The closure of convex hull compact












1














Let $A_n={x_n,x_{n+1},...}subset E$ for each $nin mathbb N$, such that $E$ is a Banach space.

If the closure of the convex hull of $A_n$ is compact i.e $overline{co}(A_n)$ compact, is $overline{co}(A_1)$ compact?



what we can say about $A_1$?






real-analysis functional-analysis compactness







share|cite|improve this question



























    1














    Let $A_n={x_n,x_{n+1},...}subset E$ for each $nin mathbb N$, such that $E$ is a Banach space.

    If the closure of the convex hull of $A_n$ is compact i.e $overline{co}(A_n)$ compact, is $overline{co}(A_1)$ compact?



    what we can say about $A_1$?






    real-analysis functional-analysis compactness







    share|cite|improve this question

























      1












      1








      1


      1





      Let $A_n={x_n,x_{n+1},...}subset E$ for each $nin mathbb N$, such that $E$ is a Banach space.

      If the closure of the convex hull of $A_n$ is compact i.e $overline{co}(A_n)$ compact, is $overline{co}(A_1)$ compact?



      what we can say about $A_1$?






      real-analysis functional-analysis compactness







      share|cite|improve this question













      Let $A_n={x_n,x_{n+1},...}subset E$ for each $nin mathbb N$, such that $E$ is a Banach space.

      If the closure of the convex hull of $A_n$ is compact i.e $overline{co}(A_n)$ compact, is $overline{co}(A_1)$ compact?



      what we can say about $A_1$?






      real-analysis functional-analysis compactness




      real-analysis functional-analysis compactness






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 at 10:32









      Motaka

      229111




      229111






















          2 Answers
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          active

          oldest

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          1














          It is well-know that the closure of the convex-hull $overline{mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $overline{A_n} subset overline{mathrm{Co}(A_n)}$. Thus, if $overline{mathrm{Co}(A_n)}$ is compact, then $overline{A_n}$ is compact too. Since $$A_1 subset {x_1,ldots, x_{n-1}} cup overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $overline{mathrm{Co}(A_1)}$ is compact. In fact, note that
          $$overline{mathrm{Co}(A_1)} subset overline{mathrm{Co}(K)}$$
          and the last set is compact (because $K$ is compact).






          share|cite|improve this answer





















          • Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
            – Motaka
            Nov 26 at 11:15






          • 1




            $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
            – p4sch
            Nov 26 at 11:20










          • I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
            – Motaka
            Nov 26 at 12:15








          • 1




            Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
            – p4sch
            Nov 26 at 12:17










          • Exactly, I forget this one , Thank you so much
            – Motaka
            Nov 26 at 12:18



















          1














          It suffices to show the following (and then use induction)



          Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0in E$, then $,overline{mathrm{co}},(Kcup{x_0})$ is also compact.



          Proof of the Claim. Let ${z_n}subset overline{mathrm{co}},(Kcup{x_0})$, we need to show that there exists a converging subsequence of ${z_n}$ with limit in
          $overline{mathrm{co}},(Kcup{x_0})$.



          First of all, there exists another sequence,
          ${w_n}subset {mathrm{co}},(Kcup{x_0})$, such that
          $$
          |z_n-w_n|<frac{1}{n}qquadtext{and}qquad w_n=t_nk_n+(1-t_n)x_0,
          $$
          where $k_nin K$ and $t_nin[0,1]$. Clearly, ${t_n}$ possesses and converging subsequence
          ${t_{n_j}}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that
          $$
          t_ellto tin[0,1]qquadtext{and}qquad k_ellto kin K.
          $$
          Then
          $$
          w_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0})
          $$
          and as $z_ell-w_ellto 0$, we also have that
          $$
          z_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0}).
          $$
          Hence, every sequence in
          $overline{mathrm{co}},(Kcup{x_0})$
          possesses a converging subsequenc in
          $overline{mathrm{co}},(Kcup{x_0})$.






