# A sequence in generalized order spaces

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 $$X^*$$?

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

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.