Equip $[0,1]$ with its usual metric topology inherited from $\mathbb{R}.$ It is well known that the covering dimension of $[0,1]$ is $1$.
Question: Given a fixed number $n$, can I construct a finite open cover $U_1, U_2, \cdots, U_n$ of $[0,1]$ such that any finite open refinement of multiplicity $2$ needs to consist of at least $k$ elements for an arbitrary number $k$?
For example, can I construct an open cover of $[0,1]$ consisting of $3$ elements such that any open refinement of multiplicity $2$ needs to have at least $7$ elements?
In general, given a compact space $X$ of covering dimension $s$ and a fixed number $n$, can I construct a finite open cover $U_1, U_2, \cdots, U_n$ of $X$ such that any finite open refinement of multiplicity $s+1$ needs to consist of at least $k$ elements for an arbitrary number $k$?
Any help would be greatly appreciated.
Definitions:
Covering dimension: A nonempty topological space $X$ is said to have the covering dimension $n$ if $n$ is the smallest non-negative integer with the property such that each finite open cover of $X$ has a finite open refinement of multiplicity at most $n+1$.
Multiplicity of a cover A cover of a topological space $X$ has multiplicity $n$ if and only if it is the smallest non-negative integer such that each point $x$ of $X$ belongs to at most $n$ elements of the cover.
Best Answer
Suppose $\mathcal{V} = \{V_j : j \in J\}$ is a multiplicity $m$ open refinement of $U_1, U_2, \ldots, U_n.$
For $1 \leq i \leq n,$ define $A_i = \{V \in \mathcal{V} : V \subseteq U_i\}$ and note that $$W_i = \bigcup \left(A_i\setminus\left(\bigcup_{k=1}^{i-1}A_k\right)\right), \qquad 1\leq i \leq n$$ defines an open refinement of $U_1,U_2,\ldots,U_n$ with at most $n$ elements and at most multiplicity $m.$
Notation: The arbitrary union of a set of sets $\mathcal{A}$ is defined by $$\displaystyle\bigcup \mathcal{A} = \bigcup_{A \in \mathcal{A}} A = \{x : \exists A \in \mathcal{A}, x \in A \in \mathcal{A}\}$$