$\{1-\frac{1}{n} \}_n$ does not converge in the Sorgenfry line

Question

I am trying to prove that the sequence $\{1-\frac{1}{n} \}_n$ does not converge in the Sorgenfry line. Below is my attempt.

Consider $\{1-\frac{1}{n}\}_n=(0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\cdots)$. If $\{1-\frac{1}{n}\}_n$ converges, then $\{1-\frac{1}{n}\}_n \rightarrow 1$, and so by definition for each $W \in \tau$ containing $1$ there is an $N \in \mathbb{N}$ such that $x_n \in W$ for all $n \geq N$.

Consider $[1,2)$. Then $1 \in [1,2)$, however $1 - \frac{1}{n} \notin [1,2)$ for all $n \in \mathbb{N}$.

Hence $\{1-\frac{1}{n} \}_n$ does not converge in the Sorgenfrey line.

Answer 1

Your claim that if the sequence converges, then it converges to $1$ requires proof. A slight variation on the argument you gave shows that the sequence does not converge to any value $L\ge 1$. Now you need an argument why the limit can't be any $L<1$. That case will be quite like showing the same fact for the standard topology.

Alternatively, the Sorgenfrey line is quasi-metrizable and you can show the sequence is not Cauchy.