Prove $O = bigcup_{j=1}^infty O_j$ and $E subset bigcup_{j=1}^infty E_j implies O-E subset...












0














Prove that



$$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$





Below is my attempted proof, I'm stuck at the last expression.



Proof



$$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$



I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.










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    0














    Prove that



    $$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$





    Below is my attempted proof, I'm stuck at the last expression.



    Proof



    $$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$



    I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.










    share|cite|improve this question



























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      Prove that



      $$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$





      Below is my attempted proof, I'm stuck at the last expression.



      Proof



      $$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$



      I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.










      share|cite|improve this question















      Prove that



      $$O = bigcup_{j=1}^infty O_j quad text{and} quad E = bigcup_{j=1}^infty E_j quad implies quad O-E subset bigcup_{j=1}^{+infty}left(O_j-E_jright)$$





      Below is my attempted proof, I'm stuck at the last expression.



      Proof



      $$O - E = left(bigcup_{j=1}^infty O_jright) - left(bigcup_{j=1}^infty E_jright) = left(bigcup_{j=1}^infty O_jright) cap left(bigcup_{j=1}^infty E_jright)^c = left(bigcup_{j=1}^infty O_jright) cap left(bigcap_{j=1}^infty E^c_jright)$$



      I'm not sure how to handle the last "intersection of intersections". But I get the feeling my approach is just confusing in general. Thank you.







      elementary-set-theory






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      edited Nov 27 '18 at 1:52

























      asked Nov 26 '18 at 1:55









      Zduff

      1,544819




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          If $x in O - E$, then there exists $j$ such that $x in O_j$.
          Then $x in O_j - E_j$ as well since $x notin E_j$.






          share|cite|improve this answer





















          • What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
            – DanielWainfleet
            Nov 26 '18 at 7:01












          • @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
            – angryavian
            Nov 26 '18 at 17:09










          • I expected it was a typo but I wanted to ask first.
            – DanielWainfleet
            Nov 28 '18 at 10:36











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          1 Answer
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          If $x in O - E$, then there exists $j$ such that $x in O_j$.
          Then $x in O_j - E_j$ as well since $x notin E_j$.






          share|cite|improve this answer





















          • What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
            – DanielWainfleet
            Nov 26 '18 at 7:01












          • @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
            – angryavian
            Nov 26 '18 at 17:09










          • I expected it was a typo but I wanted to ask first.
            – DanielWainfleet
            Nov 28 '18 at 10:36
















          1














          If $x in O - E$, then there exists $j$ such that $x in O_j$.
          Then $x in O_j - E_j$ as well since $x notin E_j$.






          share|cite|improve this answer





















          • What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
            – DanielWainfleet
            Nov 26 '18 at 7:01












          • @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
            – angryavian
            Nov 26 '18 at 17:09










          • I expected it was a typo but I wanted to ask first.
            – DanielWainfleet
            Nov 28 '18 at 10:36














          1












          1








          1






          If $x in O - E$, then there exists $j$ such that $x in O_j$.
          Then $x in O_j - E_j$ as well since $x notin E_j$.






          share|cite|improve this answer












          If $x in O - E$, then there exists $j$ such that $x in O_j$.
          Then $x in O_j - E_j$ as well since $x notin E_j$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 1:57









          angryavian

          39.2k23280




          39.2k23280












          • What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
            – DanielWainfleet
            Nov 26 '18 at 7:01












          • @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
            – angryavian
            Nov 26 '18 at 17:09










          • I expected it was a typo but I wanted to ask first.
            – DanielWainfleet
            Nov 28 '18 at 10:36


















          • What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
            – DanielWainfleet
            Nov 26 '18 at 7:01












          • @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
            – angryavian
            Nov 26 '18 at 17:09










          • I expected it was a typo but I wanted to ask first.
            – DanielWainfleet
            Nov 28 '18 at 10:36
















          What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
          – DanielWainfleet
          Nov 26 '18 at 7:01






          What if $O_j=E_j={1}$ for every $j,$ and $E=emptyset$? Then ${1}=O=O-E $ but $cup_j(O_j-E_j)=cup_jemptyset=emptyset.$
          – DanielWainfleet
          Nov 26 '18 at 7:01














          @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
          – angryavian
          Nov 26 '18 at 17:09




          @DanielWainfleet I misread $E subset bigcup_j E_j$ as $E = bigcup_j E_j$. I suspsect this is a typo, since in the first step of OP's proof attempt they use the latter.
          – angryavian
          Nov 26 '18 at 17:09












          I expected it was a typo but I wanted to ask first.
          – DanielWainfleet
          Nov 28 '18 at 10:36




          I expected it was a typo but I wanted to ask first.
          – DanielWainfleet
          Nov 28 '18 at 10:36


















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