Union of Dynkin systems is not a Dynkin system (counterexample)












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If $X$ is a set and $mathcal{D_n}subset mathcal{P}(X)$ are Dynkin systems, then the union $displaystyle bigcup_{n} mathcal{D_n}$ is not always a Dynkin system. I need to find a counterexample for that, can anyone give me a hint?










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    If $X$ is a set and $mathcal{D_n}subset mathcal{P}(X)$ are Dynkin systems, then the union $displaystyle bigcup_{n} mathcal{D_n}$ is not always a Dynkin system. I need to find a counterexample for that, can anyone give me a hint?










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      If $X$ is a set and $mathcal{D_n}subset mathcal{P}(X)$ are Dynkin systems, then the union $displaystyle bigcup_{n} mathcal{D_n}$ is not always a Dynkin system. I need to find a counterexample for that, can anyone give me a hint?










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      If $X$ is a set and $mathcal{D_n}subset mathcal{P}(X)$ are Dynkin systems, then the union $displaystyle bigcup_{n} mathcal{D_n}$ is not always a Dynkin system. I need to find a counterexample for that, can anyone give me a hint?







      measure-theory elementary-set-theory examples-counterexamples






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      edited 2 days ago









      Davide Giraudo

      125k16150260




      125k16150260










      asked Mar 8 '17 at 15:15









      dimvoltdimvolt

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          $X={1,2,3,4,5,6}$,



          $$mathcal D_1=bigl{varnothing, X, {1,2}, {3,4,5,6}, {1,3}, {2,4,5,6}bigr}$$
          $$mathcal D_2=bigl{varnothing, X, {3,4}, {1,2,5,6}, {3,5}, {1,2,4,6}bigr}$$



          $mathcal D_1cup mathcal D_2$ is not a Dynkin system since ${1,2}cup{3,4}notin mathcal D_1cup mathcal D_2$, but these are disjoint sets.






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          • You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
            – dimvolt
            Mar 8 '17 at 16:23










          • Oh, cup vs cap are so close :) Thank you, i'll correct it.
            – NCh
            Mar 8 '17 at 16:25











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          1 Answer
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          $X={1,2,3,4,5,6}$,



          $$mathcal D_1=bigl{varnothing, X, {1,2}, {3,4,5,6}, {1,3}, {2,4,5,6}bigr}$$
          $$mathcal D_2=bigl{varnothing, X, {3,4}, {1,2,5,6}, {3,5}, {1,2,4,6}bigr}$$



          $mathcal D_1cup mathcal D_2$ is not a Dynkin system since ${1,2}cup{3,4}notin mathcal D_1cup mathcal D_2$, but these are disjoint sets.






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          • You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
            – dimvolt
            Mar 8 '17 at 16:23










          • Oh, cup vs cap are so close :) Thank you, i'll correct it.
            – NCh
            Mar 8 '17 at 16:25
















          2














          $X={1,2,3,4,5,6}$,



          $$mathcal D_1=bigl{varnothing, X, {1,2}, {3,4,5,6}, {1,3}, {2,4,5,6}bigr}$$
          $$mathcal D_2=bigl{varnothing, X, {3,4}, {1,2,5,6}, {3,5}, {1,2,4,6}bigr}$$



          $mathcal D_1cup mathcal D_2$ is not a Dynkin system since ${1,2}cup{3,4}notin mathcal D_1cup mathcal D_2$, but these are disjoint sets.






          share|cite|improve this answer























          • You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
            – dimvolt
            Mar 8 '17 at 16:23










          • Oh, cup vs cap are so close :) Thank you, i'll correct it.
            – NCh
            Mar 8 '17 at 16:25














          2












          2








          2






          $X={1,2,3,4,5,6}$,



          $$mathcal D_1=bigl{varnothing, X, {1,2}, {3,4,5,6}, {1,3}, {2,4,5,6}bigr}$$
          $$mathcal D_2=bigl{varnothing, X, {3,4}, {1,2,5,6}, {3,5}, {1,2,4,6}bigr}$$



          $mathcal D_1cup mathcal D_2$ is not a Dynkin system since ${1,2}cup{3,4}notin mathcal D_1cup mathcal D_2$, but these are disjoint sets.






          share|cite|improve this answer














          $X={1,2,3,4,5,6}$,



          $$mathcal D_1=bigl{varnothing, X, {1,2}, {3,4,5,6}, {1,3}, {2,4,5,6}bigr}$$
          $$mathcal D_2=bigl{varnothing, X, {3,4}, {1,2,5,6}, {3,5}, {1,2,4,6}bigr}$$



          $mathcal D_1cup mathcal D_2$ is not a Dynkin system since ${1,2}cup{3,4}notin mathcal D_1cup mathcal D_2$, but these are disjoint sets.







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          edited Mar 8 '17 at 16:25

























          answered Mar 8 '17 at 16:17









          NChNCh

          6,2882723




          6,2882723












          • You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
            – dimvolt
            Mar 8 '17 at 16:23










          • Oh, cup vs cap are so close :) Thank you, i'll correct it.
            – NCh
            Mar 8 '17 at 16:25


















          • You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
            – dimvolt
            Mar 8 '17 at 16:23










          • Oh, cup vs cap are so close :) Thank you, i'll correct it.
            – NCh
            Mar 8 '17 at 16:25
















          You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
          – dimvolt
          Mar 8 '17 at 16:23




          You probably mean ${1,2}cup{3,4}$ . Thank you for your answer!
          – dimvolt
          Mar 8 '17 at 16:23












          Oh, cup vs cap are so close :) Thank you, i'll correct it.
          – NCh
          Mar 8 '17 at 16:25




          Oh, cup vs cap are so close :) Thank you, i'll correct it.
          – NCh
          Mar 8 '17 at 16:25


















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