On $ X = (0,1) \times (0,1) $ with anti lexicographic order topology find the character of topological space $X$ ( anti lexicographic order On $ X = (0,1) \times (0,1) $ with anti lexicographic order topology find the character of topological space $X$  ( anti lexicographic order : $(x_1,y_1) < (x_2,y_2) \iff y_1 < y_2 \text{ or } ( \ y_1 = y_2 \text{ and } \ x_1<x_2 ) $.
I could use at least a hint.
Definition of character: https://proofwiki.org/wiki/Definition:Character_of_Point_in_Topological_Space
 A: For $(x,y)\in X$ the sub-space $Y_y=(0,1)\times \{y\}$ is open in $X$ and is homeomorphic to $(0,1),$ which  is $1$st-countable. In any space $X,$ if $p\in Y\subset X$ where $Y$ is a $1$st-countable open sub-space of $X$ then $p$ has countable character in $X.$
Note : $(x,y)$ denotes an ordered pair but $(0,1)$ denotes an  interval.
A: It is easy to see $X$ is first countable (e.g. see here), which means $\chi(x, X)$ (character of $x$) $\leq \aleph_0$. Now we want to prove $\chi(x, X)\geq \aleph_0$. For the sake of contradiction, let $\mathcal B_x$ be a finite local base of $x$. The intersection $\bigcap \mathcal B_x$ is clearly open. $\bigcap \mathcal B_x\neq\{x\}$ as one-point sets are not open in $X$. If $\bigcap\mathcal B_x$ consists of at least two points, say $y\neq x$, then $X$ cannot be $T_1$. For if $U$ is an open set containing $x$ but not $y$, by definition of a local base, there exists $B_i\in \mathcal B_x$ such that $B_i\subset U$. As $\bigcap\mathcal B_x\subset B_i\subset U$, $\bigcap\mathcal B_x$ cannot contain $y$, a contradiction. We know that order topology is $T_1$ (in fact, completely normal), thus a finite local base of $x$ doesn't exist, which means $\chi(x, X)=\aleph_0$.
