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.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.