We say that two disjoint, non-empty subsets $S, T$ of an infinite cardinal $\kappa$ are neighboring if there is $\alpha\in \kappa$ such that $$S\cap\{\alpha,\alpha+1\} \neq \varnothing \neq T\cap\{\alpha, \alpha+1\}.$$ Given an infinite cardinal $\kappa$, is there a partition ${\cal B}$ of $\kappa$ with $|{\cal B}|=\kappa$ and whenever $B_1\neq B_2 \in {\cal B}$ we have that $B_1, B_2$ are neighboring?


Yes. List the pairs $(\alpha,\beta)$ with $\alpha<\beta<\kappa$ as $(\alpha_\lambda,\beta_\lambda), \lambda<\kappa$. Then construct the sets $B_\alpha\in\mathcal B, \alpha<\kappa$ as follows:

At stage 0, all $B_\alpha=\emptyset$.

At limit stages just take unions.

At successor stages $\lambda+2n, n\in\omega, n\ge 1$, choose the least $\lambda$ such that $(B_{\alpha_\lambda},B_{\beta_\lambda})$ are not yet neighboring. Put $\lambda$ into $B_{\alpha_\lambda}$ and $\lambda+1$ into $B_{\beta_\lambda}$.


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