# Proving that $\left[ 0,1 \right]$ is not the countable union of disjoint closed intervals using Baire Category Theorem

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?

• If $\{ x \}$ is isolated in $E$, wouldn't it be open? – Mike D Feb 28 at 4:06

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.