I'm trying to exhibit two formal languages $A,B ⊆ \{0,1\}^*$ such that $A^* = B^*$ and $\{0,1\}$ is contained in $A$ but not in $B$.

Finding a language for $A$ is very easy, but I get stuck on $B$, because since $A^*=B^*$, and A has $\{0,1\}$, that $B^*$ must also have $\{0,1\}$. I can't think of a language that has $\{0,1\}$ in $B^*$ but would not be in $B$. Maybe I'm missing something.

  • $\begingroup$ If $0 \in A$ then $0 \in A^* = B^*$ and so $0 \in B$. Similarly $1 \in B$. So this is impossible. Perhaps you got the question wrong? $\endgroup$ – Yuval Filmus Feb 5 at 18:06
  • $\begingroup$ A={0,1} and B={0,1,00,01,10,11} would those two work? $\endgroup$ – James Swanson Feb 5 at 18:27
  • $\begingroup$ Well, $\{0,1\}$ is contained in $B$. $\endgroup$ – Yuval Filmus Feb 5 at 18:28
  • $\begingroup$ isnt {0,1} exclusively not a part of b, because it is the set {0,1} and not the string 01 $\endgroup$ – James Swanson Feb 5 at 18:29
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    $\begingroup$ You haven't explained what "contained" means (in fact, you used a different word), but assuming the usual meaning "$\subseteq$", then $\{0,1\}$ is certainly contained in your $B$. $\endgroup$ – Yuval Filmus Feb 5 at 18:30

If $0 \in A$ then clearly $0 \in A^* = B^*$. Conversely, if $0 \in B^*$ then $0 \in B$: indeed, if $0 = w_1 \ldots w_n \in B$, then exactly one of the $w_i$ can be non-empty, and it must equal $0$. Therefore $\{0,1\} \subseteq A$ and $A^* = B^*$ implies $\{0,1\} \subseteq B$. In other words, there are no two languages $A,B$ satisfying your constraints.


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