What is an explicit bijection in combinatorics?












12












$begingroup$


A standard way of demonstrating that two collections of combinatorial objects have the same cardinality is to exhibit a bijection between them. Browsing through some examples (here, there, yonder) quickly reveals that combinatorialists call such bijections explicit, presumably to differentiate them from other less palpable kinds of bijections. Wikipedia speaks of the method of bijective proof.



It seems that we have here a typical example of an informal mathematical notion that is quite familiar to most mathematicians, however it is difficult to pin down a proper and satisfying mathematical definition. I asked the local combinatorialists and did not really get a good answer.



Question: What is a proper mathematical definition of an explicit bijection?



Often we ask for an explicit bijection between two families of combinatorial objects, i.e., bijections $b_n : A_n to B_n$, one for each $n in mathbb{N}$. Here $(A_n)_n$ and $(B_n)_n$ are two families of combinatorial objects, parametrized by $n$. The parameter need not be a single number.



Here are some unsatisfactory answers:




  1. "A bijection is explicit if it is computable." This definition is too wide, because it allows silly algorithms that order combinatorial objects according to the layout of the sequences of bits that represent them, and use the order to establish a bijection. Bit representations typically have nothing to do with the combinatorial content of the objects under consideration.


  2. "A bijection is explicit if it can be written down as an expression." This takes us back several centuries in terms of level of mathematical abstraction, and also varies a lot depending on what expressions are allowed. We really should be looking for a combinatorially meaningful notion, not a syntactic surrogate.


  3. "A bijection is explicit if we can give a constructive proof of its existence." Well, a constructive proof certainly guarantees that a computable bijection exists, and can moreover be extracted from the proof, but this still feels too permissive. For example, we can always compose an explicit bijection so obtained with a computable automorphism of one of the sets, and still have a constructive proof. But such an automorphism could completely obfuscate the combinatorial structure of the set.


  4. "Well-order $V_omega$ (as all combinatorial objects easily live in it) and take the first bijection under the well ordering." Only a set theorist would have such thoughts. Again, we should strive for a definition which will be accepted as natural by combinatorialists.



Let me also say that I would prefer to not generalize the question to "what is an explicit thing?" At least in combinatorics "explicit bijections" are a well-established and useful notion, whereas mathematicians in general do not posses a universally agreed upon notion of "explicit thing".










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    That is one silly set theorist.
    $endgroup$
    – Andrés E. Caicedo
    2 hours ago






  • 1




    $begingroup$
    You know how they are, forcing things left and right.
    $endgroup$
    – Andrej Bauer
    2 hours ago






  • 2




    $begingroup$
    in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
    $endgroup$
    – Dima Pasechnik
    2 hours ago






  • 2




    $begingroup$
    My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
    $endgroup$
    – Michael Albert
    1 hour ago












  • $begingroup$
    Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
    $endgroup$
    – Dima Pasechnik
    1 hour ago
















12












$begingroup$


A standard way of demonstrating that two collections of combinatorial objects have the same cardinality is to exhibit a bijection between them. Browsing through some examples (here, there, yonder) quickly reveals that combinatorialists call such bijections explicit, presumably to differentiate them from other less palpable kinds of bijections. Wikipedia speaks of the method of bijective proof.



It seems that we have here a typical example of an informal mathematical notion that is quite familiar to most mathematicians, however it is difficult to pin down a proper and satisfying mathematical definition. I asked the local combinatorialists and did not really get a good answer.



Question: What is a proper mathematical definition of an explicit bijection?



Often we ask for an explicit bijection between two families of combinatorial objects, i.e., bijections $b_n : A_n to B_n$, one for each $n in mathbb{N}$. Here $(A_n)_n$ and $(B_n)_n$ are two families of combinatorial objects, parametrized by $n$. The parameter need not be a single number.



Here are some unsatisfactory answers:




  1. "A bijection is explicit if it is computable." This definition is too wide, because it allows silly algorithms that order combinatorial objects according to the layout of the sequences of bits that represent them, and use the order to establish a bijection. Bit representations typically have nothing to do with the combinatorial content of the objects under consideration.


