Let $\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$ be an infinite sequence of families of compact connected sets in $\mathbb{R}^{d}$. An infinite sequence of compact connected sets $\left\{ B_{n} \right\}_{n\in \mathbb{N}}$ is called heterochromatic sequence from $\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$ if there exists an infinite sequence $\left\{ i_{n} \right\}_{n\in \mathbb{N}}$ of natural numbers satisfying the following two properties: (a) $\{i_{n}\}_{n\in \mathbb{N}}$ is a monotonically increasing sequence, and (b) for all $n \in \mathbb{N}$, we have $B_{n} \in \mathcal{F}_{i_n}$. We show that if every heterochromatic sequence from $\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$ contains $d+1$ sets that can be pierced by a single hyperplane then there exists a finite collection $\mathcal{H}$ of hyperplanes from $\mathbb{R}^{d}$ that pierces all but finitely many families from $\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$. As a direct consequence of our result, we get that if every countable subcollection from an infinite family $\mathcal{F}$ of compact connected sets in $\mathbb{R}^{d}$ contains $d+1$ sets that can be pierced by a single hyperplane then $\mathcal{F}$ can be pierced by finitely many hyperplanes. To establish the optimality of our result we show that, for all $d \in \mathbb{N}$, there exists an infinite sequence $\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$ of families of compact connected sets satisfying the following two conditions: (1) for all $n \in \mathbb{N}$, $\mathcal{F}_{n}$ is not pierceable by finitely many hyperplanes, and (2) for any $m \in \mathbb{N}$ and every sequence $\left\{B_n\right\}_{n=m}^{\infty}$ of compact connected sets in $\mathbb{R}^d$, where $B_i\in\mathcal{F}_i$ for all $i \geq m$, there exists a hyperplane in $\mathbb{R}^d$ that pierces at least $d+1$ sets in the sequence.

翻译：令$\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$为$\mathbb{R}^{d}$中紧连通集族的一个无穷序列。若存在自然数的无穷序列$\left\{ i_{n} \right\}_{n\in \mathbb{N}}$满足以下两个性质：(a) $\{i_{n}\}_{n\in \mathbb{N}}$是单调递增序列，且(b)对所有$n \in \mathbb{N}$，有$B_{n} \in \mathcal{F}_{i_n}$，则称无穷序列$\left\{ B_{n} \right\}_{n\in \mathbb{N}}$为来自$\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$的异色序列。我们证明：若来自$\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$的每个异色序列都包含$d+1$个可被单个超平面刺穿的集合，则存在一个来自$\mathbb{R}^{d}$的有限超平面集合$\mathcal{H}$，它能刺穿$\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$中除有限个族之外的所有族。作为我们结果的直接推论，我们得到：若$\mathbb{R}^{d}$中无穷族$\mathcal{F}$的每个可数子族都包含$d+1$个可被单个超平面刺穿的集合，则$\mathcal{F}$可被有限个超平面刺穿。为确立结果的最优性，我们证明：对所有$d \in \mathbb{N}$，存在紧连通集族的无穷序列$\left\{ \mathcal{F}_{n}\right\}_{n \in \mathbb{N}}$满足以下两个条件：(1)对所有$n \in \mathbb{N}$，$\mathcal{F}_{n}$不可被有限个超平面刺穿，且(2)对任意$m \in \mathbb{N}$及$\mathbb{R}^{d}$中紧连通集的任意序列$\left\{B_n\right\}_{n=m}^{\infty}$（其中对所有$i \geq m$有$B_i\in\mathcal{F}_i$），存在$\mathbb{R}^{d}$中的一个超平面，它能刺穿该序列中至少$d+1$个集合。