Consider the following lemma from "Lectures on von Neumann algebras":
I understand the proof of $(i)$ and $(ii)$. However, the proof says that $(iii)$ and $(iv)$ follow immediately from $(i)$ and $(ii)$. I can't see this. Can someone please explain this?
Since $\sigma(\mathscr E,\mathscr F)\subset \sigma(\mathscr E,\tilde{\mathscr F})$, the latter topology is stronger. Now suppose that $x_j\to x$ in $\sigma(\mathscr E,\mathscr F)$ and $x,x_j\in\mathscr E_1$ for all $j$. For any $\varphi\in\tilde{\mathscr F}$, by $(ii)$, $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous, so $\varphi(x_j)\to\varphi(x)$. Thus $x_j\to x$ in $\sigma(\mathscr E,\bar{\mathscr F})$.
If $\mathscr F$ is closed and $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous on $\mathscr E_1$, by $(ii)$ you get that $\varphi\in\mathscr F$. By $(i)$, $\varphi$ is $\sigma(\mathscr E,\mathscr F)$-continuous.