  2. "A bijection is explicit if it can be written down as an expression." This takes us back several centuries in terms of level of mathematical abstraction, and also varies a lot depending on what expressions are allowed. We really should be looking for a combinatorially meaningful notion, not a syntactic surrogate.


  3. "A bijection is explicit if we can give a constructive proof of its existence." Well, a constructive proof certainly guarantees that a computable bijection exists, and can moreover be extracted from the proof, but this still feels too permissive. For example, we can always compose an explicit bijection so obtained with a computable automorphism of one of the sets, and still have a constructive proof. But such an automorphism could completely obfuscate the combinatorial structure of the set.


  4. "Well-order $V_omega$ (as all combinatorial objects easily live in it) and take the first bijection under the well ordering." Only a set theorist would have such thoughts. Again, we should strive for a definition which will be accepted as natural by combinatorialists.



Let me also say that I would prefer to not generalize the question to "what is an explicit thing?" At least in combinatorics "explicit bijections" are a well-established and useful notion, whereas mathematicians in general do not posses a universally agreed upon notion of "explicit thing".










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    That is one silly set theorist.
    $endgroup$
    – Andrés E. Caicedo
    2 hours ago






  • 1




    $begingroup$
    You know how they are, forcing things left and right.
    $endgroup$
    – Andrej Bauer
    2 hours ago






  • 2




    $begingroup$
    in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
    $endgroup$
    – Dima Pasechnik
    2 hours ago






  • 2




    $begingroup$
    My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
    $endgroup$
    – Michael Albert
    1 hour ago












  • $begingroup$
    Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
    $endgroup$
    – Dima Pasechnik
    1 hour ago














12












12








12


1



$begingroup$


A standard way of demonstrating that two collections of combinatorial objects have the same cardinality is to exhibit a bijection between them. Browsing through some examples (here, there, yonder) quickly reveals that combinatorialists call such bijections explicit, presumably to differentiate them from other less palpable kinds of bijections. Wikipedia speaks of the method of bijective proof.



It seems that we have here a typical example of an informal mathematical notion that is quite familiar to most mathematicians, however it is difficult to pin down a proper and satisfying mathematical definition. I asked the local combinatorialists and did not really get a good answer.



Question: What is a proper mathematical definition of an explicit bijection?



Often we ask for an explicit bijection between two families of combinatorial objects, i.e., bijections $b_n : A_n to B_n$, one for each $n in mathbb{N}$. Here $(A_n)_n$ and $(B_n)_n$ are two families of combinatorial objects, parametrized by $n$. The parameter need not be a single number.



Here are some unsatisfactory answers:




  1. "A bijection is explicit if it is computable." This definition is too wide, because it allows silly algorithms that order combinatorial objects according to the layout of the sequences of bits that represent them, and use the order to establish a bijection. Bit representations typically have nothing to do with the combinatorial content of the objects under consideration.


  2. "A bijection is explicit if it can be written down as an expression." This takes us back several centuries in terms of level of mathematical abstraction, and also varies a lot depending on what expressions are allowed. We really should be looking for a combinatorially meaningful notion, not a syntactic surrogate.


  3. "A bijection is explicit if we can give a constructive proof of its existence." Well, a constructive proof certainly guarantees that a computable bijection exists, and can moreover be extracted from the proof, but this still feels too permissive. For example, we can always compose an explicit bijection so obtained with a computable automorphism of one of the sets, and still have a constructive proof. But such an automorphism could completely obfuscate the combinatorial structure of the set.


  4. "Well-order $V_omega$ (as all combinatorial objects easily live in it) and take the first bijection under the well ordering." Only a set theorist would have such thoughts. Again, we should strive for a definition which will be accepted as natural by combinatorialists.



Let me also say that I would prefer to not generalize the question to "what is an explicit thing?" At least in combinatorics "explicit bijections" are a well-established and useful notion, whereas mathematicians in general do not posses a universally agreed upon notion of "explicit thing".










share|cite|improve this question











$endgroup$




A standard way of demonstrating that two collections of combinatorial objects have the same cardinality is to exhibit a bijection between them. Browsing through some examples (here, there, yonder) quickly reveals that combinatorialists call such bijections explicit, presumably to differentiate them from other less palpable kinds of bijections. Wikipedia speaks of the method of bijective proof.



