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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?
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  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.
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