Let $X$ be a GO-space with the topology $\tau$ and $\lambda$ be the usual open interval topology on $X$. Put
$$ R= \{x\in X: [x, \rightarrow) \in \tau\setminus \lambda \} \text{ and } L= \{x\in X: (\leftarrow,x] \in \tau\setminus \lambda \}. $$
Define $X^\* \subset X\times \mathbb{Z}$ as follows:
$$X^\*=(X\times \{0\}) \cup (R \times \{k \in \mathbb{Z}: k<0\})\cup (L \times \{k \in \mathbb{Z}: k>0\}).$$
Let $X^\*$ have the open interval topology generated by the lexicographical order. Then $f: X \rightarrow X^\*$ defined by $f(x)=\langle x, 0\rangle$
is an order-preserving homeomorphism from $X$ onto the closed subspace $X\times \{0\}$ of $X^\*$. So the space $X^\*$ is a closed linearly ordered
extension of $X$.


My questions are as follows:
1. Let $\{a\_n=\langle x\_n, k\_n\rangle\}$ be a sequence of $X^\*$ and let the sequence $\{x\_n\} \subset X$ be convergent to $x \in X$. Does $\{a\_n\}$ converge to the point $\langle x, 0\rangle$ of $X^\*$?


2. Suppose that $X$ is first countable. Is also $X^\*$ first countable?



A sequence in generalized order spaces
1. Oddly enough, no: take $x\in R$ and let $a\_n=\langle x,-n\rangle$ for all $n$ (so $x\_n=x$ and $k\_n=-n$) then $\{x\_n\}$ is constant and converges to $x$, but $\{a\_n\}$ does not converge at all.
2. Yes; just consider cases, say if $x\in R\setminus L$ and $\{[x,c\_n):n\in\mathbb{N}\}$ is a local base for $X$ at $x$ then $\{(\langle x,-1\rangle,\langle c\_n,0\rangle):n\in\mathbb{N}\}$ is a local base for $X^\*$ at $\langle x,0\rangle$. The extra points are isolated so they pose no problem.


