Find a covering $\{V_i\}_i^\infty$ for every $i$ and $x\in U$ s.t only a finite number of the $V_i$ covers $x$.

Question

For any open set $U\subset \mathbb{R}^n$, find a covering $\{V_i\}_i^\infty$ such that $V_i$ is bounded for every $i$ and if $x\in U$ only a finite number of the $V_i$ covers $x$.

Attempt:

I tried to define $V_i=\{x\in U:~\text{distance}(x,\partial U)>1/i\}$.

If $V_i=B(0,1-\frac{1}{i+1})$, then $\{V_i\}_i^\infty$ covers $B(0,1)$and so $x=0$ is covered by all infinity sets.

However, this seems not to make sense

Answer 1

You want your cover locally finite. However, let $A:=\{\prod_{k=1}^n (m_k-\frac{1}{2},m_k+\frac{3}{2})|m_k\in \mathbb{Z}\}$. Let $(V_i)_{i\in\mathbb{N}}$ denote some enumeration of the elements $A$ (the set is in bijection with $\mathbb{Z}^n$). Then, each $V_i$ is clearly bounded, $\cup_{i=1}^{\infty} V_i=\mathbb{R}^n$ so, in particular, they form a cover of $U$, and any given $x$ is an element of at most $2^n$ different $V_i$.

Answer 2

All you need to do is go in the opposite direction: take $V_0 = U$, and for $i \geq 1$ $$ V_i = \{x \in U: \operatorname{dist}(x,\partial U) < 1/i\}. $$ If $U$ is not bounded, we can instead consider $W_i = V_i \cap B(0,i)$.