer












              $endgroup$



              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.







              share|cite|improve this answer















              share|cite|improve this answer




              share|cite|improve this answer








              edited Sep 15 at 14:40

























              answered Sep 15 at 14:33









              Matthew DalyMatthew Daly

              10.7k4 gold badges18 silver badges40 bronze badges




              10.7k4 gold badges18 silver badges40 bronze badges










              • 1




                $begingroup$
                This argument is correct for finite sets. See @EricTowers answer.
                $endgroup$
                – 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




                $begingroup$
                This argument is correct for finite sets. See @EricTowers answer.
                $endgroup$
                – 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.
              $endgroup$
              – 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











              1

















              $begingroup$

              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.






              share|cite|improve this answer










              $endgroup$



















                1

















                $begingroup$

                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.






                share|cite|improve this answer










                $endgroup$

















                  1















                  1











                  1







                  $begingroup$

                  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.






                  share|cite|improve this answer










                  $endgroup$



                  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.







                  share|cite|improve this answer













                  share|cite|improve this answer




                  share|cite|improve this answer










                  answered Sep 16 at 15:01









                  AcccumulationAcccumulation

                  8,1682 gold badges9 silver badges20 bronze badges




                  8,1682 gold badges9 silver badges20 bronze badges































                      draft saved

                      draft discarded















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3357414%2fwhich-version-of-the-pigeonhole-principle-is-correct-one-is-far-stronger-than-t%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown









                      Popular posts from this blog

                      Tamil (spriik) Luke uk diar | Nawigatjuun

                      Align equal signs while including text over equalitiesAMS align: left aligned text/math plus multicolumn alignmentMultiple alignmentsAligning equations in multiple placesNumbering and aligning an equation with multiple columnsHow to align one equation with another multline equationUsing \ in environments inside the begintabularxNumber equations and preserving alignment of equal signsHow can I align equations to the left and to the right?Double equation alignment problem within align enviromentAligned within align: Why are they right-aligned?

                      Where does the image of a data connector as a sharp metal spike originate from?Where does the concept of infected people turning into zombies only after death originate from?Where does the motif of a reanimated human head originate?Where did the notion that Dragons could speak originate?Where does the archetypal image of the 'Grey' alien come from?Where did the suffix '-Man' originate?Where does the notion of being injured or killed by an illusion originate?Where did the term “sophont” originate?Where does the trope of magic spells being driven by advanced technology originate from?Where did the term “the living impaired” originate?