Consider the alphabet $\Sigma = \{a,b\}$. For which languages does the Myhill–Nerode equivalence relation have exactly one class?

From what I understand about equivalence classes, each state is considered a class. So would $\{ a^n : n>0\}$ be the one class?

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    – D.W.
    Commented Apr 13, 2016 at 22:22
  • $\begingroup$ It should be: "for which languages does the Myhill–Nerode equivalence relation have exactly one class?" $\endgroup$
    – A.Schulz
    Commented Apr 14, 2016 at 15:38
  • $\begingroup$ Your example needs three states. aaa would be in an accepting state, eps and aba wouldn’t. You can move from eps to an $\endgroup$
    – gnasher729
    Commented Feb 15 at 18:50

2 Answers 2


Assuming that you mean the Myhill-Nerode relation, the Myhill-Nerode theorem states that the number of equivalence classes exactly equals the number of states in the minimal DFA accepting the language (if it's regular; otherwise there are infinitely many equivalence classes). So we have reduced the question to the following one:

Which regular languages can be accepted using a DFA having only one state?

I'm sure you can answer this one yourself.


Let $L \subseteq \{ a, b\}^*$ be a language with one Myhill-Nerode equivalence class. You can show that $L$ must be trivial, that is $L\in \{\emptyset, \{a, b\}^*\}$, either by using the Myhill-Nerode theorem as Yuval did, or by showing that directly. Indeed, as for $L$ to have a single Myhill-Nerode equivalence class, it means that for all three words $x,y$ and $z$ over $\{ a, b\}$, it holds that $$x\cdot z \in L \Longleftrightarrow y\cdot z \in L$$

In words, we cannot separate any two words $x$ and $y$, by any word $z$. Hence, $L$ must be trivial, because otherwise there are two words $x\in L$, and $y\notin L$, and for $z = \epsilon$, we get that $x \cdot z = x \in L$, but $y\cdot z = y \notin L$, and we have reached a contradiction.


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