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?



Partitioning an infinite cardinal $\kappa$ into pairwise neighboring subsets
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 $B\_{\beta\_\lambda}$.


