2
$\begingroup$

This question already has an answer here:

Any compact subset of $\mathbb{R}_{l} $ must be a countable set.

Consider the open cover $\{[n,n+1): n \in \Bbb Z\}$ of $\Bbb R $ which has no subcover. So $\Bbb R $ is not compact with respect to lower limit (or Sorgenfrey) topology.

But how to answer this question?

$\endgroup$

marked as duplicate by BCLC, José Carlos Santos, Vladhagen, Cesareo, Matthew Towers Oct 25 '18 at 20:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ To get more attention especially of mathematiciens who are not 'topologists' but know something about topology I advice you to give a definition of the 'lower limit topology'. This in your question. Not in a comment. $\endgroup$ – drhab Nov 8 '14 at 8:57
3
$\begingroup$

HINT: Supppose that $X\subseteq\Bbb R_\ell$ is uncountable; you want to show that $X$ is not compact. Show that there are an $x\in X$ and a strictly increasing sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ that converges to $x$ in the usual topology. Then consider a countable open cover of $X$ that includes sets of the form $[x_n,x_{n+1})$, among others.

Added: Let $A$ be the set of points of $X$ that are not the limit of a strictly increasing sequence in $X$.

  • Show that for each $x\in A$ there is an $\epsilon_x>0$ such that $(x-\epsilon_x,x)\cap X=\varnothing$.
  • Explain why $\{(x-\epsilon_x,x):x\in A\}$ is a pairwise disjoint family of non-empty open intervals.
  • Deduce that $A$ is countable and hence that $X\setminus A\ne\varnothing$.
$\endgroup$
  • $\begingroup$ Thank you Sir @Brian M. Scott .. But I can not proceed using your hints. Please answer in detail. $\endgroup$ – Digjoy Paul Nov 9 '14 at 6:42
  • 2
    $\begingroup$ @Rajib: I’ve greatly expanded the hint for the harder part. $\endgroup$ – Brian M. Scott Nov 9 '14 at 9:20

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