$(\Bbb R, \tau)$ with $\tau$ defined as: $U\in \tau \iff U=\emptyset \vee 0 \in U$ is not second-countable.

Question

Let $(\Bbb R,\tau)$ be a topological space with $tau$ is defined as: $$U\in \tau \iff U=\emptyset \vee 0 \in U.$$ Show that $\tau$ is not second-countable.

Attempt: The goal is to show that for any basis $\mathcal{B}$ for a topology $\tau$, then $\mathcal{B}$ is uncountable. Let $\mathcal{B}$ be any basis for topology $\tau$. By definition, for any $U \in \tau$ and for all $x \in U$, there is $B \in \mathcal{B}$ such that $x \in B \subseteq U$. But then, how to construct $U$ for which the cardinality of $\mathcal{B}$ and $\Bbb R$ are the same?

Thanks in advanced.

Answer 1

Hint:

Let $x\in \mathbb{R}$, show that any basis of $\mathbb{R}$ with this topology must contain $\{0,x\}$.