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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

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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 $\Bbb R.$

A few useful results that may help you with this problem (and are good exercises to prove) are the following:

  • 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.
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  • $\begingroup$ since discrete referes to the set being both open and closure, would I then assume that the closure is the set, the interior would also be the set since if you make the neighborhood small enough there will always be a point in the set and im not sure what they mean by the derived sets of A? @CameronBuie $\endgroup$ – cele Oct 21 '13 at 22:38
  • $\begingroup$ derived set $\endgroup$ – Tyler Oct 21 '13 at 22:44
  • $\begingroup$ @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 $A$. $\endgroup$ – Cameron Buie Oct 22 '13 at 0:45
  • $\begingroup$ @CameronBuie ok, so limit points are kind of like accumulation points? $\endgroup$ – cele Oct 22 '13 at 16:05
  • $\begingroup$ @Kathryn: Yes, many texts use the terms interchangeably. Did my additions to the answer make things any clearer/easier for you? $\endgroup$ – Cameron Buie Oct 22 '13 at 16:08

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