$$\label{disj sets}
   \mathbb{N}=\bigcup\limits_{n=1}^{\infty} A_n,\quad A_n\cap A_m=\emptyset\:\:\text{if}\:\: n\neq m,\quad \left|A_n\right|\leq K\:\:\text{for all}\:\: n,$$ such that the corresponding sequence of subspaces $\{Lin \{\phi_n\}_{n\in A_n}\}_{n=1}^{\infty}$ constitute a Riesz basis of subspaces of $H$ with uniform bound of dimensions of all subspaces not exceeding $K$. See e.g. Example 1.3 in [@Zwart] for details. In order to show that the XYZ Theorem is sharp we will prove that this situation is impossible for our construction from Theorem \[construction\]. More precisely, thereby we demonstrate that if the eigenvalues of the generator of the $C_0$-group in a Hilbert space do not satisfy (\[1\]), then the conclusion of the XYZ Theorem can be false. Furthermore, we will prove a little more.

\[sharp\] Let $k\in\mathbb{N}$ and $\{e_n\}_{n=1}^{\infty}\subset H_k\left(\{e_n\}\right)$ be a sequence from Theorem \[construction\]. Suppose that $\{A_n\}_{n=1}^{\infty}$ is an arbitrary decomposition of $\mathbb{N}$ into disjoint sets, i.e. $$\mathbb{N}=\bigcup\limits_{n=1}^{\infty} A_n,\quad A_n\cap A_m=\emptyset, n\neq m.$$ Then $\left\{\overline{Lin} \{e_j\}_{j\in A_n}\right\}_{n=1}^{\infty}$ does not form a Schauder decomposition of $H_k\left(\{e_n\}\right)$.

Fix $k\in\mathbb{N}$ and assume the opposite, i.e. let there exists a decomposition of $\mathbb{N}$ into disjoint sets, $\mathbb{N}=\bigcup\limits_{n=1}^{\infty} A_n,$ $A_n\cap A_m=\emptyset$ if $n\neq m$, such that $\left\{\mathfrak{M}_n=\overline{Lin} \{e_j\}_{j\in A_n}\right\}_{n=1}^{\infty}$ constitutes a Schauder decomposition of $H_k\left(\{e_n\}\right)$. Then, by the definition of the Schauder decomposition, every 