About refinements of order at most $d+1$ I am a little confused about the definition of covering dimension of a topological space. In particular, when defining the dimension of a space $X$, we take the minimum over all natural numbers such that every open cover of $X$ has a open refinement of order at most $n+1$.
Why at most? What is an example of a topological space of covering dimension $d$ with an open cover which has a refinement of order at most $d+1$, but not a refinement of order precisely $d+1$?
Edit: for sake of clarity, here are the relevant definitions, as stated by Munkres, word by word, in paragraph 50 of his book "Topology":
A collection $\mathcal A$ of subsets of the space $X$ is said to have order $m+1$ if some point of $X$ lies in $m+1$ elements of $\mathcal A$, and no point of $X$ lies in more than $m+1$ elements of $\mathcal A$.
A space $X$ is said to be finite dimenstional il there is some integer $m$ such that for every open covering $\mathcal A$ of $X$, there is an open covering $\mathcal B$ which refines $\mathcal A$ and has order at most $m+1$. The topological dimension of $X$ is defined to be the smallest value of $m$ for which this statement holds.
Second edit, possibly a weaker question which does not involve the notion of dimension: given a topological space $X$ and an open cover $\mathcal U$ of order $m$, is it always possible to find an open refinement of $\mathcal U$ of order $m+1$?
 A: If you take a space like $\Bbb Q \cup (0,1)$ in the topology it inherits ffrom $\Bbb R$ a cover for it will have a refinement that is order $1$ (i.e. pairwise disjoint) on the rationals outside $(0,1)$ but order $2$ on the $(0,1)$ part, making the order for the whole cover 2 as well. The "at most" clause is mostly to work for such "hybrid" spaces where the so-called "local dimension" is not constant across the space. I often get confused by the "off by 1" nature of the definition (order 1 = disjoint, order 2: at most pairwise intersecting, which can be achieved in $\Bbb R$ etc.)
A space like $\Bbb R\setminus \{0\}$ has a cover of order $1$ (obvious ) but not every open cover of it has this property.
A boring observation but it shows the need for the "at most" clause if we consider all covers.
The covering definition is due to Lebesgue (early 20th century) but he could not prove rigorously that $\dim \Bbb R^n = n$. This is what Brouwer did later (via a detour, his own inductively defined dimension, "Dimensionsgrad" (papers in German..) using his own fixed point theorem as a tool... Finally topologists had a way to prove that all $\Bbb R^n$ were not homeomorphic for different $n$ (the space-filling curves had seeded doubts: if the continuous image of $[0,1]$ can fill a solid ball continuously, what is our intuition worth?) This as a historical footnote.
