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Why is $[a,\infty)$ both closed and open in the lower limit topology $\mathbb{R}_l$, actually I do get why it is closed (because it is the complement of the open set $(-\infty, a),$ but I do not get why it is considered open.

Generally speaking if I have some set $[2,5),$ if I want to check this set is open, then do I compute its complement and see if its closed? For instance, the complement of $[2,5)$ seems to be $(-\infty,2) \cup [5,\infty)$, but I fail to see how this is closed. Could anyone give me some thorough explanation regarding this confusion? Help would be much appreciated.

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  • $\begingroup$ What is the lower limit topology? Are you working on $\mathbb{R}$? $\endgroup$ – user370967 Mar 6 '18 at 9:47
  • $\begingroup$ Yes the $\mathbb{R}_{l}$ is the topology I am talking about. I will edit the post. $\endgroup$ – Aurora Borealis Mar 6 '18 at 9:49
  • $\begingroup$ $[a, \infty)$ can be written as union of half open intervals (and those form a basis for the topology you describe). Since a topology is closed under unions, the set is open $\endgroup$ – user370967 Mar 6 '18 at 9:52
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It's open as $[a,\infty) = \bigcup_{n=0}^\infty [a+n, a+n+1)$ which is a union of open sets in the lower limit (Sorgenfrey) topology.

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  • $\begingroup$ Thank you, but I do not understand why $[a+n,a+n+1)$ is necessarily an open set? $\endgroup$ – Aurora Borealis Mar 6 '18 at 9:53
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    $\begingroup$ Look at the definition of the topology. It is the topology generated by all half open intervals. I.e., half open intervals are a basis and by definition open. $\endgroup$ – user370967 Mar 6 '18 at 9:53
  • $\begingroup$ @AuroraBorealis By definition. $\endgroup$ – Henno Brandsma Mar 6 '18 at 9:54
  • $\begingroup$ So it is just by definition open because any subset in a topology is defined to be an open subset? Is there a way to show that it actually is open? $\endgroup$ – Aurora Borealis Mar 6 '18 at 10:03
  • $\begingroup$ @AuroraBorealis being open by definition means being in the topology (a set of subsets). It’s synonymous. And half-open intervals are in the lower limit topology by definition. $\endgroup$ – Henno Brandsma Mar 6 '18 at 10:05

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