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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?


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

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  • $\begingroup$ 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

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 $X\setminus A\ne\varnothing$.
  • $\begingroup$ 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

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