If $U=\bigcup_{j=1}^{\infty}K_j$ then $U=\bigcup_{j=1}^{\infty}\mathring{K}_j$ Let $U\subset\mathbb{R}^n$ be a domain (connected open set), such that $U=\bigcup_{j=1}^{\infty}K_j$ where each $K_j$ is a compact set, and $K_j\subset K_{j+1}$ for all $j\ge 1$.
Is the following proposition true?
$U=\bigcup_{j=1}^{\infty}\mathring{K}_j$  
$\mathring{K}_j:$ interior of $K_j$
Any help would be appreciated.

No, consider $n=1$ with $U=(-1,1)$ and $K_n= [-1+1/n,-1/n]|\cup \{0\} \cup [1/n,1-1/n].$
