Existence of infinite non empty intersection of sets Let $U \subset \mathbb{R}^d$ be an open bounded set and $U_{k,n} \subset U$ be a collection of sets (can be assumed open or closed) for $n\in \mathbb{N}$ and $k\in \{1,\ldots,K\}$. If we have
$$ U = \bigcup_{k\leq K} U_{k,n}, \quad \forall n\in \mathbb{N}$$
is the following true: there exists an infinite subset $\mathcal{T} \subset \mathbb{N}$ and $k_0 \leq K$ such that
$$ \bigcap_{n \in \mathcal{T}} U_{k_0,n} \; \; \text{is of non empty interior}$$
 A: Here is a counterexample.
$U = (0,1)$.  $K = 2$.   For $n = 1,2,3,\dots$
let $A_n = \{j/n : 1 < j < n, j \text{ odd}\}$ and
$B_n = \{j/n : 1 < j < n, j \text{ even}\}$.  Define
$$
U_{1,n} = (0,1) \setminus A_n,\qquad U_{2,n} = (0,1) \setminus B_n
$$
Any interval contined in $U_{k,n}$ has length at most $2/n$.
For all $n$ we have $(0,1) = U_{1,n} \cup U_{2,n}$.
Now let $\mathcal T \subseteq \mathbb N$ be infinite, let $k_0 \in \{1,2\}$
and
$$
V = \bigcap_{n \in \mathcal T} U_{k_0,n}.
$$
I claim $V$ has empty interior.  Let $\varepsilon > 0$.  Choose $n_0$
so that $2/n_0 < \varepsilon$.  Then choose $n_1 \in \mathcal T$ so that
$n_1 > n_0$.  This means
$$
V \subseteq U_{k_0,n_1}
$$
so any interval contained in $V$ has length at most $2/n_1 < \varepsilon$.
This is true for every $\varepsilon> 0$.  So
the connected components of $V$ are single points.
A: Let $U=(0,1)$ and for each $n$ and $k\le K$ make $U_{k,n}$ as follows:
$$
U_{k,n}=\bigcup\Bigl\{\bigl(i\cdot K^{-n},(i+2)\cdot K^{-n}\bigr):0\le i< K^n-1,\ i\equiv k \pmod{K}\Bigr\}
$$
Then $U_{k,n}$ contains no intervals longer than $2K^{-n}$, so set of the form $\bigcap_{n\in T}U_{k_0,n}$ contains non-trivial intervals.