It seems that we have here a typical example of an informal mathematical notion that is quite familiar to most mathematicians, however it is difficult to pin down a proper and satisfying mathematical definition. I asked the local combinatorialists and did not really get a good answer.



Question: What is a proper mathematical definition of an explicit bijection?



Often we ask for an explicit bijection between two families of combinatorial objects, i.e., bijections $b_n : A_n to B_n$, one for each $n in mathbb{N}$. Here $(A_n)_n$ and $(B_n)_n$ are two families of combinatorial objects, parametrized by $n$. The parameter need not be a single number.



Here are some unsatisfactory answers:




  1. "A bijection is explicit if it is computable." This definition is too wide, because it allows silly algorithms that order combinatorial objects according to the layout of the sequences of bits that represent them, and use the order to establish a bijection. Bit representations typically have nothing to do with the combinatorial content of the objects under consideration.


  2. "A bijection is explicit if it can be written down as an expression." This takes us back several centuries in terms of level of mathematical abstraction, and also varies a lot depending on what expressions are allowed. We really should be looking for a combinatorially meaningful notion, not a syntactic surrogate.


  3. "A bijection is explicit if we can give a constructive proof of its existence." Well, a constructive proof certainly guarantees that a computable bijection exists, and can moreover be extracted from the proof, but this still feels too permissive. For example, we can always compose an explicit bijection so obtained with a computable automorphism of one of the sets, and still have a constructive proof. But such an automorphism could completely obfuscate the combinatorial structure of the set.


  4. "Well-order $V_omega$ (as all combinatorial objects easily live in it) and take the first bijection under the well ordering." Only a set theorist would have such thoughts. Again, we should strive for a definition which will be accepted as natural by combinatorialists.



Let me also say that I would prefer to not generalize the question to "what is an explicit thing?" At least in combinatorics "explicit bijections" are a well-established and useful notion, whereas mathematicians in general do not posses a universally agreed upon notion of "explicit thing".







co.combinatorics foundations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







Andrej Bauer

















asked 2 hours ago









Andrej BauerAndrej Bauer

30.7k478168




30.7k478168








  • 1




    $begingroup$
    That is one silly set theorist.
    $endgroup$
    – Andrés E. Caicedo
    2 hours ago






  • 1




    $begingroup$
    You know how they are, forcing things left and right.
    $endgroup$
    – Andrej Bauer
    2 hours ago






  • 2




    $begingroup$
    in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
    $endgroup$
    – Dima Pasechnik
    2 hours ago






  • 2




    $begingroup$
    My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
    $endgroup$
    – Michael Albert
    1 hour ago












  • $begingroup$
    Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
    $endgroup$
    – Dima Pasechnik
    1 hour ago














  • 1




    $begingroup$
    That is one silly set theorist.
    $endgroup$
    – Andrés E. Caicedo
    2 hours ago






  • 1




    $begingroup$
    You know how they are, forcing things left and right.
    $endgroup$
    – Andrej Bauer
    2 hours ago






  • 2




    $begingroup$
    in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
    $endgroup$
    – Dima Pasechnik
    2 hours ago






  • 2




    $begingroup$
    My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
    $endgroup$
    – Michael Albert
    1 hour ago












  • $begingroup$
    Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
    $endgroup$
    – Dima Pasechnik
    1 hour ago








1




1




$begingroup$
That is one silly set theorist.
$endgroup$
– Andrés E. Caicedo
2 hours ago




$begingroup$
That is one silly set theorist.
$endgroup$
– Andrés E. Caicedo
2 hours ago




1




1




$begingroup$
You know how they are, forcing things left and right.
$endgroup$
– Andrej Bauer
2 hours ago




$begingroup$
You know how they are, forcing things left and right.
$endgroup$
– Andrej Bauer
2 hours ago




2




2




$begingroup$
in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
$endgroup$
– Dima Pasechnik
2 hours ago




