The following picture is extracted from Munkres' "Topology" P305.

http://i3.6.cn/cvbnm/dd/ef/6c/11096e...178584a807.jpg

I have problem when the space $\displaystyle X$ is the empty set $\displaystyle \emptyset$. In this case, the only two open covering is $\displaystyle \emptyset$ and $\displaystyle \{\emptyset\}$. For $\displaystyle \emptyset$, the only open refinement is $\displaystyle \emptyset$, which is also an open covering of $\displaystyle \emptyset$. For $\displaystyle \{\emptyset\}$, its only open refinements are $\displaystyle \emptyset$ and $\displaystyle \{\emptyset\}$, both of them are open coverings of $\displaystyle \emptyset$. Now the problem comes: what is the order of $\displaystyle \emptyset$ and $\displaystyle \{\emptyset\}$, respectively? or, what is the "some point" in the definition of?order

Here are some of my understandings:

(1)The only topology of $\displaystyle \emptyset$ is $\displaystyle \{\emptyset\}$, which is discrete. According to Ex 1. of this section(P315), it seems that dim $\displaystyle \emptyset$ should be 0?

(2)$\displaystyle \cup\emptyset=\emptyset$, so $\displaystyle \emptyset$ is a covering of $\displaystyle \emptyset$.

(3)The first-order logic formulation of "$\displaystyle \cal B$ is a refinement of $\displaystyle \cal A$" is: $\displaystyle \forall x(x\in{\cal B}\rightarrow{x}$ is a subset of some element of $\displaystyle \cal A)$. For $\displaystyle \cal B=\emptyset$, the left part of $\displaystyle \rightarrow$ would always be false, then this implication is vacuously true, and in turn the whole wff is true, so $\displaystyle \emptyset$ is a refinement of $\displaystyle \cal A$. Note: this logic is just like what we use to prove that the empty set is contained in any set.

(4)"open refinement" means that every element in the collection is open. Express it as a wff: $\displaystyle \forall x(x\in{\cal B}\rightarrow{x}$ is open). The same discussion as (3) shows that $\displaystyle \emptyset$ is an open refinement of $\displaystyle \cal A$. Additionally, $\displaystyle \emptyset$ is an open covering of $\displaystyle \emptyset$ for the same reason.

(5)Expressed in first-order logic wff, a natural number $\displaystyle m+1$ is the order of $\displaystyle \cal A$ if the following is true: $\displaystyle [\exists x(x\in{X}\wedge n(x)=m+1)]\wedge[\neg\exists x(x\in{X}{\wedge}n(x)>m+1)]$, where $\displaystyle n(x)$ is a function giving the number of elements of $\displaystyle \cal A$ that contains $\displaystyle x$, i.e., $\displaystyle n(x)=|\{A\in{\cal A}|x\in{A}\}|$. When $\displaystyle X=\emptyset, \exists x(x\in{X}\wedge ...)$ is always false, then no natural number can satisfy the definition of order. This is where my problem lies in.

Another book written by Engelking, "General Topology", says in P385 explicitly that dim $\displaystyle \emptyset$=-1. I was totally confused.

Please help me explain (in rigorous logic) what is the topological dimension of $\displaystyle \emptyset$, if it exists. Thanks!