Express the set of all paths with infinite $1$'s using set theoretical representation and elements of another...












1












$begingroup$


Define the collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ as
$$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
Let all infinite length paths be denoted by
$$omega=(omega_n)_{nge1}$$
Then, define $A$, the set of all paths with infinitely many up(${1}$) moves by
$$A=left{omega:summathbb{I}_{{omega_n=1}}=inftyright}$$




Express $A$ using sets of $mathscr{A}$ and 'countably many', $cup$, 'complementation', $cap$... (set theoretic representation)





My approach:

Initially, I proceeded by defining $A_1,A_2...inmathscr{A}$ as $A_k=S_ktimesOmega$, where $S_k={1,1,...$ $_k$ $_{times}$$}$ and using $underset{k=1}{overset{infty}{bigcap}}A_k={1,1,1...}$.

But, this excludes the cases the paths with finite $0$'s.

However, if I take $underset{k=1}{overset{infty}{bigcup}}A_k={1,1,1...}$, this also includes paths that do not satisfy the infinte $1$'s condition.


How must I do this?






probability-theory elementary-set-theory random-walk







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Define the collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ as
    $$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
    Let all infinite length paths be denoted by
    $$omega=(omega_n)_{nge1}$$
    Then, define $A$, the set of all paths with infinitely many up(${1}$) moves by
    $$A=left{omega:summathbb{I}_{{omega_n=1}}=inftyright}$$




    Express $A$ using sets of $mathscr{A}$ and 'countably many', $cup$, 'complementation', $cap$... (set theoretic representation)





    My approach:

    Initially, I proceeded by defining $A_1,A_2...inmathscr{A}$ as $A_k=S_ktimesOmega$, where $S_k={1,1,...$ $_k$ $_{times}$$}$ and using $underset{k=1}{overset{infty}{bigcap}}A_k={1,1,1...}$.

    But, this excludes the cases the paths with finite $0$'s.

    However, if I take $underset{k=1}{overset{infty}{bigcup}}A_k={1,1,1...}$, this also includes paths that do not satisfy the infinte $1$'s condition.


    How must I do this?






    probability-theory elementary-set-theory random-walk







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Define the collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ as
      $$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
      Let all infinite length paths be denoted by
      $$omega=(omega_n)_{nge1}$$
      Then, define $A$, the set of all paths with infinitely many up(${1}$) moves by
      $$A=left{omega:summathbb{I}_{{omega_n=1}}=inftyright}$$




      Express $A$ using sets of $mathscr{A}$ and 'countably many', $cup$, 'complementation', $cap$... (set theoretic representation)





      My approach:

      Initially, I proceeded by defining $A_1,A_2...inmathscr{A}$ as $A_k=S_ktimesOmega$, where $S_k={1,1,...$ $_k$ $_{times}$$}$ and using $underset{k=1}{overset{infty}{bigcap}}A_k={1,1,1...}$.

      But, this excludes the cases the paths with finite $0$'s.

      However, if I take $underset{k=1}{overset{infty}{bigcup}}A_k={1,1,1...}$, this also includes paths that do not satisfy the infinte $1$'s condition.


      How must I do this?






      probability-theory elementary-set-theory random-walk







      share|cite|improve this question











      $endgroup$




      Define the collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ as
      $$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
      Let all infinite length paths be denoted by
      $$omega=(omega_n)_{nge1}$$
      Then, define $A$, the set of all paths with infinitely many up(${1}$) moves by
      $$A=left{omega:summathbb{I}_{{omega_n=1}}=inftyright}$$




      Express $A$ using sets of $mathscr{A}$ and 'countably many', $cup$, 'complementation', $cap$... (set theoretic representation)





      My approach:

      Initially, I proceeded by defining $A_1,A_2...inmathscr{A}$ as $A_k=S_ktimesOmega$, where $S_k={1,1,...$ $_k$ $_{times}$$}$ and using $underset{k=1}{overset{infty}{bigcap}}A_k={1,1,1...}$.

      But, this excludes the cases the paths with finite $0$'s.

