Question about union of set difference vs difference of set unions

This is from Royden's Analysis 4th ed page 41. I was wondering why $\bigcup_{k=1}^{\infty}O_k\setminus E$ is just a subset of and not in fact equal to $\bigcup_{k=1}^{\infty}[O_k\setminus E_k]$. The direction highlighted is obvious, but isn't the other direction also true?

Solution 1:

I don't know the full details of the context, but I can explain why the equation

$$\bigcup_k (O_k \setminus E_k) = (\bigcup_k O_k) \setminus \bigcup_k E_k$$

does not always hold.

For example, take $O_1 = [0, 1]$ and $O_i = \emptyset$ for $i \neq 1$.

Take $E_2 = [0, 1]$ and $E_i = \emptyset$ for $i \neq 2$.

Then $\bigcup\limits_{k \in \mathbb{N}} O_k = \bigcup\limits_{k \in \mathbb{N}} E_k = [0, 1]$. Therefore, $(\bigcup\limits_{k \in \mathbb{N}} O_k) \setminus \bigcup\limits_{k \in \mathbb{N}} E_k = [0, 1] \setminus [0, 1] = \emptyset$.

On the other hand, we have $O_1 \setminus E_1 = [0, 1] \setminus \emptyset = [0, 1]$. For $i \neq 1$, we have $O_i \setminus E_i = \emptyset \setminus E_i = \emptyset$. Therefore, we have $\bigcup\limits_{k \in \mathbb{N}} (O_k \setminus E_k) = [0, 1]$.

So the two are not equal.