Say that $S$ is an infinite collection of non-empty open intervals in $\mathbb{R}$ whose union contains $[0,1]$. $$[0,1] \subset \bigcup_\mathrm{I \in S} I$$
For each $x \in (0, 1]$, if there is a finite set of intervals $\{I_i\}^n_\mathrm{i=1} \subset S$ whose union contains $[0,x]$, we will call $x$ an arbitrary condition like "true"
Say $T$ is the set of "true" numbers in $(0,1]$ and $t = supT$.
I'm trying to prove that $t$ is "true".
I guess this must be proving that the supremum of a set is in the set, which isn't the case for all sets with supremum.
My guess would be that if all values of $x$ are "true" on this interval, that the supremum would be 1, and 1 is a "true" number as a member of the interval.