Is the Empty metric space, complete?

I have two questions.

1. Can the empty set be formed into a metric space?

2. If it exists, is it complete?

I have thought that the empty set is a complete metric space, since we can let $d:\emptyset\times\emptyset\to\mathbb{R}$ to be the empty metric on the empty set, and since there does not exists a Cauchy sequence such that it does not converge on the emptyset.

But, as I was reading a book, I found that a complete metric space cannot be written as a countable union of nowhere dense subsets... By this theorem, I found that the empty set cannot be complete, since the empty set is a nowhere dense subset of itself.

So, do I have to conclude that the empty set cannot be formed into a metric space?

1. It's a matter of convention, but I agree with Qiaochu that the "right" answer is yes. For instance, we want any subset of a metric space to be a metric space. As you noticed, we can just take the metric to be $d: \varnothing \times \varnothing \to \varnothing$, and it certainly satisfies the required axioms: $\forall x \forall y \; d(x,y) = d(y,x)$; $\forall x \; d(x,x) = 0$; $\forall x \forall y \forall z \; d(x,z) \le d(x,y) + d(y,z)$.

   There are of course some technicalities, e.g. the diameter of this space. Here is a related question with no answer yet.

2. The empty metric space is complete.

   Firstly, as you noticed, every Cauchy sequence converges (since there are no Cauchy sequences).

   Secondly, while it's true that the empty metric space can be written as a countable union of nowhere dense sets ($\varnothing$ is nowhere dense, or more easily we can just take the empty union), it looks like this is a special case. Look at the statement of the Baire category theorem here. From the definition $\varnothing$ checks out as a Baire space, so BCT1 holds. BCT3 is stated as

   BCT3. A non-empty complete metric space is NOT the countable union of nowhere-dense closed sets.

   Notice that the space is stipulated to be non-empty, so BCT3 holds vacuously as well. (There must be something in the proof that BCT1 and BCT3 are equivalent that causes the non-empty condition to arise.)