I am trying to prove whether $L = \{s : s \text{ contains exactly 1 a}\}$ for the alphabet $\Sigma = \{a, b\}$is regular or not using the Pumping Lemma. I think it is regular because I can construct a simple deterministic finite automata that accepts it.
However, if $L$ is regular, then, by the Pumping Lemma, for any string $s \in L$ of length at least $p$, $xy^iz \in L$ for any $i \ge 0$. If $y$ happens to contain the string with that one $a$, then $xy^0z = xz \not\in L$. And... that makes $L$ not pumpable?
Where did I err here?
For any long enough word, there exists an initial section of that word which contains a "middle" section which is pumpable. For this particular language, that middle section can never contain the letter $a$.