          share|cite|improve this answer





















          • Thank you @Yiorgos S. Smyrlis
            – Motaka
            Nov 26 at 13:58











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          2 Answers
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          2 Answers
          2






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          active

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          active

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          1














          It is well-know that the closure of the convex-hull $overline{mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $overline{A_n} subset overline{mathrm{Co}(A_n)}$. Thus, if $overline{mathrm{Co}(A_n)}$ is compact, then $overline{A_n}$ is compact too. Since $$A_1 subset {x_1,ldots, x_{n-1}} cup overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $overline{mathrm{Co}(A_1)}$ is compact. In fact, note that
          $$overline{mathrm{Co}(A_1)} subset overline{mathrm{Co}(K)}$$
          and the last set is compact (because $K$ is compact).






          share|cite|improve this answer





















          • Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
            – Motaka
            Nov 26 at 11:15






          • 1




            $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
            – p4sch
            Nov 26 at 11:20










          • I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
            – Motaka
            Nov 26 at 12:15








          • 1




            Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
            – p4sch
            Nov 26 at 12:17










          • Exactly, I forget this one , Thank you so much
            – Motaka
            Nov 26 at 12:18
















          1














          It is well-know that the closure of the convex-hull $overline{mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $overline{A_n} subset overline{mathrm{Co}(A_n)}$. Thus, if $overline{mathrm{Co}(A_n)}$ is compact, then $overline{A_n}$ is compact too. Since $$A_1 subset {x_1,ldots, x_{n-1}} cup overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $overline{mathrm{Co}(A_1)}$ is compact. In fact, note that
          $$overline{mathrm{Co}(A_1)} subset overline{mathrm{Co}(K)}$$
          and the last set is compact (because $K$ is compact).






          share|cite|improve this answer





















          • Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
            – Motaka
            Nov 26 at 11:15






          • 1




            $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
            – p4sch
            Nov 26 at 11:20










          • I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
            – Motaka
            Nov 26 at 12:15








          • 1




            Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
            – p4sch
            Nov 26 at 12:17










          • Exactly, I forget this one , Thank you so much
            – Motaka
            Nov 26 at 12:18














          1












          1








          1






          It is well-know that the closure of the convex-hull $overline{mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $overline{A_n} subset overline{mathrm{Co}(A_n)}$. Thus, if $overline{mathrm{Co}(A_n)}$ is compact, then $overline{A_n}$ is compact too. Since $$A_1 subset {x_1,ldots, x_{n-1}} cup overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $overline{mathrm{Co}(A_1)}$ is compact. In fact, note that
          $$overline{mathrm{Co}(A_1)} subset overline{mathrm{Co}(K)}$$
          and the last set is compact (because $K$ is compact).






          share|cite|improve this answer












          It is well-know that the closure of the convex-hull $overline{mathrm{Co}(K)}$ is compact, if $K$ is compact and $E$ is a Banach space. Note that $overline{A_n} subset overline{mathrm{Co}(A_n)}$. Thus, if $overline{mathrm{Co}(A_n)}$ is compact, then $overline{A_n}$ is compact too. Since $$A_1 subset {x_1,ldots, x_{n-1}} cup overline{A_n} =:K$$ and the latter set $K$ is compact, we get that $overline{mathrm{Co}(A_1)}$ is compact. In fact, note that
          $$overline{mathrm{Co}(A_1)} subset overline{mathrm{Co}(K)}$$
          and the last set is compact (because $K$ is compact).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 10:59









          p4sch

          4,800217




          4,800217












          • Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
            – Motaka
            Nov 26 at 11:15






          • 1




            $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
            – p4sch
            Nov 26 at 11:20










          • I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
            – Motaka
            Nov 26 at 12:15








          • 1




            Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
            – p4sch
            Nov 26 at 12:17










          • Exactly, I forget this one , Thank you so much
            – Motaka
            Nov 26 at 12:18


















          • Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
            – Motaka
            Nov 26 at 11:15






