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Prove that

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


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

Proof

$$O - E = \left(\bigcup_{j=1}^\infty O_j\right) - \left(\bigcup_{j=1}^\infty E_j\right) = \left(\bigcup_{j=1}^\infty O_j\right) \cap \left(\bigcup_{j=1}^\infty E_j\right)^c = \left(\bigcup_{j=1}^\infty O_j\right) \cap \left(\bigcap_{j=1}^\infty E^c_j\right)$$

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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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 $x \notin E_j$.

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  • $\begingroup$ What if $O_j=E_j=\{1\}$ for every $j,$ and $E=\emptyset$? Then $\{1\}=O=O-E $ but $\cup_j(O_j-E_j)=\cup_j\emptyset=\emptyset.$ $\endgroup$ – DanielWainfleet Nov 26 '18 at 7:01
  • $\begingroup$ @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. $\endgroup$ – angryavian Nov 26 '18 at 17:09
  • $\begingroup$ I expected it was a typo but I wanted to ask first. $\endgroup$ – DanielWainfleet Nov 28 '18 at 10:36

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