Exercise in Just/Weese (amoeba forcing) (1/2) I solved the following exercise (18.3): 

Can you tell me if I got it right? Thanks:
18.3.(a) The inclusion "$\supseteq$" follows immediately from the definition. For "$\subseteq$" let $U \in \mathbb A (\varepsilon)$. Then $U$ is open and $\mu(U) < \varepsilon$. Since $\mu(U \setminus F) = \mu (U) - \mu (\bigcup F)$ we can rewrite the second condition in $\mathbb A(\varepsilon)_F$ as $\mu (U) < \frac{\varepsilon + \delta_F}{2}$. We distinguish three cases: 
If $\mu(U) < \varepsilon / 2$: then $U \in \mathbb A (\varepsilon)_\varnothing$.
If $\mu(U) = \varepsilon / 2$: If $B_n$ are basic open sets with $\bigcup_n B_n = U$ then pick any $B_n$ with $\mu(B_n) > 0$ and let $F = \{B_n\}$. Then $U \in \mathbb A(\varepsilon)_F$.
If $\mu(U) > \varepsilon / 2$: Again assume $B_n$ are basic open sets with $U = \bigcup B_n$. Then $\mu (B_n) \to 0$ since the sequence is absolutely convergent. Fix $n_0$ such that $\sum_{n \le n_0} \mu (B_n) \ge \frac{\mu (U) + \varepsilon / 2}{2} = K \ge \varepsilon / 2$. Then $\mu (U) < \varepsilon = \varepsilon / 2 + \varepsilon / 2 \le \varepsilon / 2 + K$. Let $n_1$ be such that  $\sum_{n \le n_0} \mu (B_n) \ge K / 2$. Then $U \in \mathbb A (\varepsilon)_{B_0, B_1, \dots, B_{n_1}}$.

18.3. (b): Let $U,V \in \mathbb A(\varepsilon)_F$. Then $\mu ( U \cup V) \le \mu (U \setminus \bigcup F) + \mu (\bigcup F) + \mu ( V \setminus \bigcup F) < \varepsilon$.
Thanks for your help. 
 A: For (a) you really don't have to distinguish between cases (and your third case seems needlessly complicated).  Overall I cannot find any problems with the proof.  Here is my solution:

Given any nonempty $U \in \mathbb{A} (\varepsilon)$, since $U$ is open there is a sequence $\langle U_i \rangle_{i \in \omega}$ in $\mathcal{B}$ such that $U = \bigcup_{i \in \omega} U_i$.  For each $n \in \omega$ let $\delta_n = \mu_1 ( \bigcup_{i < n} U_i )$.  Since $\langle \delta_n \rangle_{n \in \omega}$ is non-decreasing with limit $\mu_1 ( U ) < \varepsilon$, there is an $n$ such that $$\mu_1 (U) - \delta_n < \varepsilon - \mu_1 ( U ),$$ but, as you have basically noted, $\mu_1 (U) = \delta_n + \mu_1 ( U \setminus {\textstyle \bigcup_{i \leq n}} U_i )$, so by rearranging we have
  $$
\mu_1 ( U \setminus {\textstyle \bigcup_{i < n}} U_i ) = \mu_1 ( U )  - \delta_n < \varepsilon - \mu_1 (U) = \varepsilon - \delta_n - \mu_1 ( U \setminus {\textstyle \bigcup_{i < n}} U_i ).
$$
  Thus $F = \{ U_i : i < n \}$ is as required.

For (b), your solution seems perfect (except to be pedantic one should note that $U \cap V$ is open).
