# How can complements of open sets in the lower limit topology be closed?

If we take a typical open set in the lower limit topology [a,b), how can its complement be closed when the definition of a closed set is one that contains all of its limit points. In this case the missing limit point is a.

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You have a very nonstandard definition of a closed subset. They are typically defined as complements to open sets. – studiosus Apr 26 '14 at 21:42

Recall that given a topological space $X$, an $A \subseteq X$ and $x \in A$, we say that $x$ is a limit point of $A$ (in $X$) if every open neighbourhood of $x$ (in $X$) meets $A$.
As such, being a limit point of a set is wholly dependent on the underlying topology you are currently using/interested in. While $a$ is a limit point of $A =( - \infty , a ) \cup [ b , + \infty )$ in the usual metric topology on $\mathbb{R}$, this need not be true when you use a different topology. For instance, the set $[a,b)$ is an open neighbourhood of $a$ in the lower limit topology — it is an open set in the lower limit topology which contains $a$ — which is disjoint from $A$, and so $a$ cannot be a limit point of $A$ in the lower limit topology.