Infimum of partitions

Question

Let $X\neq\emptyset$ be a set. A partition is a subset $P\subseteq {\cal P}(X)\setminus \{\emptyset\}$ such that $\bigcup P = X$ and any distinct members of $P$ are disjoint. We denote by $\text{Part}(X)$ the set of all partitions of $X$.

We order $\text{Part}(X)$ by the refinement relation and we write $P\leq_r Q$ if $Q$ is a refinement of $P$. So $(\text{Part}(X), \leq_r)$ is a complete lattice.

For $P,Q \in \text{Part}(X)$ set $P\triangleleft Q$ if $\text{card}(P\setminus Q) \leq \text{card}(Q\setminus P)$. Note that $P\leq_r Q$ implies $P\triangleleft Q$, but not vice versa in general.

Pick $P_0\in \text{Part}(X)$ and suppose that $\mathcal{C}\subseteq \text{Part}(X)$ is a chain in $\text{Part}(X)$ with respect to $\leq_r$ such that for all $C\in \mathcal{C}$ we have $C\triangleleft P_0$. Does this imply that $$\text{sup}(\mathcal{C}) \triangleleft P_0?$$ (The partition $\text{sup}(\mathcal{C})$ is the supremum of $\cal C$ in the complete lattice $(\text{Part}(X), \leq_r)$.)

Answer 1

Counterexample: Let $X=[0,1)$; let $\mathcal C=\{C_n:n\in\mathbb N\}$ where $C_n=\{[\frac{k-1}{2^n},\frac k{2^n}):1\le k\le2^n\}$;and let $P_0=\{[\frac{n-1}n,\frac n{n+1}):n\in\mathbb N\}$.