$\tau$ be the topology on $\mathbb R$ generated by $\{(a,b] : a,b \in \mathbb R\}$ ; then is any basis for $(\mathbb R,\tau)$ must be uncountable ?

Question

Let $\tau$ be the topology on $\mathbb R$ generated by the basis $\{(a,b] : a,b \in \mathbb R\}$ , then is it true that any basis for $(\mathbb R,\tau)$ must be uncountable ?

@LandonCarter: are you sure? The lower limit topology is not second-countable (see here). — Watson, Sep 19 '16 at 15:13

score 1 · Answer 1 · answered Sep 19 '16 at 15:22

Let $B$ be such a countable basis. For each $x \in \mathbb{R}$, consider $( - \infty, x]$. Since $B$ is a basis, for each such interval there exists a basis element containing $x$ that is contained in such interval. Notice that for any 2 distinct intervals from our collection the basis elements associated with them are distinct (can you see why?). Thus, we have an injective map from a subset of our basis to an uncountable set, contradicting the fact that our basis was countable.

$\tau$ be the topology on $\mathbb R$ generated by $\{(a,b] : a,b \in \mathbb R\}$ ; then is any basis for $(\mathbb R,\tau)$ must be uncountable ?

1 Answers1