why separable normal space has only continuum many different open subsets? I cannot prove the fact in the title. Please help!
I am reading the handbook of set theoretic topology.
And I found this fact in a proof in the book.
Thank you.
 A: This is false: $X = 2^{2^{\omega}}$ under the usual product topology is a counterexample. It is obviously normal being compact Hausdorff. Checking that $X$ is separable is a fun exercise. Finally, $X$ has more than continuum many points and hence more than continuum many open sets. The correct upper bound is $2^{2^{\omega}}$ and you can prove this by showing that $\{\{x \in X: f(x) \neq 0\} \mid f:X \to \mathbb{R}$ continuous$\}$ is a basis of size at most continuum. The above example shows that this bound is optimal.
Addendum: I looked at chapter 18 and noticed that the proof of Jones' lemma on page 784 contains the pharse "Since $X$ is separable, it has only continuum many different open sets ..." which is incorrect. The correct argument goes as follows: Given $F$ discrete closed in $X$, for each $K \subseteq F$, get an open set $U_K$ such that $K \subseteq U_K, \overline{U_K} \cap (F \setminus K) = \phi$. Let $D$ be countable dense in $X$. Suppose $K, K'$ are distinct subsets of $F$ and say $x \in K \setminus K'$. Let $x \in V \subseteq U_{K}$ where $V$ is open and disjoint with $U_{K'}$. Then $D \cap V \subseteq U_K \backslash U_{K'}$. Hence the map $K \mapsto U_K \cap D$ is injective. So $2^{|F|} \leq 2^{|D|} = 2^{\aleph_0}$.
