Not open set looks like the disjoint union of countably many open intervals.

Question

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.

Answer 1

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 $A$.