Examples of proofs by making reduction to a finite setWhy are planar graphs so exceptional?Graphs of Tangent SpheresChromatic number of graphs of tangent closed ballsWhat are some good examples of non-monotone graph properties?Examples of two different descriptions of a set that are not obviously equivalent?Beraha numbers and zeros of the chromatic polynomial of planar graphsAlgebraic proof of Five-Color Theorem using chromatic polynomials by Birkhoff and Lewis in 1946Do you know a faster algorithm to color planar graphs?A direct proof that every $r$-colored complete graph on $n=(r+1)m-(r-1)$ vertices has a monochromatic matching of size $m$?

Examples of proofs by making reduction to a finite set


Why are planar graphs so exceptional?Graphs of Tangent SpheresChromatic number of graphs of tangent closed ballsWhat are some good examples of non-monotone graph properties?Examples of two different descriptions of a set that are not obviously equivalent?Beraha numbers and zeros of the chromatic polynomial of planar graphsAlgebraic proof of Five-Color Theorem using chromatic polynomials by Birkhoff and Lewis in 1946Do you know a faster algorithm to color planar graphs?A direct proof that every $r$-colored complete graph on $n=(r+1)m-(r-1)$ vertices has a monochromatic matching of size $m$?













5












$begingroup$


This is a very abstract question, I hope this is appropriate.

Suppose $T$ is some claim over some infinite set $A$, for example, let $A$ be the set of all loopless planar graphs, and $T$ be the claim "for every element $a in A$, the chromatic number of $a$'s dual graph is $leq 4$" (this is known as "Four-color Theorem"); or, let $A = mathbbN$, and $T$ be the claim "there are no 3 elements $x, y, z in A$ such that $x^5 + y^5 =z^5$" (a specific case of Fermat's last theorem).

In the first example, it is possible to prove the claim $T$ by testing some claim $T'$ over finite set $A' subset A$, see proof by computer section. This is done by a series of reductions, showing that if all the elements in some finite set satisfy a property, then the (original) claim holds.

In some sense, mathematical induction is similar: we test a claim on finite set ("the base case"), then proving $a_n rightarrow a_n+1$, which shows the claim is correct for all space.



Are there more known cases like that? i.e. proving (a combinatorial) claim by reduction to finite cases?










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  • 1




    $begingroup$
    I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
    $endgroup$
    – LSpice
    10 hours ago







  • 4




    $begingroup$
    There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
    $endgroup$
    – user36212
    10 hours ago















5












$begingroup$


This is a very abstract question, I hope this is appropriate.

Suppose $T$ is some claim over some infinite set $A$, for example, let $A$ be the set of all loopless planar graphs, and $T$ be the claim "for every element $a in A$, the chromatic number of $a$'s dual graph is $leq 4$" (this is known as "Four-color Theorem"); or, let $A = mathbbN$, and $T$ be the claim "there are no 3 elements $x, y, z in A$ such that $x^5 + y^5 =z^5$" (a specific case of Fermat's last theorem).

In the first example, it is possible to prove the claim $T$ by testing some claim $T'$ over finite set $A' subset A$, see proof by computer section. This is done by a series of reductions, showing that if all the elements in some finite set satisfy a property, then the (original) claim holds.

In some sense, mathematical induction is similar: we test a claim on finite set ("the base case"), then proving $a_n rightarrow a_n+1$, which shows the claim is correct for all space.



Are there more known cases like that? i.e. proving (a combinatorial) claim by reduction to finite cases?










share|cite|improve this question









New contributor



SomeoneHAHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 1




    $begingroup$
    I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
    $endgroup$
    – LSpice
    10 hours ago







  • 4




    $begingroup$
    There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
    $endgroup$
    – user36212
    10 hours ago













5












5








5


1



$begingroup$


This is a very abstract question, I hope this is appropriate.

Suppose $T$ is some claim over some infinite set $A$, for example, let $A$ be the set of all loopless planar graphs, and $T$ be the claim "for every element $a in A$, the chromatic number of $a$'s dual graph is $leq 4$" (this is known as "Four-color Theorem"); or, let $A = mathbbN$, and $T$ be the claim "there are no 3 elements $x, y, z in A$ such that $x^5 + y^5 =z^5$" (a specific case of Fermat's last theorem).

In the first example, it is possible to prove the claim $T$ by testing some claim $T'$ over finite set $A' subset A$, see proof by computer section. This is done by a series of reductions, showing that if all the elements in some finite set satisfy a property, then the (original) claim holds.

In some sense, mathematical induction is similar: we test a claim on finite set ("the base case"), then proving $a_n rightarrow a_n+1$, which shows the claim is correct for all space.



Are there more known cases like that? i.e. proving (a combinatorial) claim by reduction to finite cases?










share|cite|improve this question









New contributor



SomeoneHAHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




This is a very abstract question, I hope this is appropriate.

