Proving a topological space is first-countable and investigating if it is separable So i have $X=\mathbb{R^2}$ and let $\mathcal{T}$ be a topology generated by lexicographical order.
And i need to prove it's first-countable, which means that every point x has a countable local base.
And also i need to find out if it's separable or not(if it has a countable dense subset).
I would be grateful for any help.
 A: There clearly is no maximum nor a minimum, so basic neighbourhoods of $(x,y)$ look like intervals $\left((a,b), (c,d)\right)$ where $(a,b) <_l (x,y) <_l (c,d)$. So either $a < x$ or $a=x$ and $b < y$. In the latter case $$(a,b) <_l (x,b) <_l (x,y)$$ as well, and in the former case $$(a,b) <_l (x,y-1) <_l (x,y)\text{.}$$ Also either $x < c$ ,and then 
$$(x,y) <_l (x,y+1) <_l (c,d)$$ or $x=c$ and $y < d$ and then
$$(x,y) <_l (x,d’) <_l (c,d)$$ for any $d’$ in $(y,d)$
So any open neighbourhood of $(x,y)$ contains a neighbourhood of the form $((x,a’), (x,b’))_l = \{x\} \times (a’, b’)$ for  some $a’< y< b’$. It then follows that the set $$\{(\{x\} \times (y-\frac{1}{n}, y+\frac{1}{n}): n \in \mathbb{N}\}$$ is a countable neighbourhood base at $(x,y)$.
$\mathbb{R}^2$ is thus essentially a bunch of vertical copies of $\mathbb{R}$, stacked side by side, and these copies hardly interact, we have a so-called topological sum here. 
If $D$ is dense then as $\{x\} \times \mathbb{R}$ is open, it must intersect $D$.
So for any $x \in \mathbb{R}$ there is some $(x,d_x) \in D$ and these are all different for distinct $x$, so $|D| \ge |\mathbb{R}|$.
A: $\{ \{ x\}  \times (y-\frac {1}{n},y+\frac {1}{n}): n\in \Bbb N\}$ is a countable local base at $(x,y)\in \Bbb R^2.$
$F=\{\{x\}\times \Bbb R: x\in \Bbb R\}$ is an uncountable family of pair-wise disjoint open sets, and a dense subset of $\Bbb R^2$ must intersect every member of $F.$  
