Below is the Pumping lemma as stated in Automata and Computability by (Dexter C. Kozen)

Let $$A$$ be a regular set. Then the following property holds of $$A$$:

There exist $$k≥0$$ such that for any string $$x,y,z$$ with $$xyz \in A$$ and $$|y|≥k$$, there exist strings $$u,v,w$$ such that $$y=uvw$$,$$v\not=ε$$, and for all $$i≥0$$, the string $$xuv^iwx\in A$$

Here are my questions:

Let $$∑=\{1\}$$ $$A=\{1\}$$. Clearly $$A$$, is regular. Now, applying the pumping lemma, let $$x=1,y=ε,z=ε$$ and so $$1εε=1 \in A$$, and $$|y|=0$$ so $$k=0$$ but there is no way to split $$y$$ such that $$v \not = ε$$. So, how is the pumping lemma valid for this regular set?

Also, we can think of $$k$$ as being the number of states, but a DFA cannot have $$0$$ states. So why is $$k$$ not required to be strictly greater than 0.

Perhaps you can help my understanding by applying the pumping lemma above to the language $$\{1\}$$ as well the language of the empty string. What do value does k have in these situations? What about x,y,z and u,v,w?

Suppose that $$A$$ is regular. The pumping lemma says that there is a $$k$$ such if $$w\in A$$ has length at least $$k$$, then $$w$$ can be pumped; it does not say that $$A$$ necessarily has any words of length $$k$$ or more. In fact it’s clear that if $$A$$ actually does have a word of length at least $$k$$, then pumping it will produce infinitely many words. Thus, if $$A$$ is finite, every word of $$A$$ must be shorter than $$k$$. In particular, for $$A=\{1\}$$ we can take $$k$$ to be any integer greater than $$1$$: $$A$$ will then satisfy the conclusion of the pumping lemma vacuously, since $$A$$ has no words of length $$k$$ or more. More generally, if $$A$$ is finite, and $$k>\max\{|w|:w\in A\}$$, then $$A$$ vacuously satisfies the conclusion of the pumping lemma, since it has no words long enough to be pumped.
The pumping lemma is really only interesting when $$A$$ is infinite.