Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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