Suppose $T$ is some claim over some infinite set $A$, for example, let $A$ be the set of all loopless planar graphs, and $T$ be the claim "for every element $a in A$, the chromatic number of $a$'s dual graph is $leq 4$" (this is known as "Four-color Theorem"); or, let $A = mathbbN$, and $T$ be the claim "there are no 3 elements $x, y, z in A$ such that $x^5 + y^5 =z^5$" (a specific case of Fermat's last theorem).

In the first example, it is possible to prove the claim $T$ by testing some claim $T'$ over finite set $A' subset A$, see proof by computer section. This is done by a series of reductions, showing that if all the elements in some finite set satisfy a property, then the (original) claim holds.

In some sense, mathematical induction is similar: we test a claim on finite set ("the base case"), then proving $a_n rightarrow a_n+1$, which shows the claim is correct for all space.



Are there more known cases like that? i.e. proving (a combinatorial) claim by reduction to finite cases?







nt.number-theory co.combinatorics graph-theory lo.logic big-list






share|cite|improve this question









New contributor



SomeoneHAHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago









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Check out our Code of Conduct.












  • 1




    $begingroup$
    I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
    $endgroup$
    – LSpice
    10 hours ago







  • 4




    $begingroup$
    There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
    $endgroup$
    – user36212
    10 hours ago












  • 1




    $begingroup$
    I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
    $endgroup$
    – LSpice
    10 hours ago







  • 4




    $begingroup$
    There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
    $endgroup$
    – user36212
    10 hours ago







1




1




$begingroup$
I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
$endgroup$
– LSpice
10 hours ago





$begingroup$
I think that this is a very interesting question, but I think it doesn't fit the MO mission of focussed questions with a definite answer. MO has some tolerance for big-list questions, so maybe it would be appropriate if made community wiki. (You can do this by flagging your own post for moderator attention, if you agree.)
$endgroup$
– LSpice
10 hours ago





4




4




$begingroup$
There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
$endgroup$
– user36212
10 hours ago




$begingroup$
There are certainly a lot of proofs which are vaguely similar to the Four Colour Theorem; search for 'discharging method' which is a (the most common but certainly not the only) way to reduce a general graph problem to a finite set of cases to check. For a rather different example, in some sense Helfgott's proof of the weak Goldbach conjecture is of this form: he reduces the problem to checking finitely many cases and then does the check. I think this is an example of what you don't want to see (because without the finite check Helfgott still proves something; that's not true for 4CT).
$endgroup$
– user36212
10 hours ago










3 Answers
3






active

oldest

votes


















2














$begingroup$

In 1972, I have developed a general theory of isometric embeddings of finite metric spaces into finite-dimensional Banach spaces, in particular, the injective case, i.e. into metrically injective spaces $ mathbb R^n $ with metrics given by $ max.$ An announcement has appeared a couple of years later in AMS Notices. One of the theorems fits the present thread:



For every natural $ k n $ there is an EXPLICIT natural constant $ mu(k n) $ such that the following two statements are equivalent:







(i) every $k$-element metric space can be isometrically embedded
into $ mathbb R^n$ with the distance given by $ max;$



(ii) every $k$-element metric space with integer distances, and of diameter
$ le mu(k n), $ can be isometrically embedded into subspace
$, 0 ldots mu(k n)^n $ of $ mathbb R^n$ with the distance
given by $ max$.







For each natural $ n $ there exists a maximal $ u(n)=k $ as above; and also
$ U(n) $ similar to $ u(n) $ but for ALL $ n$-dimensional Banach spaces TOGETHER (for each fixed dimension $ n. $) Then in the non-trivial case of $ nge 2 $ we get



$$ n+2 le u(n) le U(n) le biglceilfrac3cdot n2bigrceil + 1 $$



We see from (i)+(ii) that finding $ u(n) $ got reduced to a finite computation.






share|cite|improve this answer











$endgroup$






















    1














    $begingroup$

    It is very common, when finding solutions to Diophantine equations, to use Baker's method of linear forms in logarithms to reduce the problem to a finite computation (that is, to find an upper bound for the solutions).






    share|cite|improve this answer









    $endgroup$






















      1














      $begingroup$

      For the sake of completeness, let me mention the direction which is opposite to the OP's Question.



      There are combinatorial optimizations problems which deal with a finite (but large) input set at the start, and the point is to optimize a real function over the input. Such problems are often messy.



      Then, some time ago, Hungarian mathematicians started to embed the input set into a Euclidean n-space, and they'd extend the said function to a linear or convex function over the convex hull of the input. Since the optimum over the whole hull is reached at a vertex then... etc.



