Show that $R^{+} \equiv R \leftrightarrow L(RR) \subset L(R)$

Question

Show that $R^{+} \equiv R \leftrightarrow L(RR) \subset L(R)$

sigma is any alphabet. R is a regular expression.

How can L(RR) even be a subset or equal to L(R)?

Answer 1

Let me first answer your second question. Take $R = a^+$, or $R = a^*$. In both cases, $R^2 \subseteq R$.

Coming back to your first question, first recall that $R^+ = \sum_{n > 0}R^n$. Thus if $R^+ = R$, then $R^n \subseteq R$ for all $n > 0$ and in particular $R^2 \subseteq R$.

Suppose now that $R^2 \subseteq R$. I give you a hint to prove that $R^+ = R$.

Hint. Show that $R^n \subseteq R$ for all $n > 0$ and conclude that $R^+ \subseteq R$. Deduce that $R^+ = R$.