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  1. for the set of rational numbers, what would be its interior?

  2. And what is its interior's closure?

  3. If one says that a set's closure has empty interior, what does it mean? So it means that all elements of a set are limit points?

  4. By "limit points", how are they exactly defined? Can't real number be also limit point?

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HINTS for (1) and (2) and answers to (3) and (4):

  1. Every non-empty open set in $\Bbb R$ is a union of open intervals. Does $\Bbb Q$ contain any non-empty open interval?

  2. What is $\operatorname{cl}\varnothing$?

  3. $\operatorname{int}\operatorname{cl}A=\varnothing$ if and only if $\operatorname{cl}A$ contains no non-empty open set. An example of such an $A$ is $\left\{\frac1n:n\in\Bbb Z^+\right\}$: the closure of this set is $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, which contains no non-empty open interval. Another way to say this: every point of $\operatorname{cl}A$ is a limit point of $\Bbb R\setminus\operatorname{cl}A$, the complement of $\operatorname{cl}A$.

  4. If $A\subseteq\Bbb R$, a point $x\in\Bbb R$ is a limit point of $A$ if and only if every open neighborhood of $x$ contains a point of $A\setminus\{x\}$. If you’re working in $\Bbb R$, as you seem to be, real numbers are the only things that can be limit points of $A$: they’re the only things that are even points in your space.

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    $\begingroup$ but you answered (1) in your second hint! $\endgroup$ – Jimmy R Feb 11 '13 at 7:25
  • $\begingroup$ @JimmyR: Implicitly, but long experience assures me that there’s no guarantee that any given student will recognize this unless someone points it out $-$ as you just did. (Indeed, even that’s no guarantee, though it markedly increases the chances.) $\endgroup$ – Brian M. Scott Feb 11 '13 at 7:34
  • $\begingroup$ One last question to help my understanding: for a set of rational numbers, what would be its closure? $\endgroup$ – Common Knowledge Feb 11 '13 at 8:59
  • $\begingroup$ @CommonKnowledge: If you mean an arbitrary set of rational numbers, that could depends on the set. If you mean the closure of $\Bbb Q$, you should be able to answer that yourself: is there any real number $x$ with an open neighborhood that does not contain any rational number? $\endgroup$ – Brian M. Scott Feb 11 '13 at 9:24
  • $\begingroup$ @BrianM.Scott I now completely get it. Thanks. $\endgroup$ – Common Knowledge Feb 11 '13 at 10:00
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To say that a set has empty interior is to say that it doesn't contain an open ball. In $\mathbb R$ this means it doesn't contain an open interval and indeed between any two rational points is an irrational point so the rationals have empty interior.

The interior is empty and the empty set is closed so the closure of the interior is the empty set.

A limit point of a set $S$ is a point $x$ such that every neighborhood of $x$ contains a point of $S$ other than $x$. Every open set contains a rational point so every neighborhood of every point intersects the rationals. This means every point in $\mathbb R$ is a limit point for the set of rationals.

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