I am in particular interested in families of compact sets given by $\{K+z_j:j\in \mathcal{J} \}$ such that $\mathbb{R}^d = \bigcup_{j\in \mathcal{J}}K+z_j$ where $K\subset \mathbb{R}^d$ is compact and $\{z_j:j\in \mathcal{J} \}\subset\mathbb{R}^d$ is just a sequence of points in $\mathbb{R}^d.$ How do I show that $$ \sup_{x\in \mathbb{R}^d} \# \{j\in\mathcal{J}: x\in K+z_j \} = \sup_{i\in \mathcal{J}} \# \{i\in\mathcal{J}: (K+z_i) \cap (k+z_j) \neq \emptyset \}? $$
For $x\in \mathbb{R}^d,$ $i\in \mathcal{J}$ we can define $B_x := \{j\in\mathcal{J}: x\in K+z_j \}$ and $Q_i:= \{i\in\mathcal{J}: (K+z_i) \cap (k+z_j) \neq \emptyset \}$. I was able to show that $\sup_{x\in \mathbb{R}^d} \#B_x \leq \sup_{i\in \mathcal{J}}\# Q_i$ by noting that for any $x\in \mathbb{R}^d$ we can always find an $i\in \mathcal{J}$ such that $x\in K+z_i,$ so $\#B_x \leq \#Q_i$. The reverse inequality remains elusive to me, and I suspect it has something to do with the compactness of $K$. Anybody knows if this is actually doable without extra hypotheses? If it is, can you give me some hints?
For some context, I am studying Amalgam Spaces (Introduction to Weighted Wiener Amalgams by C. Heil), and this equality came up without proof in the definition of BUPUs.
It is not true. In fact we always have $\sup_{x\in \mathbb{R}^d} \#B_x \leq \sup_{i\in \mathcal{J}}\# Q_i$. This is true because $\sup_{x \in K + z_i} \#B_x \le \#Q_i$ so that
$$\sup_{x\in \mathbb{R}^d} \#B_x = \sup_{i \in \mathcal{J}} \sup_{x \in K + z_i} \#B_x \le \sup_{i \in \mathcal{J}} \#Q_i .$$
Now consider the following example for $d=2$. Let $K = [0,1] \times [0,1]$ and let $z_j$ be all points having the form $(m,n)$ with $m, n \in \mathbb{Z}$ and $m$ even and $(m,n+\frac{1}{2})$ with $m, n \in \mathbb{Z}$ and $m$ odd. Then it easy to see that $\sup_{x\in \mathbb{R}^d} \#B_x = 3$ and $\sup_{i \in \mathcal{J}} \#Q_i = 7$.