Pumping lemma for regular languages

For a deterministic finite automaton with $N$ states, every accepted word $w$ with $|w|\ge N$ has a decomposition $w=xyz$ such that $|xy|\le N$, $|y|\ge 1$, and $x{y}^{i}z$ is accepted for every integer $i\ge 0$.

Over an alphabet containing $a,b$, the language $\{w^n:n\ge 2\}$ is not regular language. Intersecting it with the regular language $a^{+}b{a}^{+}b$ gives exactlywhich fails the pumping lemma for regular languages.

Let $K\subseteq \mathbb{N}$ be undecidable andPumping its first symbol always stays in $L$, so it satisfies the pumping lemma with pumping length one. Yet $L\cap a{b}^{\ast }=\{a{b}^n:n\in K\}$ is not regular, because a finite automaton for it would decide $K$.

The base-two representations of powers of two form the regular language $10^{\ast }$. Their base-ten representations are not regular. Pumping a fixed decimal substring of length $d$ would produce powers of two $a_i$ satisfying $a_{i+1}-10^d{a}_i=C$. After division by $a_i$ and passage to large $i$, integrality would force a power of two to equal $10^d$, contradicting unique factorization.