I already showed that that set of all end points of the intervals are closed. Letting $E$ be this set, we know that $(E,d)$ is a complete metric space. Ultimately I want to use the Baire category theorem to reach a contradiction. Since $E$ is the countable union of singleton sets, it is enough to show that each $x \in E$ is nowhere dense. I understand why $\{ x \}$ is closed in $E$, but I'm struggling to show that $\{ x \}$ has empty interior. Maybe I'm just getting confused with subspace topologies, since I know that $\{ x \}$ has empty interior in $\left[0,1\right]$, but I'm not sure if that holds in $E$. Any tips on how to proceed with this argument?
Proving that $\left[ 0,1 \right] $ is not the countable union of disjoint closed intervals using Baire Category Theorem
baire-categoryreal-analysis
Best Answer
If the interior of $\{x\}$ is not empty, then it must be $\{x\}$ itself. In other words, $\{x\}$ would be an open set in the subspace topology of $E$. That means there must exist a set $U$ which is open in $[0,1]$ such that $U \cap E = \{x\}$. Now, knowing what open sets look like in $[0,1]$, try to reach a contradiction. (Hint: $x$ is either the left or the right endpoint of one of your closed intervals...)
Note that you will encounter a problem when $x=0$ or $1$; indeed, in those cases $\{x\}$ could in fact be open in $E$. Explain why this does not wreck your argument.