      However, if I take $underset{k=1}{overset{infty}{bigcup}}A_k={1,1,1...}$, this also includes paths that do not satisfy the infinte $1$'s condition.


      How must I do this?






      probability-theory elementary-set-theory random-walk




      probability-theory elementary-set-theory random-walk






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 22:26









      Davide Giraudo

      127k17154268




      127k17154268










      asked Jan 6 at 4:58









      Za IraZa Ira

      161115




      161115






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Let $A_k:=left{left(omega_nright)_{ngeqslant 1}in Omegamid omega_k=1right}$. Then the equality
          $$
          A=bigcap_{jgeqslant 1}bigcup_{kgeqslant j}A_k
          $$
          holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $mathcal A$, write this as $S_ktimesOmega$ where $S_k:=left{left(omega_iright)_{i=1}^kin {0,1}^kmid omega_k=1right}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
            $endgroup$
            – Za Ira
            Jan 6 at 12:13








          • 1




            $begingroup$
            I mean that $A_kinmathcal A$.
            $endgroup$
            – Davide Giraudo
            Jan 6 at 18:40










          • $begingroup$
            I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
            $endgroup$
            – Za Ira
            Jan 14 at 14:36












          • $begingroup$
            Also, could you please explain the construction of that set a little. I am a little lost.
            $endgroup$
            – Za Ira
            Jan 14 at 14:37






          • 1




            $begingroup$
            @ZaIra I edited in order to include a link and correcting the typo.
            $endgroup$
            – Davide Giraudo
            Jan 16 at 16:32











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          active

          oldest

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          1












          $begingroup$

          Let $A_k:=left{left(omega_nright)_{ngeqslant 1}in Omegamid omega_k=1right}$. Then the equality
          $$
          A=bigcap_{jgeqslant 1}bigcup_{kgeqslant j}A_k
          $$
          holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $mathcal A$, write this as $S_ktimesOmega$ where $S_k:=left{left(omega_iright)_{i=1}^kin {0,1}^kmid omega_k=1right}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
            $endgroup$
            – Za Ira
            Jan 6 at 12:13








          • 1




            $begingroup$
            I mean that $A_kinmathcal A$.
            $endgroup$
            – Davide Giraudo
            Jan 6 at 18:40










          • $begingroup$
            I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
            $endgroup$
            – Za Ira
            Jan 14 at 14:36












          • $begingroup$
            Also, could you please explain the construction of that set a little. I am a little lost.
            $endgroup$
            – Za Ira
            Jan 14 at 14:37






          • 1




            $begingroup$
            @ZaIra I edited in order to include a link and correcting the typo.
            $endgroup$
            – Davide Giraudo
            Jan 16 at 16:32
















          1












          $begingroup$

          Let $A_k:=left{left(omega_nright)_{ngeqslant 1}in Omegamid omega_k=1right}$. Then the equality
          $$
          A=bigcap_{jgeqslant 1}bigcup_{kgeqslant j}A_k
          $$
          holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $mathcal A$, write this as $S_ktimesOmega$ where $S_k:=left{left(omega_iright)_{i=1}^kin {0,1}^kmid omega_k=1right}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
            $endgroup$
            – Za Ira
            Jan 6 at 12:13








          • 1




            $begingroup$
            I mean that $A_kinmathcal A$.
            $endgroup$
            – Davide Giraudo
            Jan 6 at 18:40










          • $begingroup$
            I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
            $endgroup$
            – Za Ira
            Jan 14 at 14:36












          • $begingroup$
            Also, could you please explain the construction of that set a little. I am a little lost.
            $endgroup$
            – Za Ira
            Jan 14 at 14:37






          • 1




            $begingroup$
            @ZaIra I edited in order to include a link and correcting the typo.
            $endgroup$
            – Davide Giraudo
            Jan 16 at 16:32














