Suppose we have a set $I=\bigcup_{n=1}^{\infty} R_n$ where each $R_n$ is a closed set. Then $I$ is a countable Union of closed sets. Can we assert that $ I$ doesn't contain any non-empty open intervals?
Open interval in countable Union of closed sets
real-analysis
Related Question
- Real Analysis – Why Use Open Interval as Open Sets for Real Line Topology
- [Math] Every open set in $\mathbb{R}$ is a countable union of closed sets
- Which subsets of $\Bbb R$ are a countable union of open sets and countable sets
- [Math] Is the infinite (countable or uncountable) union of disjoint closed sets closed
- Representing an open set in countable union of closed sets
Best Answer
Suppose your first interval is $[0, 1]$. Then the union contains $(0, 1)$.
For global considerations, suppose the $n$th set is $[1/n, 1 - 1/n]$. Then then union exactly is the interval $(0, 1)$, which is open.