Witness languages for the proper inclusions in the Chomsky hierarchy

The language $L_{ab}$ is context-free via the grammar $S\to aSb\mid \epsilon$. If it were regular, [L2] would give a pumping length $p$, but pumping the word $a^p{b}^p$ inside its first $p$ symbols changes only the number of $a$'s. So $L_{ab}$ witnesses $\text{regular}⊊\text{context-free}\mathrm{.}$

The language $L_{abc}$ is decidable by a machine that first checks the form $a^{\ast }{b}^{\ast }{c}^{\ast }$ and then counts the three block lengths and compares them. If it were context-free, [L3] would give a pumping length $p$ for the word $a^p{b}^p{c}^p$, but any factor $vwx$ with $\mathrm{\mid }vwx\mathrm{\mid }\le p$ lies in at most two consecutive blocks, so pumping $v$ and $x$ changes at most two symbol counts and destroys equality of all three counts. Thus $L_{abc}$ witnesses $\text{context-free}⊊\text{decidable}\mathrm{.}$

By [L4] and [L5], $A_{TM}$ is computably enumerable but not decidable. So it witnesses $\text{decidable}⊊\text{computably enumerable}\mathrm{.}$

These are exactly the three strict containments proved in [L1]. The point of the example is that one should keep the witnesses separate rather than expecting one language to witness every gap at once.