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

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.

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 $$x \notin E_j$$.
• What if $O_j=E_j=\{1\}$ for every $j,$ and $E=\emptyset$? Then $\{1\}=O=O-E$ but $\cup_j(O_j-E_j)=\cup_j\emptyset=\emptyset.$ – 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