$begingroup$
in math.ucla.edu/~pak/papers/ICM-paper9.pdf one finds some studies on the subject.
$endgroup$
– Dima Pasechnik
2 hours ago




2




2




$begingroup$
My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
$endgroup$
– Michael Albert
1 hour ago






$begingroup$
My feeling is that when one writes "We define an explicit bijection ..." one means a bijection whose computation depends only on the description of an individual object in the domain -- in particular the main dichotomy for me is between explicit and recursively defined bijections. Having said all that I'll also say that I'd be happier if "explicit bijection" were left as an informal notion! Especially since I suspect that I've been entirely inconsistent in its use.
$endgroup$
– Michael Albert
1 hour ago














$begingroup$
Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
$endgroup$
– Dima Pasechnik
1 hour ago




$begingroup$
Pak's paper I cited argues that an algorithm that computes a bijection as in 1) has to be "fast" in a well-defined sense.
$endgroup$
– Dima Pasechnik
1 hour ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

This is not at all intended as a complete answer to the question, but one criterion that feels important is that for a bijection $f$ to count as explicit, one shouldn't need to know in advance that there exists a bijection in order to prove that $f$ is a well-defined bijection. So for example if you order the elements of two sets $A$ and $B$ in some way that has nothing to do with why $|A|=|B|$, then you need to know that $|A|=|B|$ in order to conclude that the bijection that maps the $k$th element of $A$ to the $k$th element of $B$ is indeed a well-defined bijection.



I think this criterion rules out 1 and 4 (or would do if one could make it more formal, which might itself not be wholly easy).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
    $endgroup$
    – darij grinberg
    1 hour ago












  • $begingroup$
    I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
    $endgroup$
    – gowers
    1 hour ago










  • $begingroup$
    I don't, but I know it exists :)
    $endgroup$
    – darij grinberg
    1 hour ago










  • $begingroup$
    Ah, I didn't express my criterion clearly. I'll edit it now.
    $endgroup$
    – gowers
    1 hour ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
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/3.0/"u003ecc by-sa 3.0 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%2fmathoverflow.net%2fquestions%2f323779%2fwhat-is-an-explicit-bijection-in-combinatorics%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

This is not at all intended as a complete answer to the question, but one criterion that feels important is that for a bijection $f$ to count as explicit, one shouldn't need to know in advance that there exists a bijection in order to prove that $f$ is a well-defined bijection. So for example if you order the elements of two sets $A$ and $B$ in some way that has nothing to do with why $|A|=|B|$, then you need to know that $|A|=|B|$ in order to conclude that the bijection that maps the $k$th element of $A$ to the $k$th element of $B$ is indeed a well-defined bijection.



I think this criterion rules out 1 and 4 (or would do if one could make it more formal, which might itself not be wholly easy).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
    $endgroup$
    – darij grinberg
    1 hour ago












  • $begingroup$
    I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
    $endgroup$
    – gowers
    1 hour ago










  • $begingroup$
    I don't, but I know it exists :)
    $endgroup$
    – darij grinberg
    1 hour ago










  • $begingroup$
    Ah, I didn't express my criterion clearly. I'll edit it now.
    $endgroup$
    – gowers
    1 hour ago
















5












$begingroup$

This is not at all intended as a complete answer to the question, but one criterion that feels important is that for a bijection $f$ to count as explicit, one shouldn't need to know in advance that there exists a bijection in order to prove that $f$ is a well-defined bijection. So for example if you order the elements of two sets $A$ and $B$ in some way that has nothing to do with why $|A|=|B|$, then you need to know that $|A|=|B|$ in order to conclude that the bijection that maps the $k$th element of $A$ to the $k$th element of $B$ is indeed a well-defined bijection.



