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

Question

Let $T\subseteq[0,1]$ be a subset containing $1$. Now we know that $T$ satisfies the following property:

For every $t\in [0,1)$, if there exists a decreasing sequence $\{t_n\}_{n\ge 1}\subset T$ such that $t_n\to t$ as $n\to\infty$, then one has $t\in T$.

Now I would like a characterization of $T$. Denote by $\mathring{T}$ the interior of $T$, thus $\mathring{T}$ could be represented as an union of at most countable disjoint open intervals, i.e.

$$\mathring{T}=\bigcup_{n\ge 1}(s_n,t_n).$$

But I don't know what I should do next. My question is whether $T$ has the following expression:

$$T=\bigcup_{n\ge 1}[s_n,t_n),$$

where every two different intervals are disjoint. If not (that is what I belive), how to characterize $T$? Many thx for the reply!

Answer 1

As already pointed out in the comments, your conjectured representation is not correct (take any closed $T$ with empty interior). However, we can run a slightly modified version of the usual argument that displays an open subset as an at most countable union of disjoint open intervals (= its connected components) to obtain such a representation of the complement of $T$:

For $t\notin T$, let $I_t$ be the maximal subinterval containing $t$ and contained in $T^c$. Then $I_t=[a,b)$ or $I_t=(a,b)$. (More formally, define $a=\inf\{x\in[0,t): [x,t]\cap T=\emptyset\}$, $b=\sup\{ x: [t,x]\cap T=\emptyset\}$; observe that $b>t$, so we do obtain a non-degenerate interval.)

Any two such intervals are identical or disjoint; also, if $I_t=[a,b)$ is such an interval, then $a$ is not a right endpoint of an $I_{t'}$. Thus $$ T^c = \bigcup I_{t_n} , $$ and this union is disjoint in the strong sense just spelled out. Conversely, any $T$ whose complement has such a representation satisfies your condition.