Short quiz about Myhill-Nerode theorem Given is DFA M: 





1.) abb and bbb

2.) baa and ab

3.) a and ε (where ε is the empty word)

4.) abb and baa

If I understood correctly, x and y are equivalent to each other (in terms of Nerode), if you add the same suffix for both such that they can be read by the DFA, e.g. they are in L(M)

Thus 1.) is false because abb throws us into the "endless" state 5, there is no escape and thus no suffix.

2.) Is true, simply add suffix aa and both words will still be accepted.

3.) True, take a as suffix.

4.) False, same reason as 1.) you get into the endless state and there is no escape, thus no suffix.



Can you confirm if this is correct? Or did I understand the entire concept wrong? :o
You are mistaken. Strings $x$ and $y$ are equivalent modulo the language $L$ iff for  all suffixes added to the words the results are both in or both outside of the language:

$x\sim_L y$ iff $xw\in L \iff yw\in L$ for every $w$.

In particular, when the strings $x$ and $y$ end in the same state for the (deterministic) automaton $M$, then $x\sim_{L(M)} y$, as they can be extended by exactly the same strings. This also holds for the dead-end state where no strings can extend the original string to acceptance.