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For two collections of sets $\{A_i\}_{i\in\mathbb{N}}$ and $\{B_i\}_{i\in\mathbb{N}}$ the question is to prove that $$\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i\subset\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$$ My attempt: $$\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]=(\bigcup\limits_{i\in\mathbb{N}}[A_i\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne i}B_n\}])\ \cup\ [\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i]$$ So $\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$ is the union of two things, one of which is the thing we want prove that it is included in $\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$

Can sombody confirm or disprove what I did please?

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    $\begingroup$ Why do you believe that the union you've given is correct? $\endgroup$ – B. Mehta Feb 20 '18 at 19:38
  • $\begingroup$ @B.Mehta the first thing I did was draw some Venn diagrams and observe the differences between the two things. Then observe that the thing on the left cannot have any element from any of the $B_i$'s but the one on the right can $\endgroup$ – John Cataldo Feb 20 '18 at 19:43
  • $\begingroup$ Venn diagrams over infinitely many sets are hard to draw and error prone, better to avoid them then. $\endgroup$ – SK19 Feb 20 '18 at 19:48
  • $\begingroup$ @SK19 well they give you a certain idea and you don't need to draw infinitely many sets to see the picture $\endgroup$ – John Cataldo Feb 20 '18 at 19:49
  • $\begingroup$ Unfortunately "see the picture" is sometimes not enough. I have edited my answer to show that your equation is wrong. $\endgroup$ – SK19 Feb 20 '18 at 20:00
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HINT

In such cases it is often useful to work with elements of the relevant sets. Show that for any $x\in\bigcup A_i\setminus \bigcup B_i$ holds $x\in\bigcup (A_i\setminus B_i)$.

Note that $x\in A\setminus B \Leftrightarrow x\in A \wedge x\notin B$ and $$x\in\bigcup_{i\in\mathbb{N}} C_i \Leftrightarrow \exists\; k\in\mathbb{N}:x\in C_k$$

EDIT: On popular demand, I will show why

$$\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]=(\bigcup\limits_{i\in\mathbb{N}}[A_i\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne i}B_n\}])\ \cup\ [\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i]$$

does not hold. (For me, $\mathbb{N}=\{1,2,3,\ldots\}$. Not relevant for the proof, just noticed it to avoid confusion.)

Set $A_1:=\{0\}$, $A_2=A_3=A_4=\ldots:=\emptyset$ and $B_1=B_2=B_3=\ldots:=\{0\}$. Clearly $\bigcup (A_i\setminus B_i)=\emptyset$ and $\bigcup A_i \setminus \bigcup B_i=\emptyset$ but $$\bigcup_{n\in\mathbb{N},n\neq i}B_n=\{0\}$$ for all $i\in\mathbb{N}$, therefore $$A_1\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne 1}B_n\}=\{0\}$$ and so the right hand side of the equation is $\{0\}$ while the LHS is $\emptyset$, contradiction.

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  • $\begingroup$ Thanks for the hint but I was rather looking for criticism regarding what I've already done (just trying to learn form errors and possibly not start from scratch just yet) $\endgroup$ – John Cataldo Feb 20 '18 at 19:47

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