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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 ?

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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 $\mathscr U=\{U_{p,q}\mid p<x_2<q,p,q\in \mathbb Q\}$ 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 $(x_1,x_2)$.

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{ (r,x-1/n), (r,x+1/n)) : n in N } is a local base for (r,x).

Hint for not 2nd countable.
Any base has to include sets of the form ((r,a), (r.b)).

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