2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Compliment $\leadsto$ complement. $\endgroup$ – Clement C. Apr 29 '17 at 22:11
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $A$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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