Proving theorem on indexing sets.












0












$begingroup$


Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:




  1. $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$


I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.



My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.



Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.






real-analysis elementary-set-theory proof-explanation







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  • 4




    $begingroup$
    That's not De Morgan's laws (no negations anywhere); it is just distributivity.
    $endgroup$
    – Henning Makholm
    Jan 8 at 2:41










  • $begingroup$
    So, in order to prove this identity holds true, how can I begin?
    $endgroup$
    – Ryan
    Jan 8 at 5:59
















0












$begingroup$


Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:




  1. $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$


I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.



My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.



Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.






real-analysis elementary-set-theory proof-explanation







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    That's not De Morgan's laws (no negations anywhere); it is just distributivity.
    $endgroup$
    – Henning Makholm
    Jan 8 at 2:41










  • $begingroup$
    So, in order to prove this identity holds true, how can I begin?
    $endgroup$
    – Ryan
    Jan 8 at 5:59














0












0








0





$begingroup$


Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:




  1. $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$


I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.



My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.



Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.






real-analysis elementary-set-theory proof-explanation







share|cite|improve this question









$endgroup$




Let $Lambda$ be an arbitrary indexing set and let {$B_alpha$}$_{alphain Lambda}$ be a collection of sets indexed by $Lambda$. Let $C$ be any set. Then the following identities hold:




  1. $Ccapbigcup_{alphainLambda}B_alpha = bigcup_{alphainLambda}(Ccap B_{alpha})$


I haven't got much into it at all, but would the notion of DeMorgan's Laws apply to this theorem at all? Just poking ideas at it.



My attempt at reading this, which is probably incorrect: "The intersection of a set $C$ and the union over $alphainLambda$ of the $B_{alpha}$'s is equal to the union over $alphainLambda$ on the set $C$'s intersection over $B_{alpha}$'s.



Apologies if I completely butchered it - this is the first time seeing these notations presented in this form. Also, we are supposed to prove why that theorem (1) holds true.






real-analysis elementary-set-theory proof-explanation




real-analysis elementary-set-theory proof-explanation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 2:33









RyanRyan

1156




1156








  • 4




    $begingroup$
    That's not De Morgan's laws (no negations anywhere); it is just distributivity.
    $endgroup$
    – Henning Makholm
    Jan 8 at 2:41










  • $begingroup$
    So, in order to prove this identity holds true, how can I begin?
    $endgroup$
    – Ryan
    Jan 8 at 5:59














  • 4




    $begingroup$
    That's not De Morgan's laws (no negations anywhere); it is just distributivity.
    $endgroup$
    – Henning Makholm
    Jan 8 at 2:41










  • $begingroup$
    So, in order to prove this identity holds true, how can I begin?
    $endgroup$
    – Ryan
    Jan 8 at 5:59








4




4




$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41




$begingroup$
That's not De Morgan's laws (no negations anywhere); it is just distributivity.
$endgroup$
– Henning Makholm
Jan 8 at 2:41












$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59




$begingroup$
So, in order to prove this identity holds true, how can I begin?
$endgroup$
– Ryan
Jan 8 at 5:59










2 Answers
2






active

oldest

votes


















1












$begingroup$

The general case is probably a bit distracting. Try a simple example first. Prove that



$C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$



That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Appreciate your feedback, given the errors in my original question.
    $endgroup$
    – Ryan
    Jan 9 at 5:06










  • $begingroup$
    You're welcome. I'm glad I could help!
    $endgroup$
    – Metric
    Jan 9 at 5:06



















0












$begingroup$

Tough it out point by point.

x in C $cap$ $cup${ B$_a$ : a in A }

iff x in C and exists a in A with x in B$_a$

iff ....






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The general case is probably a bit distracting. Try a simple example first. Prove that



    $C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$



    That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! Appreciate your feedback, given the errors in my original question.
      $endgroup$
      – Ryan
      Jan 9 at 5:06










    • $begingroup$
      You're welcome. I'm glad I could help!
      $endgroup$
      – Metric
      Jan 9 at 5:06
















    1












    $begingroup$

    The general case is probably a bit distracting. Try a simple example first. Prove that



    $C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$



    That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! Appreciate your feedback, given the errors in my original question.
      $endgroup$
      – Ryan
      Jan 9 at 5:06










    • $begingroup$
      You're welcome. I'm glad I could help!
      $endgroup$
      – Metric
      Jan 9 at 5:06














    1












    1








    1





    $begingroup$

    The general case is probably a bit distracting. Try a simple example first. Prove that



    $C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$



    That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.






    share|cite|improve this answer









    $endgroup$



    The general case is probably a bit distracting. Try a simple example first. Prove that



    $C cap (B_1 cup B_2) = (C cap B_1) cup (C cap B_2)$



    That is, show that each set is a subset of the other. If you can get this, then the ideas involved in the proof to your general case is essentially identical.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 9 at 5:05









    MetricMetric

    91839




    91839












    • $begingroup$
      Thanks! Appreciate your feedback, given the errors in my original question.
      $endgroup$
      – Ryan
      Jan 9 at 5:06










    • $begingroup$
      You're welcome. I'm glad I could help!
      $endgroup$
      – Metric
      Jan 9 at 5:06


















    • $begingroup$
      Thanks! Appreciate your feedback, given the errors in my original question.
      $endgroup$
      – Ryan
      Jan 9 at 5:06










    • $begingroup$
      You're welcome. I'm glad I could help!
      $endgroup$
      – Metric
      Jan 9 at 5:06
















    $begingroup$
    Thanks! Appreciate your feedback, given the errors in my original question.
    $endgroup$
    – Ryan
    Jan 9 at 5:06




    $begingroup$
    Thanks! Appreciate your feedback, given the errors in my original question.
    $endgroup$
    – Ryan
    Jan 9 at 5:06












    $begingroup$
    You're welcome. I'm glad I could help!
    $endgroup$
    – Metric
    Jan 9 at 5:06




    $begingroup$
    You're welcome. I'm glad I could help!
    $endgroup$
    – Metric
    Jan 9 at 5:06











    0












    $begingroup$

    Tough it out point by point.

    x in C $cap$ $cup${ B$_a$ : a in A }

    iff x in C and exists a in A with x in B$_a$

    iff ....






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Tough it out point by point.

      x in C $cap$ $cup${ B$_a$ : a in A }

      iff x in C and exists a in A with x in B$_a$

      iff ....






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Tough it out point by point.

        x in C $cap$ $cup${ B$_a$ : a in A }

        iff x in C and exists a in A with x in B$_a$

        iff ....






        share|cite|improve this answer









        $endgroup$



        Tough it out point by point.

        x in C $cap$ $cup${ B$_a$ : a in A }

        iff x in C and exists a in A with x in B$_a$

        iff ....







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 4:06









        William ElliotWilliam Elliot

        7,7322720




        7,7322720






























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