Counterexample: a pair of linearly ordered sets that are isomorphic to subsets of the other, but not isomorphic between themInitial segments of well-ordered sets are isomorphicOrdered sets $langle mathbbN times mathbbQ, le_lex rangle$ and $langle mathbbQ times mathbbN, le_lex rangle$ not isomorphic$langle mathbbR times mathbbR, le_textlex rangle$ and $langle mathbbR times mathbbQ, le_textlex rangle$ are not isomorphicEquinumerous well ordered sets are isomorphicProve that $mathbbZtimesmathbbN$ is not isomorphic to $mathbbZtimesmathbbZ$ (both strictly ordered): is my proof correct?If $(A_1,<_1)$ and $(A_2,<_2)$ are linearly ordered sets and $|A_1|=|A_2|<aleph_0$. Then $(A_1,<_1)$ and $(A_2,<_2)$ are isomorphicLet $(P,<),(Q,prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(P,<),(Q,prec)$ are order-isomorphicThe unique isomorphism between well-ordered setsCounterexample: linearly ordered sets for which there exists more than one isomorphismIn an infinite linearly ordered set every initial section is finite. ¿Is it isomorphic to $langlemathbbN,,,text<_mathbbNrangle$?

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Counterexample: a pair of linearly ordered sets that are isomorphic to subsets of the other, but not isomorphic between them


Initial segments of well-ordered sets are isomorphicOrdered sets $langle mathbbN times mathbbQ, le_lex rangle$ and $langle mathbbQ times mathbbN, le_lex rangle$ not isomorphic$langle mathbbR times mathbbR, le_textlex rangle$ and $langle mathbbR times mathbbQ, le_textlex rangle$ are not isomorphicEquinumerous well ordered sets are isomorphicProve that $mathbbZtimesmathbbN$ is not isomorphic to $mathbbZtimesmathbbZ$ (both strictly ordered): is my proof correct?If $(A_1,<_1)$ and $(A_2,<_2)$ are linearly ordered sets and $|A_1|=|A_2|<aleph_0$. Then $(A_1,<_1)$ and $(A_2,<_2)$ are isomorphicLet $(P,<),(Q,prec)$ be countable, dense, and linearly ordered sets without endpoints. Then $(P,<),(Q,prec)$ are order-isomorphicThe unique isomorphism between well-ordered setsCounterexample: linearly ordered sets for which there exists more than one isomorphismIn an infinite linearly ordered set every initial section is finite. ¿Is it isomorphic to $langlemathbbN,,,text<_mathbbNrangle$?













4












$begingroup$


I have encountered myself with the following exercise:




Let $langle A, <_Rrangle$ and $langle B, <_Srangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'subseteq A$ and $B'subseteq B$ such that:



$$langle A,<_Rrangleconglangle B',<_Scap(B'times B')rangleqquad&qquadlangle B,<_Srangleconglangle A',<_Rcap(A'times A')rangle$$



Is it necessarily true that $langle A,<_Rrangleconglangle B,<_Srangle$?




The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $langle B',<_Scap(B'times B')rangle$ is not isomorphic to $langle B,<_Srangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $langle B,<_Srangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.



However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?



I have tried with many examples between subsets of $mathbbR$ and other subsets of $mathbbR$, but it seems that they are all isomorphic to each other.



I ran out of ideas. Are there any counterexamples to this statement? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?



Thanks in advance for your time.



P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?










share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    I have encountered myself with the following exercise:




    Let $langle A, <_Rrangle$ and $langle B, <_Srangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'subseteq A$ and $B'subseteq B$ such that:



    $$langle A,<_Rrangleconglangle B',<_Scap(B'times B')rangleqquad&qquadlangle B,<_Srangleconglangle A',<_Rcap(A'times A')rangle$$



    Is it necessarily true that $langle A,<_Rrangleconglangle B,<_Srangle$?




    The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $langle B',<_Scap(B'times B')rangle$ is not isomorphic to $langle B,<_Srangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $langle B,<_Srangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.



    However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?



    I have tried with many examples between subsets of $mathbbR$ and other subsets of $mathbbR$, but it seems that they are all isomorphic to each other.



    I ran out of ideas. Are there any counterexamples to this statement? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?



    Thanks in advance for your time.