          • 1




            $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
            – p4sch
            Nov 26 at 11:20










          • I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
            – Motaka
            Nov 26 at 12:15








          • 1




            Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
            – p4sch
            Nov 26 at 12:17










          • Exactly, I forget this one , Thank you so much
            – Motaka
            Nov 26 at 12:18
















          Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
          – Motaka
          Nov 26 at 11:15




          Great, thank you. Now we know that $A_1$ is relatively compact. in the paper that i'm reading, they say that ${x_n}$ has a convergent subsequence, do you have any idea why?
          – Motaka
          Nov 26 at 11:15




          1




          1




          $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
          – p4sch
          Nov 26 at 11:20




          $overline{A_1}$ is compact. Since the notation of compactness and sequential compactness is equivalent in metric spaces, your statement is a direct consequence of compactness.
          – p4sch
          Nov 26 at 11:20












          I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
          – Motaka
          Nov 26 at 12:15






          I will rephrase my statement. $overline{A_1}=overline{{x1,x_2,...}}$ is compact so its sequential compact, but why $A_1$ is sequential compact
          – Motaka
          Nov 26 at 12:15






          1




          1




          Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
          – p4sch
          Nov 26 at 12:17




          Any sequence in $A_1$ also also a sequence in $overline{A_1}$. Thus it has a convergent subsequence. However, the limes lies in $overline{A_1}$ and can be not in $A_1$. (The paper doesn't say that the limes is in $A_1$!)
          – p4sch
          Nov 26 at 12:17












          Exactly, I forget this one , Thank you so much
          – Motaka
          Nov 26 at 12:18




          Exactly, I forget this one , Thank you so much
          – Motaka
          Nov 26 at 12:18











          1














          It suffices to show the following (and then use induction)



          Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0in E$, then $,overline{mathrm{co}},(Kcup{x_0})$ is also compact.



          Proof of the Claim. Let ${z_n}subset overline{mathrm{co}},(Kcup{x_0})$, we need to show that there exists a converging subsequence of ${z_n}$ with limit in
          $overline{mathrm{co}},(Kcup{x_0})$.



          First of all, there exists another sequence,
          ${w_n}subset {mathrm{co}},(Kcup{x_0})$, such that
          $$
          |z_n-w_n|<frac{1}{n}qquadtext{and}qquad w_n=t_nk_n+(1-t_n)x_0,
          $$
          where $k_nin K$ and $t_nin[0,1]$. Clearly, ${t_n}$ possesses and converging subsequence
          ${t_{n_j}}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that
          $$
          t_ellto tin[0,1]qquadtext{and}qquad k_ellto kin K.
          $$
          Then
          $$
          w_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0})
          $$
          and as $z_ell-w_ellto 0$, we also have that
          $$
          z_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0}).
          $$
          Hence, every sequence in
          $overline{mathrm{co}},(Kcup{x_0})$
          possesses a converging subsequenc in
          $overline{mathrm{co}},(Kcup{x_0})$.






          share|cite|improve this answer





















          • Thank you @Yiorgos S. Smyrlis
            – Motaka
            Nov 26 at 13:58
















          1














          It suffices to show the following (and then use induction)



          Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0in E$, then $,overline{mathrm{co}},(Kcup{x_0})$ is also compact.



          Proof of the Claim. Let ${z_n}subset overline{mathrm{co}},(Kcup{x_0})$, we need to show that there exists a converging subsequence of ${z_n}$ with limit in
          $overline{mathrm{co}},(Kcup{x_0})$.



