2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ If $\{ x \}$ is isolated in $E$, wouldn't it be open? $\endgroup$ – Mike D Feb 28 at 4:06
2
$\begingroup$

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.

$\endgroup$

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.