    P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?










    share|cite|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      I have encountered myself with the following exercise:




      Let $langle A, <_Rrangle$ and $langle B, <_Srangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'subseteq A$ and $B'subseteq B$ such that:



      $$langle A,<_Rrangleconglangle B',<_Scap(B'times B')rangleqquad&qquadlangle B,<_Srangleconglangle A',<_Rcap(A'times A')rangle$$



      Is it necessarily true that $langle A,<_Rrangleconglangle B,<_Srangle$?




      The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $langle B',<_Scap(B'times B')rangle$ is not isomorphic to $langle B,<_Srangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $langle B,<_Srangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.



      However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?



      I have tried with many examples between subsets of $mathbbR$ and other subsets of $mathbbR$, but it seems that they are all isomorphic to each other.



      I ran out of ideas. Are there any counterexamples to this statement? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?



      Thanks in advance for your time.



      P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?










      share|cite|improve this question











      $endgroup$




      I have encountered myself with the following exercise:




      Let $langle A, <_Rrangle$ and $langle B, <_Srangle$ be two linearly ordered sets so that each one is isomorphic to a subset of the other, that is, there exists $A'subseteq A$ and $B'subseteq B$ such that:



      $$langle A,<_Rrangleconglangle B',<_Scap(B'times B')rangleqquad&qquadlangle B,<_Srangleconglangle A',<_Rcap(A'times A')rangle$$



      Is it necessarily true that $langle A,<_Rrangleconglangle B,<_Srangle$?




      The original statement talked about well-ordered sets, but that case is pretty easy, because supposing that $langle B',<_Scap(B'times B')rangle$ is not isomorphic to $langle B,<_Srangle$, the theorem of comparison between well-ordered sets assures that $B'$ with its restricted relation $<_S$ is isomorphic to an initial section of $langle B,<_Srangle$. But then, composing the unique isomorphism between $A$ and $B'$ with the restriction to $B'$ of the only isomorphism between $B$ and $A'$, we find that $A$ is isomorphic to a subset of $A$ with strict upper bounds (namely, the image of the element of $B$ that defines the initial section of $B'$ via the isomorphism between $B$ and $A'$) in the sense of $<_R$, which is absurd.



      However, when the sets are not well-ordered, can we find a couple of linearly ordered sets that, although verifying the stated property, are not isomorphic to each other?



      I have tried with many examples between subsets of $mathbbR$ and other subsets of $mathbbR$, but it seems that they are all isomorphic to each other.



      I ran out of ideas. Are there any counterexamples to this statement? Or does this property actually characterize when two ordered structures are isomorphic? In case the second part holds, why is that the case?



      Thanks in advance for your time.



      P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?







      elementary-set-theory examples-counterexamples order-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Andrés E. Caicedo

      66.3k8160252




      66.3k8160252










      asked 1 hour ago









      AkerbeltzAkerbeltz

      378216




      378216




















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ with the usual ordering are order isomorphic to subsets of one another:




          • $I$, being a subset of $J$, is obviously order isomorphic to itself


          • $J$ is order isomorphic to $I' = [-frac12, frac12] subset I$ by the scaling map $omega : J ni j mapsto frac12j in I'$

          However, $I$ and $J$ are not isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.





          Indeed, order isomorphisms must preserve the existence of least and greatest elements.





          For if $j_textmin$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $omega$, then $omega(j_textmin)$ must be the least element of $I$. To see this, let $i in I$ be any element. Then the unique preimage $j = omega^-1(i) in J$ of $i$ must be greater than $j_textmin$ by definition of $j_textmin$ being the least element of $J$. That is, $j_textmin leq_J j$. So as $phi$ is order preserving, this implies $omega(j_textmin) leq_I omega(j) = i$. So, $omega(j_textmin)$ is indeed smaller than every element of $I$.




          P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?




          Using the exact same idea as before, you can show this to be true. Any closed interval of $mathbbR$ must have a least element; however, $mathbbR$ does not have a least element.






          share|cite|improve this answer











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            active

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            6












            $begingroup$

            This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ with the usual ordering are order isomorphic to subsets of one another:




            • $I$, being a subset of $J$, is obviously order isomorphic to itself


            • $J$ is order isomorphic to $I' = [-frac12, frac12] subset I$ by the scaling map $omega : J ni j mapsto frac12j in I'$

            However, $I$ and $J$ are not isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.





            Indeed, order isomorphisms must preserve the existence of least and greatest elements.





