A doubt

Question

This question might be silly,

Suppose an automata A accepts a language L. Can I say that the Automata also accepts a language L' such that L' ⊂ L ?

For example, consider the language L = ab* + (ab)*, L₁ = ab*, L₂ = (ab)* , L₁ ⊂ L and L₂ ⊂ L. Since this is regular, a DFA would exist for this. Can we say that the L₁ is also accepted by the DFA ?

Thanks in advance!

Answer 1

No. Regular set is not closed under subset operation. For example the irregular language $\{a^nb^n\mid n >0\}$ is a subset of the regular language $a^*b^*$. Also any language including not even recursively enumerable is a subset of the regular language $\Sigma^*$

Comments

Ok, I think I have to frame my question in a different way.

We say that a Language L is accepted by a Automaton A if for all strings w in L, a set of state transitions in the Automaton leads to the final state. For the example I've given in the question, this is true for L₁ as well as L₂. So can we say that the DFA, in the example, accepts L₁ ?