The set is the complement of $A$, where

$A = \left\{\frac{n+1}{n} \, | \, n \in \{1,2,...\}\right\} \subset \mathbb{R}$

The complement of $A$ is not open in $\mathbb{R}$ because any ball with centre $1$, contains elements of $A$.

But we proved a theorem in class showing that any open subset of $\mathbb{R}$ is the union of a countable collection of intervals.

$A$ looks like its countable distinct points, so the compliment would be the union of a countable collection of disjoint open intervals.

But then the complement of $A$ would be open, which it isn't.

I'm very confused. I would greatly appreciate any clarification you could offer me.

  • $\begingroup$ Compliment $\leadsto$ complement. $\endgroup$ – Clement C. Apr 29 '17 at 22:11
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    $\begingroup$ The complement of $A$ is a countable union of disjoin intervals, but one of these intervals is not open! Precisely, one of the intervals composing $A^c$ is $(- \infty ; 1]$ which is not an open interval. $\endgroup$ – Crostul Apr 29 '17 at 22:13
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    $\begingroup$ Have you tried drawing the complement of $A$? It's not the union of countably many disjoint open intervals. It's the union $(- \infty , 1 ] \cup (1,2) \cup (2,3) \cup \dots $ $\endgroup$ – Mauro Apr 29 '17 at 22:13
  • $\begingroup$ $(\infty, 1]$ is a subset of $A^c$. $(\infty, 1]$ is a *closed interval. That keeps $A^c$ being a countable union of open intervals $\endgroup$ – fleablood Apr 29 '17 at 22:14
  • $\begingroup$ Thank you so much!!! I was being very silly. $\endgroup$ – emidude Apr 29 '17 at 22:16

The complement is not a countable union of open intervals because of the problem at $1$. You could try writing it like $$ (-\infty,1) \cup (2,\infty) \cup \bigcup_{n =1}^\infty \left( \frac{n+2}{n+1}-\frac{n+1}{n}\right)$$ but this excludes the point $1$. And there is no open interval containing $1$ which we could add to this union to get $A^C$, because as you said every such interval contains a point of $A$.

  • $\begingroup$ Yes! Thank you! $\endgroup$ – emidude Apr 29 '17 at 22:27

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