      We see that sometimes situations which are strictly finite (looking for an exact answer) get hm-reduced to infinite situations.






      share|cite|improve this answer











      $endgroup$

















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        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        $begingroup$

        In 1972, I have developed a general theory of isometric embeddings of finite metric spaces into finite-dimensional Banach spaces, in particular, the injective case, i.e. into metrically injective spaces $ mathbb R^n $ with metrics given by $ max.$ An announcement has appeared a couple of years later in AMS Notices. One of the theorems fits the present thread:



        For every natural $ k n $ there is an EXPLICIT natural constant $ mu(k n) $ such that the following two statements are equivalent:







        (i) every $k$-element metric space can be isometrically embedded
        into $ mathbb R^n$ with the distance given by $ max;$



        (ii) every $k$-element metric space with integer distances, and of diameter
        $ le mu(k n), $ can be isometrically embedded into subspace
        $, 0 ldots mu(k n)^n $ of $ mathbb R^n$ with the distance
        given by $ max$.







        For each natural $ n $ there exists a maximal $ u(n)=k $ as above; and also
        $ U(n) $ similar to $ u(n) $ but for ALL $ n$-dimensional Banach spaces TOGETHER (for each fixed dimension $ n. $) Then in the non-trivial case of $ nge 2 $ we get



        $$ n+2 le u(n) le U(n) le biglceilfrac3cdot n2bigrceil + 1 $$



        We see from (i)+(ii) that finding $ u(n) $ got reduced to a finite computation.






        share|cite|improve this answer











        $endgroup$



















          2














          $begingroup$

          In 1972, I have developed a general theory of isometric embeddings of finite metric spaces into finite-dimensional Banach spaces, in particular, the injective case, i.e. into metrically injective spaces $ mathbb R^n $ with metrics given by $ max.$ An announcement has appeared a couple of years later in AMS Notices. One of the theorems fits the present thread:



          For every natural $ k n $ there is an EXPLICIT natural constant $ mu(k n) $ such that the following two statements are equivalent:







          (i) every $k$-element metric space can be isometrically embedded
          into $ mathbb R^n$ with the distance given by $ max;$



          (ii) every $k$-element metric space with integer distances, and of diameter
          $ le mu(k n), $ can be isometrically embedded into subspace
          $, 0 ldots mu(k n)^n $ of $ mathbb R^n$ with the distance
          given by $ max$.







          For each natural $ n $ there exists a maximal $ u(n)=k $ as above; and also
          $ U(n) $ similar to $ u(n) $ but for ALL $ n$-dimensional Banach spaces TOGETHER (for each fixed dimension $ n. $) Then in the non-trivial case of $ nge 2 $ we get



          $$ n+2 le u(n) le U(n) le biglceilfrac3cdot n2bigrceil + 1 $$



          We see from (i)+(ii) that finding $ u(n) $ got reduced to a finite computation.






          share|cite|improve this answer











          $endgroup$

















            2














            2










            2







            $begingroup$

            In 1972, I have developed a general theory of isometric embeddings of finite metric spaces into finite-dimensional Banach spaces, in particular, the injective case, i.e. into metrically injective spaces $ mathbb R^n $ with metrics given by $ max.$ An announcement has appeared a couple of years later in AMS Notices. One of the theorems fits the present thread:



            For every natural $ k n $ there is an EXPLICIT natural constant $ mu(k n) $ such that the following two statements are equivalent:







            (i) every $k$-element metric space can be isometrically embedded
            into $ mathbb R^n$ with the distance given by $ max;$



            (ii) every $k$-element metric space with integer distances, and of diameter
            $ le mu(k n), $ can be isometrically embedded into subspace
            $, 0 ldots mu(k n)^n $ of $ mathbb R^n$ with the distance
            given by $ max$.







            For each natural $ n $ there exists a maximal $ u(n)=k $ as above; and also
            $ U(n) $ similar to $ u(n) $ but for ALL $ n$-dimensional Banach spaces TOGETHER (for each fixed dimension $ n. $) Then in the non-trivial case of $ nge 2 $ we get



            $$ n+2 le u(n) le U(n) le biglceilfrac3cdot n2bigrceil + 1 $$



            We see from (i)+(ii) that finding $ u(n) $ got reduced to a finite computation.






            share|cite|improve this answer











            $endgroup$



            In 1972, I have developed a general theory of isometric embeddings of finite metric spaces into finite-dimensional Banach spaces, in particular, the injective case, i.e. into metrically injective spaces $ mathbb R^n $ with metrics given by $ max.$ An announcement has appeared a couple of years later in AMS Notices. One of the theorems fits the present thread:



            For every natural $ k n $ there is an EXPLICIT natural constant $ mu(k n) $ such that the following two statements are equivalent:







            (i) every $k$-element metric space can be isometrically embedded
            into $ mathbb R^n$ with the distance given by $ max;$



            (ii) every $k$-element metric space with integer distances, and of diameter
            $ le mu(k n), $ can be isometrically embedded into subspace
            $, 0 ldots mu(k n)^n $ of $ mathbb R^n$ with the distance
            given by $ max$.







