If $f^-1[cap_iin iB_i]subseteq f^-1[B_i]Rightarrow f^-1[cap_iin iB_i]subseteq cap_iin If^-1[B_i]$

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I may be missing something trivial, I manage to prove that
for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$



Now, why can I take the intersection and say that:



$$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$







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    up vote
    0
    down vote

    favorite












    I may be missing something trivial, I manage to prove that
    for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$



    Now, why can I take the intersection and say that:



    $$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I may be missing something trivial, I manage to prove that
      for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$



      Now, why can I take the intersection and say that:



      $$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$







      share|cite|improve this question














      I may be missing something trivial, I manage to prove that
      for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$



      Now, why can I take the intersection and say that:



      $$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 15 at 12:23









      pointguard0

      1,240821




      1,240821










      asked Aug 15 at 11:49









      newhere

      792310




      792310




















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          If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.



          Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$






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          • You clear and precise explation did the trick, thanks!
            – newhere
            Aug 15 at 15:31






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            @newhere I'm glad I could help.
            – José Carlos Santos
            Aug 15 at 16:37










          Your Answer




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          1 Answer
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          active

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          1 Answer
          1






          active

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          oldest

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          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.



          Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$






          share|cite|improve this answer




















          • You clear and precise explation did the trick, thanks!
            – newhere
            Aug 15 at 15:31






          • 1




            @newhere I'm glad I could help.
            – José Carlos Santos
            Aug 15 at 16:37














          up vote
          2
          down vote



          accepted










          If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.



          Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$






          share|cite|improve this answer




















          • You clear and precise explation did the trick, thanks!
            – newhere
            Aug 15 at 15:31






          • 1




            @newhere I'm glad I could help.
            – José Carlos Santos
            Aug 15 at 16:37












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.



          Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$






          share|cite|improve this answer












          If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.



          Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 15 at 11:55









          José Carlos Santos

          116k1699178




          116k1699178











          • You clear and precise explation did the trick, thanks!
            – newhere
            Aug 15 at 15:31






          • 1




            @newhere I'm glad I could help.
            – José Carlos Santos
            Aug 15 at 16:37
















          • You clear and precise explation did the trick, thanks!
            – newhere
            Aug 15 at 15:31






          • 1




            @newhere I'm glad I could help.
            – José Carlos Santos
            Aug 15 at 16:37















          You clear and precise explation did the trick, thanks!
          – newhere
          Aug 15 at 15:31




          You clear and precise explation did the trick, thanks!
          – newhere
          Aug 15 at 15:31




          1




          1




          @newhere I'm glad I could help.
          – José Carlos Santos
          Aug 15 at 16:37




          @newhere I'm glad I could help.
          – José Carlos Santos
          Aug 15 at 16:37












           

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