If $(E,\tau)$ is a topological space $E_2$, then $(E,\tau)$ is Lindelöf and separable.

Question

I can't prove this statement:

If $(E,\tau)$ is a topological space $E_2$ (There is a countable basis for $\tau$), then $(E,\tau)$ is Lindelöf and separable.

I tried to prove Lindelöf first. Let $C = \{ C_\lambda \}$ be an open cover for $(E, \tau)$. Then I need to prove that for some $k \in \mathbb{R}$ there is $C_k = \{ C_{\lambda,k} \}$ a countable subcover for $(E,\tau)$. How can I use the fact that there is a countable basis for finding this subcover?

After my failure, I tried to prove separable. I need to prove that there is a countable dense subset of $(E,\tau)$. But I don't know how. Maybe I'm lacking creativity?

Any help will be appreciated. Thanks!

Answer 1

HINT: Let $\mathscr{B}$ be a countable base for $\tau$.

For separability pick a point $x_B\in B$ for each $B\in\mathscr{B}$ and consider the set $\{x_B:B\in\mathscr{B}\}$.

For the Lindelöf property, let $\mathscr{U}$ be an open cover of $E$.

• For each $x\in E$ there are a $U_x\in\mathscr{U}$ and a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq U_x$. Why?

Let $\mathscr{B}_0=\{B_x:x\in E\}$.

• For each $B\in\mathscr{B}_0$ choose a $U_B\in\mathscr{U}$ such that $B\subseteq U_B$. Why is this possible?

• Show that $\{U_B:B\in\mathscr{B}_0\}$ is a countable subcover of $\mathscr{U}$.