Establish Archimedean property of a vector-lattice












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I am trying to find ways to prove the Archimedian property of a certain vector lattice and got stuck on the following type of problem.



I feel the statement below (or in fact weaker versions) should be provable, but I am not being very successful so far. I am posting this question here to get suggestions and/or counterexamples, thanks!



Statement: Let $A$ be a vector lattice (wikipedia entry) and let $Bsubseteq A$ a subset of $A$ which:



1) B is closed under vector spaces operations (of $A$), i.e.: (i) if $b_1,b_2in B$ then $b_1 + b_2 in B$, and (ii) if $bin B$ then $r bin B$ for all $rinmathbb{R}$. In other words, $B$ is a partially orderd vector subspace of $A$, but $B$ is not necessarily a lattice.



2) $B$ generates $A$, i.e., every element $ain A$ is expressible as a finite combination of meet and joins from $B$: $a = bigvee_i bigwedge_j b_{i,j}$.



3) $B$ is Archimedean, in the sense that it does not have infinitesimals (except $0$): for all $b,b^primegeq0$ in $B$, if for all $ninmathbb{N}$ it holds that $nb leq b^prime$, then $b=0$.



Under these assumptions it follows that $A$ is Archimedean as a vector lattice. I.e., for all $a,a^primegeq0$ in $A$, if for all $n$, $n a leq a^prime$ then $a=0$.



end of Statement



As I said, I have not been able to prove this so far. Perhaps there is some counterexample?










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    0












    $begingroup$


    I am trying to find ways to prove the Archimedian property of a certain vector lattice and got stuck on the following type of problem.



    I feel the statement below (or in fact weaker versions) should be provable, but I am not being very successful so far. I am posting this question here to get suggestions and/or counterexamples, thanks!



    Statement: Let $A$ be a vector lattice (wikipedia entry) and let $Bsubseteq A$ a subset of $A$ which:



    1) B is closed under vector spaces operations (of $A$), i.e.: (i) if $b_1,b_2in B$ then $b_1 + b_2 in B$, and (ii) if $bin B$ then $r bin B$ for all $rinmathbb{R}$. In other words, $B$ is a partially orderd vector subspace of $A$, but $B$ is not necessarily a lattice.



    2) $B$ generates $A$, i.e., every element $ain A$ is expressible as a finite combination of meet and joins from $B$: $a = bigvee_i bigwedge_j b_{i,j}$.



    3) $B$ is Archimedean, in the sense that it does not have infinitesimals (except $0$): for all $b,b^primegeq0$ in $B$, if for all $ninmathbb{N}$ it holds that $nb leq b^prime$, then $b=0$.



    Under these assumptions it follows that $A$ is Archimedean as a vector lattice. I.e., for all $a,a^primegeq0$ in $A$, if for all $n$, $n a leq a^prime$ then $a=0$.



    end of Statement



    As I said, I have not been able to prove this so far. Perhaps there is some counterexample?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to find ways to prove the Archimedian property of a certain vector lattice and got stuck on the following type of problem.



      I feel the statement below (or in fact weaker versions) should be provable, but I am not being very successful so far. I am posting this question here to get suggestions and/or counterexamples, thanks!



      Statement: Let $A$ be a vector lattice (wikipedia entry) and let $Bsubseteq A$ a subset of $A$ which:



      1) B is closed under vector spaces operations (of $A$), i.e.: (i) if $b_1,b_2in B$ then $b_1 + b_2 in B$, and (ii) if $bin B$ then $r bin B$ for all $rinmathbb{R}$. In other words, $B$ is a partially orderd vector subspace of $A$, but $B$ is not necessarily a lattice.



      2) $B$ generates $A$, i.e., every element $ain A$ is expressible as a finite combination of meet and joins from $B$: $a = bigvee_i bigwedge_j b_{i,j}$.



      3) $B$ is Archimedean, in the sense that it does not have infinitesimals (except $0$): for all $b,b^primegeq0$ in $B$, if for all $ninmathbb{N}$ it holds that $nb leq b^prime$, then $b=0$.



      Under these assumptions it follows that $A$ is Archimedean as a vector lattice. I.e., for all $a,a^primegeq0$ in $A$, if for all $n$, $n a leq a^prime$ then $a=0$.



      end of Statement



      As I said, I have not been able to prove this so far. Perhaps there is some counterexample?










      share|cite|improve this question









      $endgroup$




      I am trying to find ways to prove the Archimedian property of a certain vector lattice and got stuck on the following type of problem.



      I feel the statement below (or in fact weaker versions) should be provable, but I am not being very successful so far. I am posting this question here to get suggestions and/or counterexamples, thanks!



      Statement: Let $A$ be a vector lattice (wikipedia entry) and let $Bsubseteq A$ a subset of $A$ which:



      1) B is closed under vector spaces operations (of $A$), i.e.: (i) if $b_1,b_2in B$ then $b_1 + b_2 in B$, and (ii) if $bin B$ then $r bin B$ for all $rinmathbb{R}$. In other words, $B$ is a partially orderd vector subspace of $A$, but $B$ is not necessarily a lattice.



      2) $B$ generates $A$, i.e., every element $ain A$ is expressible as a finite combination of meet and joins from $B$: $a = bigvee_i bigwedge_j b_{i,j}$.



      3) $B$ is Archimedean, in the sense that it does not have infinitesimals (except $0$): for all $b,b^primegeq0$ in $B$, if for all $ninmathbb{N}$ it holds that $nb leq b^prime$, then $b=0$.



      Under these assumptions it follows that $A$ is Archimedean as a vector lattice. I.e., for all $a,a^primegeq0$ in $A$, if for all $n$, $n a leq a^prime$ then $a=0$.



      end of Statement



      As I said, I have not been able to prove this so far. Perhaps there is some counterexample?







      functional-analysis vector-spaces lattice-orders vector-lattices banach-lattices






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      asked Jan 7 at 13:36









      makkiatomakkiato

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