# Prove that topological space $\mathbb{R^2}$ with dictionary order topology is first countable, but not second countable.

Prove that topological space $$\mathbb{R^2}$$ with dictionary order topology is first countable, but not second countable.

I am a bit stuck. Some hints would help. For first countability I am having trouble finding a local base for each $$(x,y) \in \mathbb{R^2}$$. For second countability can I for example look at the first quadrant and write it as: $$\cup_{x \in \mathbb{R^{+}} } ((x,0), (x, + \infty ))$$ which is a disjunct union of uncountably many uncountable sets so there can't be a countable base ?

Assume $$\mathbb R^2$$ is second countable, then for every real number $$x\in \mathbb R$$, the open set $$\{x\}\times\mathbb R$$ contains a basis element $$U_x$$. The map $$x\mapsto U_x$$ is clearly injective, hence $$\mathbb R$$ is countable, a contradiction.
For first countability, let $$(x_1,x_2)\in \mathbb R^2$$. Let $$\mathscr U=\{U_{p,q}\mid p where $$U_{p,q}=\{x\in\mathbb R^2\mid (x_1,p)<_{dic}x<_{dic}(x_1,q)\}$$. It's easy to verify that $$\mathscr U$$ is a countable local basis of $$(x_1,x_2)$$.