          First of all, there exists another sequence,
          ${w_n}subset {mathrm{co}},(Kcup{x_0})$, such that
          $$
          |z_n-w_n|<frac{1}{n}qquadtext{and}qquad w_n=t_nk_n+(1-t_n)x_0,
          $$
          where $k_nin K$ and $t_nin[0,1]$. Clearly, ${t_n}$ possesses and converging subsequence
          ${t_{n_j}}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that
          $$
          t_ellto tin[0,1]qquadtext{and}qquad k_ellto kin K.
          $$
          Then
          $$
          w_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0})
          $$
          and as $z_ell-w_ellto 0$, we also have that
          $$
          z_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0}).
          $$
          Hence, every sequence in
          $overline{mathrm{co}},(Kcup{x_0})$
          possesses a converging subsequenc in
          $overline{mathrm{co}},(Kcup{x_0})$.






          share|cite|improve this answer





















          • Thank you @Yiorgos S. Smyrlis
            – Motaka
            Nov 26 at 13:58














          1












          1








          1






          It suffices to show the following (and then use induction)



          Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0in E$, then $,overline{mathrm{co}},(Kcup{x_0})$ is also compact.



          Proof of the Claim. Let ${z_n}subset overline{mathrm{co}},(Kcup{x_0})$, we need to show that there exists a converging subsequence of ${z_n}$ with limit in
          $overline{mathrm{co}},(Kcup{x_0})$.



          First of all, there exists another sequence,
          ${w_n}subset {mathrm{co}},(Kcup{x_0})$, such that
          $$
          |z_n-w_n|<frac{1}{n}qquadtext{and}qquad w_n=t_nk_n+(1-t_n)x_0,
          $$
          where $k_nin K$ and $t_nin[0,1]$. Clearly, ${t_n}$ possesses and converging subsequence
          ${t_{n_j}}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that
          $$
          t_ellto tin[0,1]qquadtext{and}qquad k_ellto kin K.
          $$
          Then
          $$
          w_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0})
          $$
          and as $z_ell-w_ellto 0$, we also have that
          $$
          z_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0}).
          $$
          Hence, every sequence in
          $overline{mathrm{co}},(Kcup{x_0})$
          possesses a converging subsequenc in
          $overline{mathrm{co}},(Kcup{x_0})$.






          share|cite|improve this answer












          It suffices to show the following (and then use induction)



          Claim. If $K$ is a convex and compact subset of a Banach space $E$ and $x_0in E$, then $,overline{mathrm{co}},(Kcup{x_0})$ is also compact.



          Proof of the Claim. Let ${z_n}subset overline{mathrm{co}},(Kcup{x_0})$, we need to show that there exists a converging subsequence of ${z_n}$ with limit in
          $overline{mathrm{co}},(Kcup{x_0})$.



          First of all, there exists another sequence,
          ${w_n}subset {mathrm{co}},(Kcup{x_0})$, such that
          $$
          |z_n-w_n|<frac{1}{n}qquadtext{and}qquad w_n=t_nk_n+(1-t_n)x_0,
          $$
          where $k_nin K$ and $t_nin[0,1]$. Clearly, ${t_n}$ possesses and converging subsequence
          ${t_{n_j}}$ and $k_{n_j}$ possesses also a converging subsequence. For simplicity, and economy of indices, say that
          $$
          t_ellto tin[0,1]qquadtext{and}qquad k_ellto kin K.
          $$
          Then
          $$
          w_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0})
          $$
          and as $z_ell-w_ellto 0$, we also have that
          $$
          z_ellto tk+(1-k)x_0in overline{mathrm{co}},(Kcup{x_0}).
          $$
          Hence, every sequence in
          $overline{mathrm{co}},(Kcup{x_0})$
          possesses a converging subsequenc in
          $overline{mathrm{co}},(Kcup{x_0})$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 11:13









          Yiorgos S. Smyrlis

          62.4k1383162




          62.4k1383162












          • Thank you @Yiorgos S. Smyrlis
            – Motaka
            Nov 26 at 13:58


















          • Thank you @Yiorgos S. Smyrlis
            – Motaka
            Nov 26 at 13:58
















          Thank you @Yiorgos S. Smyrlis
          – Motaka
          Nov 26 at 13:58




          Thank you @Yiorgos S. Smyrlis
          – Motaka
          Nov 26 at 13:58


















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          Tribalistas

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            Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
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          Listed building

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          Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
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