I think this criterion rules out 1 and 4 (or would do if one could make it more formal, which might itself not be wholly easy).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
    $endgroup$
    – darij grinberg
    1 hour ago












  • $begingroup$
    I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
    $endgroup$
    – gowers
    1 hour ago










  • $begingroup$
    I don't, but I know it exists :)
    $endgroup$
    – darij grinberg
    1 hour ago










  • $begingroup$
    Ah, I didn't express my criterion clearly. I'll edit it now.
    $endgroup$
    – gowers
    1 hour ago














5












5








5





$begingroup$

This is not at all intended as a complete answer to the question, but one criterion that feels important is that for a bijection $f$ to count as explicit, one shouldn't need to know in advance that there exists a bijection in order to prove that $f$ is a well-defined bijection. So for example if you order the elements of two sets $A$ and $B$ in some way that has nothing to do with why $|A|=|B|$, then you need to know that $|A|=|B|$ in order to conclude that the bijection that maps the $k$th element of $A$ to the $k$th element of $B$ is indeed a well-defined bijection.



I think this criterion rules out 1 and 4 (or would do if one could make it more formal, which might itself not be wholly easy).






share|cite|improve this answer











$endgroup$



This is not at all intended as a complete answer to the question, but one criterion that feels important is that for a bijection $f$ to count as explicit, one shouldn't need to know in advance that there exists a bijection in order to prove that $f$ is a well-defined bijection. So for example if you order the elements of two sets $A$ and $B$ in some way that has nothing to do with why $|A|=|B|$, then you need to know that $|A|=|B|$ in order to conclude that the bijection that maps the $k$th element of $A$ to the $k$th element of $B$ is indeed a well-defined bijection.



I think this criterion rules out 1 and 4 (or would do if one could make it more formal, which might itself not be wholly easy).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 2 hours ago









gowersgowers

19.4k22121167




19.4k22121167












  • $begingroup$
    Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
    $endgroup$
    – darij grinberg
    1 hour ago












  • $begingroup$
    I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
    $endgroup$
    – gowers
    1 hour ago










  • $begingroup$
    I don't, but I know it exists :)
    $endgroup$
    – darij grinberg
    1 hour ago










  • $begingroup$
    Ah, I didn't express my criterion clearly. I'll edit it now.
    $endgroup$
    – gowers
    1 hour ago


















  • $begingroup$
    Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
    $endgroup$
    – darij grinberg
    1 hour ago












  • $begingroup$
    I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
    $endgroup$
    – gowers
    1 hour ago










  • $begingroup$
    I don't, but I know it exists :)
    $endgroup$
    – darij grinberg
    1 hour ago










  • $begingroup$
    Ah, I didn't express my criterion clearly. I'll edit it now.
    $endgroup$
    – gowers
    1 hour ago
















$begingroup$
Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
$endgroup$
– darij grinberg
1 hour ago






$begingroup$
Hmm, what about "map the $k$-th smallest element of $A$ to the $k$-th smallest element of $B$ or to the largest one if there is no $k$-th smallest one"? I am being somewhat tongue-in-cheek here, as we are clearly looking for a rule to follow in spirit rather than in letter. On the other hand, if we take this idea too far in the other direction, then a bijection $f : A to B$ whose bijectivity is only proven using the pigeonhole principle (i.e., by showing that it is injective or surjective, and that $left|Aright| = left|Bright|$) should not count as explicit either. (Perhaps rightfully!)
$endgroup$
– darij grinberg
1 hour ago














$begingroup$
I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
$endgroup$
– gowers
1 hour ago




$begingroup$
I'm not sure I follow the first part of your comment: how do we know that the map you define is a bijection?
$endgroup$
– gowers
1 hour ago












$begingroup$
I don't, but I know it exists :)
$endgroup$
– darij grinberg
1 hour ago




$begingroup$
I don't, but I know it exists :)
$endgroup$
– darij grinberg
1 hour ago












$begingroup$
Ah, I didn't express my criterion clearly. I'll edit it now.
$endgroup$
– gowers
1 hour ago




$begingroup$
Ah, I didn't express my criterion clearly. I'll edit it now.
$endgroup$
– gowers
1 hour ago


















draft saved

draft discarded




















































Thanks for contributing an answer to MathOverflow!


  • 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%2fmathoverflow.net%2fquestions%2f323779%2fwhat-is-an-explicit-bijection-in-combinatorics%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

Ponta tanko

Tantalo (mitologio)

Erzsébet Schaár