Strong Folner condition(SFC) implies the existence of a left Følner sequence.












2














I got stuck with this problem while reading Density in Arbitrary Semigroups by Hindman and Strauss. It says:



Problem: If $S$ is a countable semigroup. Then SFC on $S$ implies the existence of a left Følner sequence.



Definition of SFC: $forall; Hin mathscr{P}_f(S),; forall; epsilon>0,; exists; Kin mathscr{P}_f(S)$ such that $forall ; sin H, ; |KDelta sK|<epsilon|K|$.



Definition of left Folner sequence: Let $S$ be a semigroup. A left Følner sequence in $mathscr{P}_f(S)$ is a sequence ${F_n}_{ninmathbb{N}}$ in $mathscr{P}_f(S)$ such that for each $sin S, lim_{nrightarrowinfty}frac{|sF_nDelta F_n|}{|F_n|}=0$



($mathscr{P}_f(S)$ stands for the set of all finite subsets of $S$).



Thanks for any help.










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    2














    I got stuck with this problem while reading Density in Arbitrary Semigroups by Hindman and Strauss. It says:



    Problem: If $S$ is a countable semigroup. Then SFC on $S$ implies the existence of a left Følner sequence.



    Definition of SFC: $forall; Hin mathscr{P}_f(S),; forall; epsilon>0,; exists; Kin mathscr{P}_f(S)$ such that $forall ; sin H, ; |KDelta sK|<epsilon|K|$.



    Definition of left Folner sequence: Let $S$ be a semigroup. A left Følner sequence in $mathscr{P}_f(S)$ is a sequence ${F_n}_{ninmathbb{N}}$ in $mathscr{P}_f(S)$ such that for each $sin S, lim_{nrightarrowinfty}frac{|sF_nDelta F_n|}{|F_n|}=0$



    ($mathscr{P}_f(S)$ stands for the set of all finite subsets of $S$).



    Thanks for any help.










    share|cite|improve this question

























      2












      2








      2







      I got stuck with this problem while reading Density in Arbitrary Semigroups by Hindman and Strauss. It says:



      Problem: If $S$ is a countable semigroup. Then SFC on $S$ implies the existence of a left Følner sequence.



      Definition of SFC: $forall; Hin mathscr{P}_f(S),; forall; epsilon>0,; exists; Kin mathscr{P}_f(S)$ such that $forall ; sin H, ; |KDelta sK|<epsilon|K|$.



      Definition of left Folner sequence: Let $S$ be a semigroup. A left Følner sequence in $mathscr{P}_f(S)$ is a sequence ${F_n}_{ninmathbb{N}}$ in $mathscr{P}_f(S)$ such that for each $sin S, lim_{nrightarrowinfty}frac{|sF_nDelta F_n|}{|F_n|}=0$



      ($mathscr{P}_f(S)$ stands for the set of all finite subsets of $S$).



      Thanks for any help.










      share|cite|improve this question













      I got stuck with this problem while reading Density in Arbitrary Semigroups by Hindman and Strauss. It says:



      Problem: If $S$ is a countable semigroup. Then SFC on $S$ implies the existence of a left Følner sequence.



      Definition of SFC: $forall; Hin mathscr{P}_f(S),; forall; epsilon>0,; exists; Kin mathscr{P}_f(S)$ such that $forall ; sin H, ; |KDelta sK|<epsilon|K|$.



      Definition of left Folner sequence: Let $S$ be a semigroup. A left Følner sequence in $mathscr{P}_f(S)$ is a sequence ${F_n}_{ninmathbb{N}}$ in $mathscr{P}_f(S)$ such that for each $sin S, lim_{nrightarrowinfty}frac{|sF_nDelta F_n|}{|F_n|}=0$



      ($mathscr{P}_f(S)$ stands for the set of all finite subsets of $S$).



      Thanks for any help.







      ergodic-theory semigroups ramsey-theory amenability






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      asked Nov 29 '18 at 15:19









      SurajitSurajit

      5689




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          Hint Since $S$ is countable, you can construct an increasing sequence $H_1 subset H_2 subset ... subset H_n subset ...$ of finite subsets such that
          $$S= bigcup_n H_n$$



          For each $H_n$ you can find some $F_n$ such that
          $$
          frac{|sF_nDelta F_n|}{|F_n|} < frac{1}{n} qquad forall s in H_n
          $$



          Show that $F_n$ is a left Følner sequence.






          share|cite|improve this answer





















          • I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
            – Surajit
            Nov 29 '18 at 15:37













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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

          votes









          2














          Hint Since $S$ is countable, you can construct an increasing sequence $H_1 subset H_2 subset ... subset H_n subset ...$ of finite subsets such that
          $$S= bigcup_n H_n$$



          For each $H_n$ you can find some $F_n$ such that
          $$
          frac{|sF_nDelta F_n|}{|F_n|} < frac{1}{n} qquad forall s in H_n
          $$



          Show that $F_n$ is a left Følner sequence.






          share|cite|improve this answer





















          • I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
            – Surajit
            Nov 29 '18 at 15:37


















          2














          Hint Since $S$ is countable, you can construct an increasing sequence $H_1 subset H_2 subset ... subset H_n subset ...$ of finite subsets such that
          $$S= bigcup_n H_n$$



          For each $H_n$ you can find some $F_n$ such that
          $$
          frac{|sF_nDelta F_n|}{|F_n|} < frac{1}{n} qquad forall s in H_n
          $$



          Show that $F_n$ is a left Følner sequence.






          share|cite|improve this answer





















          • I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
            – Surajit
            Nov 29 '18 at 15:37
















          2












          2








          2






          Hint Since $S$ is countable, you can construct an increasing sequence $H_1 subset H_2 subset ... subset H_n subset ...$ of finite subsets such that
          $$S= bigcup_n H_n$$



          For each $H_n$ you can find some $F_n$ such that
          $$
          frac{|sF_nDelta F_n|}{|F_n|} < frac{1}{n} qquad forall s in H_n
          $$



          Show that $F_n$ is a left Følner sequence.






          share|cite|improve this answer












          Hint Since $S$ is countable, you can construct an increasing sequence $H_1 subset H_2 subset ... subset H_n subset ...$ of finite subsets such that
          $$S= bigcup_n H_n$$



          For each $H_n$ you can find some $F_n$ such that
          $$
          frac{|sF_nDelta F_n|}{|F_n|} < frac{1}{n} qquad forall s in H_n
          $$



          Show that $F_n$ is a left Følner sequence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 15:27









          N. S.N. S.

          102k5111207




          102k5111207












          • I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
            – Surajit
            Nov 29 '18 at 15:37




















          • I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
            – Surajit
            Nov 29 '18 at 15:37


















          I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
          – Surajit
          Nov 29 '18 at 15:37






          I almost tried in this way. But for each $H_m$, I was considering a Folner sequence $F_n^{(m)}$, and I thought, may be $F_n^{(n)}$ would work. Anyway, thank you very much for your help. Upvoted and accepted.
          – Surajit
          Nov 29 '18 at 15:37




















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