A non-empty perfect set has two disjoint subsets that are both perfect.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $F \subseteq \Bbb R$ be a perfect set. Let $\alpha = \inf F$ if $F$ is bounded from below and $\alpha = -\infty$ otherwise. It follows that $\alpha \in \Bbb R \implies \alpha \in F$. Let $\beta= \sup F$ if $F$ is bounded from above and $\alpha = +\infty$ otherwise. It follows that $\beta\in \Bbb R \implies \beta\in F$.
- $(\alpha, \beta) \subseteq F$
Since $\Bbb Q$ is dense in $\Bbb R$, there exist $r,s \in \Bbb Q$ such that $\alpha < r < s < \beta$. Let $A = F \cap (-\infty,r]$ and $B = F \cap [s,+\infty)$. If $\alpha \in \Bbb R$, then $A = [\alpha, r]$. If $\alpha = \infty$, then $A = (-\infty,r]$. In either case, $A$ is clearly perfect. Similarly, $B$ is perfect. Moreover, $A \cap B = \emptyset$.
- $(\alpha, \beta) \not \subseteq F$
Then there exists $a \in (\alpha, \beta)$ such that $a \not \in F$. Since $F$ is closed, its complement is open and thus there exists $\delta>0$ such that $(a-\delta,a+\delta) \cap F = \emptyset$. Since $\Bbb Q$ is dense in $\Bbb R$, there exist $r,s \in \Bbb Q$ such that $a-\delta < r < a<s < a+\delta$.
Let $A = F \cap (-\infty,r] = F \cap (-\infty,a) = F \cap (-\infty,a+\delta)$. It is clear that $A$ is closed. Assume the contrary that $b \in A$ is an isolated point of $A$. Then there exists $0 <\delta' < \delta$ such that $x \neq b$ and $|x-b| < \delta'$ implies $x \notin A$. But for such $x$, $x < b + \delta' < a +\delta$, so $x \notin F$, showing that $b$ is also an isolated point of $F$. This contradicts the fact that $F$ is perfect. Hence $A$ is perfect. Similarly, $B$ is perfect. It is clear that $A \cap B = \emptyset$.
We know $X$ is $T_3$. So the perfect $F$ has two distinct points with relatively open neighbourhoods with disjoint closures. It’s easy to see that for open $O$ in $F$, the closure of $O$ also does not have isolated points. So the disjoint closures are as required. The structure of the reals is superfluous here. It’s much simpler.