# Characterization of a subset of [0,1] $III$

I have a question related to the previous one.

Characterization of a subset of [0,1] $II$

Let $T\subseteq [0,1]$ be some subset closed under lower limit topology, i.e. $t_n$ is said to converge to $t$ iff $t_n\ge t$ and $t_n\to t$. See

https://en.wikipedia.org/wiki/Lower_limit_topology

Now let $T_0\subset T$ is the subset consisting of all condensation points under lower limit topology, i.e. a condensation point is the point all of whose nbds are uncountable. Now my question is whether we can show the complement $T\backslash T_c$ is at most countable? For example, if $T$ is countable, then it is obviously true; if $T$ has the form $T=\cup_{n}[t_n, t_n+\varepsilon_n)$, then $T\backslash T_c$ is empty, so it is also true. But I can not prove it and neither find a counterexample. If $T$ is a Cantor set, again we get $T\backslash T_c$ is empty. Could someone give a counterexample or prove it? Thx a lot for the reply!

The complement is always at most countable. Assume otherwise. Let $\{t_{\alpha}\mid \alpha<\omega_1\}$ (with $t_{\alpha}\not=t_{\beta}$ for $\alpha\not=\beta$) be an uncountable subset of $T$ consisting entirely of points which are not condensation points. Then we can choose, for each $\alpha<\omega_1$, a value $n(\alpha)\in\mathbb{N}\setminus\{0\}$ such that $[t_{\alpha},t_{\alpha}+\frac{1}{n(\alpha)})\cap T$ is countable. At least one of the fibers of the function $\omega_1\rightarrow\mathbb{N},\alpha\mapsto n(\alpha),$ is uncountable, so we can assume w.l.o.g. that there exists $n\in\mathbb{N}\setminus\{0\}$ such that for all $\alpha<\omega_1$, $[t_{\alpha},t_{\alpha}+\frac{1}{n})\cap T$ is countable.
I now claim that for each $\alpha<\omega_1$, there are only countably many $\beta<\omega_1$ such that $[t_{\alpha},t_{\alpha}+\frac{1}{n})\cap[t_{\beta},t_{\beta}+\frac{1}{n})\not=\emptyset$. This leads to a contradiction, since it implies that even for any countable subset $M\subseteq\omega_1$, the set of all $\beta<\omega_1$ such that for at least one $\alpha\in M$, we have $[t_{\alpha},t_{\alpha}+\frac{1}{n})\cap[t_{\beta},t_{\beta}+\frac{1}{n})\not=\emptyset$, is countable, which would imply that $[0,1]$ has a countably infinite family of pairwise disjoint subsets of the form $[x,x+\frac{1}{n})$.
Now to show the claim, assume otherwise. Denote the set of all $\beta<\omega_1$ such that the intersection is nonempty by $X$. For all $\beta\in X$, we either have $t_{\beta}\in[t_{\alpha},t_{\alpha}+\frac{1}{n})$ (which happens only countably often) or $t_{\beta}\in(t_{\alpha}-\frac{1}{n},t_{\alpha})$. Define $t:=\mathrm{inf}\{t_{\beta}\mid (\beta\in X)\wedge(t_{\beta}<t_{\alpha})\}$. Then $t\in T$, and $[t,t+\frac{1}{n})\cap T$ is still countable (because it is a subset of the union of the $[t_{\beta_k},t_{\beta_k}+\frac{1}{n})\cap T$ for some sequence $(t_{\beta_k})_{k\in\mathbb{N}}$ with values in $\{t_{\beta}\mid \beta\in X\}$ converging to $t$ from above), so we must have $t>t_{\alpha}-\frac{1}{n}$ and $(t_{\alpha}-\frac{1}{n},t)\cap\{t_{\beta}\mid \beta\in X\}$ is uncountable, in particular nonempty. But for any $t_{\beta}$ from this set, we have $t_{\beta}<t$, although by definition of $t$, $t\leq t_{\beta}$, a contradiction.
• Very good answer, maybe you can replace $<$ by $\in$ and $\wedge$ by $\cap$ to make your proof clearer Oct 2, 2014 at 18:14