# find closure, interior, and derived sets with respect to topologies

I have a midterm coming up and I'm trying to understand this problem. I understand what closure and interior mean but the different topologies are a little confusing to me.

Let $\mathbb{R}$ be the set of real numbers and $A= \{ x: x \mathrm{\, is \, rational \}}$.

Find the closure, interior, and derived sets of $A$ with respect to the discrete topology, the indiscrete topology and the topology formed by defining a set to be open if it contains all but at most countably many points

Hints: In the discrete topology, every subset of $\Bbb R$ is both closed and open. In the indiscrete topology, only the empty set and all of $\Bbb R$ are open (or closed). In the cocountable topology (the typical name for that last one), note that the closed subsets of $\Bbb R$ are $\Bbb R$ itself, and every at most countable subset of $\Bbb R.$
• The closure of $A$ is the smallest closed set $C$ such that $A\subseteq C$. In particular, then, $A$ is closed if and only if it is equal to its closure.
• The interior of $A$ is the largest open set $U$ such that $U\subseteq A.$ In particular, then, $A$ is open if and only if it is equal to its interior.
• @Kathryn: You're exactly right on the first two counts, but your reasoning is a bit shaky. (I will expand my answer to give an alternative approach.) The derived set of $A$ is the set of limit points of $A$--those points $x\in\Bbb R$ such that for any neighborhood $U$ of $x,$ $U$ contains a point of $A$ that is distinct from $x$. Put another way, a point $x\in\Bbb R$ is not in the derived set of $A$ if and only if there is a neighborhood $U$ of $x$ such that $U$ contains at most one point of $A$. Points in the derived set of $A$ may or may not be points of $A$. – Cameron Buie Oct 22 '13 at 0:45