            For each natural $ n $ there exists a maximal $ u(n)=k $ as above; and also
            $ U(n) $ similar to $ u(n) $ but for ALL $ n$-dimensional Banach spaces TOGETHER (for each fixed dimension $ n. $) Then in the non-trivial case of $ nge 2 $ we get



            $$ n+2 le u(n) le U(n) le biglceilfrac3cdot n2bigrceil + 1 $$



            We see from (i)+(ii) that finding $ u(n) $ got reduced to a finite computation.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 2 hours ago









            Wlod AAWlod AA

            1,3804 silver badges14 bronze badges




            1,3804 silver badges14 bronze badges
























                1














                $begingroup$

                It is very common, when finding solutions to Diophantine equations, to use Baker's method of linear forms in logarithms to reduce the problem to a finite computation (that is, to find an upper bound for the solutions).






                share|cite|improve this answer









                $endgroup$



















                  1














                  $begingroup$

                  It is very common, when finding solutions to Diophantine equations, to use Baker's method of linear forms in logarithms to reduce the problem to a finite computation (that is, to find an upper bound for the solutions).






                  share|cite|improve this answer









                  $endgroup$

















                    1














                    1










                    1







                    $begingroup$

                    It is very common, when finding solutions to Diophantine equations, to use Baker's method of linear forms in logarithms to reduce the problem to a finite computation (that is, to find an upper bound for the solutions).






                    share|cite|improve this answer









                    $endgroup$



                    It is very common, when finding solutions to Diophantine equations, to use Baker's method of linear forms in logarithms to reduce the problem to a finite computation (that is, to find an upper bound for the solutions).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    Gerry MyersonGerry Myerson

                    31.7k6 gold badges146 silver badges190 bronze badges




                    31.7k6 gold badges146 silver badges190 bronze badges
























                        1














                        $begingroup$

                        For the sake of completeness, let me mention the direction which is opposite to the OP's Question.



                        There are combinatorial optimizations problems which deal with a finite (but large) input set at the start, and the point is to optimize a real function over the input. Such problems are often messy.



                        Then, some time ago, Hungarian mathematicians started to embed the input set into a Euclidean n-space, and they'd extend the said function to a linear or convex function over the convex hull of the input. Since the optimum over the whole hull is reached at a vertex then... etc.



                        We see that sometimes situations which are strictly finite (looking for an exact answer) get hm-reduced to infinite situations.






                        share|cite|improve this answer











                        $endgroup$



















                          1














                          $begingroup$

                          For the sake of completeness, let me mention the direction which is opposite to the OP's Question.



                          There are combinatorial optimizations problems which deal with a finite (but large) input set at the start, and the point is to optimize a real function over the input. Such problems are often messy.



                          Then, some time ago, Hungarian mathematicians started to embed the input set into a Euclidean n-space, and they'd extend the said function to a linear or convex function over the convex hull of the input. Since the optimum over the whole hull is reached at a vertex then... etc.



                          We see that sometimes situations which are strictly finite (looking for an exact answer) get hm-reduced to infinite situations.






                          share|cite|improve this answer











                          $endgroup$

















                            1














                            1










                            1







                            $begingroup$

                            For the sake of completeness, let me mention the direction which is opposite to the OP's Question.



                            There are combinatorial optimizations problems which deal with a finite (but large) input set at the start, and the point is to optimize a real function over the input. Such problems are often messy.



                            Then, some time ago, Hungarian mathematicians started to embed the input set into a Euclidean n-space, and they'd extend the said function to a linear or convex function over the convex hull of the input. Since the optimum over the whole hull is reached at a vertex then... etc.



                            We see that sometimes situations which are strictly finite (looking for an exact answer) get hm-reduced to infinite situations.






                            share|cite|improve this answer











                            $endgroup$



                            For the sake of completeness, let me mention the direction which is opposite to the OP's Question.



                            There are combinatorial optimizations problems which deal with a finite (but large) input set at the start, and the point is to optimize a real function over the input. Such problems are often messy.



                            Then, some time ago, Hungarian mathematicians started to embed the input set into a Euclidean n-space, and they'd extend the said function to a linear or convex function over the convex hull of the input. Since the optimum over the whole hull is reached at a vertex then... etc.



                            We see that sometimes situations which are strictly finite (looking for an exact answer) get hm-reduced to infinite situations.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 1 hour ago

























                            answered 1 hour ago









                            Wlod AAWlod AA

                            1,3804 silver badges14 bronze badges




                            1,3804 silver badges14 bronze badges
























                                SomeoneHAHA is a new contributor. Be nice, and check out our Code of Conduct.









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