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

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!

• Let $B$ be a countable basis of $\tau$. Consider the family $B_C = \{ U \in B : (\exists \lambda)(U \subset C_{\lambda})\}$. Then $B_C$ is a countable family. Can you use that to find a countable subfamily of $\{ C_{\lambda} : \lambda \in \Lambda\}$ that covers $E$? – Daniel Fischer Jan 12 '17 at 21:04
• I can't... what is $B_C$? It isn't a cover right? Is it just a countable family that I'll use to find the countable subcover for $E$? – user286485 Jan 12 '17 at 21:24
• It is a cover. It's related to the $\mathscr{B}_0$ in Brian's answer. It's a larger family than $\mathscr{B}_0$ in general, but that's not important. One can do with $B_C$ exactly what Brian proposes to do with $\mathscr{B}_0$. – Daniel Fischer Jan 12 '17 at 21:29

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 $\mathscr{U}$.

• The first "why" can be answered from the fact (I already proved this) $E_2 \Rightarrow E_1$, right? – user286485 Jan 12 '17 at 21:29
• @Alnitak: Yes, that’s one way. You can also use directly the fact that $\mathscr{B}$ is a base for $\tau$. $\mathscr{U}$ covers $X$, so there is a $U_x\in\mathscr{U}$ such that $x\in U_x$. And $U_x$ is a union of members of the base $\mathscr{B}$, so at least one of those members must contain $x$. – Brian M. Scott Jan 12 '17 at 21:31
• "why is this possible?": could it be because of what is stated in the first "why?"? since the elements of $B_0$ in particular are $B_x$ for some $x$, then there will be some open set in the open cover $U$. – user286485 Jan 12 '17 at 23:07
• @Alnitak: Yes: each element of $\mathscr{B}_0$ is by definition a subset of at least one member of $\mathscr{U}$, and we just choose one such member of $\mathscr{U}$ to be $U_B$. – Brian M. Scott Jan 12 '17 at 23:11
• @Alnitak: More accurately, $\mathscr{B}_0$ is countable because it's a subset of the countable set $\mathscr{B}$. $\{U_B:B\in\mathscr{B}_0\}$ covers $E$ because $\mathscr{B}_0$ does. – Brian M. Scott Jan 12 '17 at 23:15