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?

  • 2
    $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.