            For if $j_textmin$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $omega$, then $omega(j_textmin)$ must be the least element of $I$. To see this, let $i in I$ be any element. Then the unique preimage $j = omega^-1(i) in J$ of $i$ must be greater than $j_textmin$ by definition of $j_textmin$ being the least element of $J$. That is, $j_textmin leq_J j$. So as $phi$ is order preserving, this implies $omega(j_textmin) leq_I omega(j) = i$. So, $omega(j_textmin)$ is indeed smaller than every element of $I$.




            P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?




            Using the exact same idea as before, you can show this to be true. Any closed interval of $mathbbR$ must have a least element; however, $mathbbR$ does not have a least element.






            share|cite|improve this answer











            $endgroup$

















              6












              $begingroup$

              This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ with the usual ordering are order isomorphic to subsets of one another:




              • $I$, being a subset of $J$, is obviously order isomorphic to itself


              • $J$ is order isomorphic to $I' = [-frac12, frac12] subset I$ by the scaling map $omega : J ni j mapsto frac12j in I'$

              However, $I$ and $J$ are not isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.





              Indeed, order isomorphisms must preserve the existence of least and greatest elements.





              For if $j_textmin$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $omega$, then $omega(j_textmin)$ must be the least element of $I$. To see this, let $i in I$ be any element. Then the unique preimage $j = omega^-1(i) in J$ of $i$ must be greater than $j_textmin$ by definition of $j_textmin$ being the least element of $J$. That is, $j_textmin leq_J j$. So as $phi$ is order preserving, this implies $omega(j_textmin) leq_I omega(j) = i$. So, $omega(j_textmin)$ is indeed smaller than every element of $I$.




              P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?




              Using the exact same idea as before, you can show this to be true. Any closed interval of $mathbbR$ must have a least element; however, $mathbbR$ does not have a least element.






              share|cite|improve this answer











              $endgroup$















                6












                6








                6





                $begingroup$

                This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ with the usual ordering are order isomorphic to subsets of one another:




                • $I$, being a subset of $J$, is obviously order isomorphic to itself


                • $J$ is order isomorphic to $I' = [-frac12, frac12] subset I$ by the scaling map $omega : J ni j mapsto frac12j in I'$

                However, $I$ and $J$ are not isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.





                Indeed, order isomorphisms must preserve the existence of least and greatest elements.





                For if $j_textmin$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $omega$, then $omega(j_textmin)$ must be the least element of $I$. To see this, let $i in I$ be any element. Then the unique preimage $j = omega^-1(i) in J$ of $i$ must be greater than $j_textmin$ by definition of $j_textmin$ being the least element of $J$. That is, $j_textmin leq_J j$. So as $phi$ is order preserving, this implies $omega(j_textmin) leq_I omega(j) = i$. So, $omega(j_textmin)$ is indeed smaller than every element of $I$.




                P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?




                Using the exact same idea as before, you can show this to be true. Any closed interval of $mathbbR$ must have a least element; however, $mathbbR$ does not have a least element.






                share|cite|improve this answer











                $endgroup$



                This should work: the real intervals $I = (-1, 1)$ and $J = [-1, 1]$ with the usual ordering are order isomorphic to subsets of one another:




                • $I$, being a subset of $J$, is obviously order isomorphic to itself


                • $J$ is order isomorphic to $I' = [-frac12, frac12] subset I$ by the scaling map $omega : J ni j mapsto frac12j in I'$

                However, $I$ and $J$ are not isomorphic to each other because $J$ has a least element $-1$ and a greatest element $1$ while $I$ has neither.





                Indeed, order isomorphisms must preserve the existence of least and greatest elements.





                For if $j_textmin$ were the least element of $J$ and $J$ were order isomorphic to $I$ by an order isomorphism $omega$, then $omega(j_textmin)$ must be the least element of $I$. To see this, let $i in I$ be any element. Then the unique preimage $j = omega^-1(i) in J$ of $i$ must be greater than $j_textmin$ by definition of $j_textmin$ being the least element of $J$. That is, $j_textmin leq_J j$. So as $phi$ is order preserving, this implies $omega(j_textmin) leq_I omega(j) = i$. So, $omega(j_textmin)$ is indeed smaller than every element of $I$.




                P.S.: I have thought, for instance, that a closed interval of $mathbbR$ shouldn't be isomorphic to the whole $mathbbR$. How can we prove this assertion, if true?




                Using the exact same idea as before, you can show this to be true. Any closed interval of $mathbbR$ must have a least element; however, $mathbbR$ does not have a least element.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 32 mins ago

























                answered 1 hour ago









                ZeroXLRZeroXLR

                1,965620




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