          1












          1








          1





          $begingroup$

          Let $A_k:=left{left(omega_nright)_{ngeqslant 1}in Omegamid omega_k=1right}$. Then the equality
          $$
          A=bigcap_{jgeqslant 1}bigcup_{kgeqslant j}A_k
          $$
          holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $mathcal A$, write this as $S_ktimesOmega$ where $S_k:=left{left(omega_iright)_{i=1}^kin {0,1}^kmid omega_k=1right}$.






          share|cite|improve this answer











          $endgroup$



          Let $A_k:=left{left(omega_nright)_{ngeqslant 1}in Omegamid omega_k=1right}$. Then the equality
          $$
          A=bigcap_{jgeqslant 1}bigcup_{kgeqslant j}A_k
          $$
          holds (see this thread for a better understanding of the limit superior and inferior of a sequence of sets). In order to see that $A_k$ belongs to $mathcal A$, write this as $S_ktimesOmega$ where $S_k:=left{left(omega_iright)_{i=1}^kin {0,1}^kmid omega_k=1right}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 16:31

























          answered Jan 6 at 10:18









          Davide GiraudoDavide Giraudo

          127k17154268




          127k17154268












          • $begingroup$
            do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
            $endgroup$
            – Za Ira
            Jan 6 at 12:13








          • 1




            $begingroup$
            I mean that $A_kinmathcal A$.
            $endgroup$
            – Davide Giraudo
            Jan 6 at 18:40










          • $begingroup$
            I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
            $endgroup$
            – Za Ira
            Jan 14 at 14:36












          • $begingroup$
            Also, could you please explain the construction of that set a little. I am a little lost.
            $endgroup$
            – Za Ira
            Jan 14 at 14:37






          • 1




            $begingroup$
            @ZaIra I edited in order to include a link and correcting the typo.
            $endgroup$
            – Davide Giraudo
            Jan 16 at 16:32


















          • $begingroup$
            do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
            $endgroup$
            – Za Ira
            Jan 6 at 12:13








          • 1




            $begingroup$
            I mean that $A_kinmathcal A$.
            $endgroup$
            – Davide Giraudo
            Jan 6 at 18:40










          • $begingroup$
            I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
            $endgroup$
            – Za Ira
            Jan 14 at 14:36












          • $begingroup$
            Also, could you please explain the construction of that set a little. I am a little lost.
            $endgroup$
            – Za Ira
            Jan 14 at 14:37






          • 1




            $begingroup$
            @ZaIra I edited in order to include a link and correcting the typo.
            $endgroup$
            – Davide Giraudo
            Jan 16 at 16:32
















          $begingroup$
          do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
          $endgroup$
          – Za Ira
          Jan 6 at 12:13






          $begingroup$
          do you mean $A_k$ in $mathcal{A}$ or $A_k$ in $mathscr{A}$. The proof makes sense, if t's the latter.
          $endgroup$
          – Za Ira
          Jan 6 at 12:13






          1




          1




          $begingroup$
          I mean that $A_kinmathcal A$.
          $endgroup$
          – Davide Giraudo
          Jan 6 at 18:40




          $begingroup$
          I mean that $A_kinmathcal A$.
          $endgroup$
          – Davide Giraudo
          Jan 6 at 18:40












          $begingroup$
          I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
          $endgroup$
          – Za Ira
          Jan 14 at 14:36






          $begingroup$
          I think it should be $underset{jge 1}{bigcap}underset{kge j}{bigcup}{A_k}$ and not $A_j$.
          $endgroup$
          – Za Ira
          Jan 14 at 14:36














          $begingroup$
          Also, could you please explain the construction of that set a little. I am a little lost.
          $endgroup$
          – Za Ira
          Jan 14 at 14:37




          $begingroup$
          Also, could you please explain the construction of that set a little. I am a little lost.
          $endgroup$
          – Za Ira
          Jan 14 at 14:37




          1




          1




          $begingroup$
          @ZaIra I edited in order to include a link and correcting the typo.
          $endgroup$
          – Davide Giraudo
          Jan 16 at 16:32




          $begingroup$
          @ZaIra I edited in order to include a link and correcting the typo.
          $endgroup$
          – Davide Giraudo
          Jan 16 at 16:32


















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