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So we're taking languages, and I understand regular expressions from a computer science perspective. I know that L* is basically L repeated any number of times from 0 to infinity (So if L = {a}, then L* is {empty, a, aa, aaa ...}.

But while doing proofs, a number of times there has been mention of L = L* or L = L+ or L = LL. (With L NOT being empty)

I don't get how this can be possible? There's even a question that states L = L* iff LL = L.

I have already written down my answer but I still don't understand it intuitively.

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Assume $L=a^*$. Can you compute $L^*$ ? More generally, can you compute $(L^*)^*$ ?

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    Ohhh... one of our problems was proving that $L^* = (L^*)^*$. So in this case for $L = L^*$ does that mean that $L$ itself is $\{\epsilon, a, aa,....\}$. $\\$**OK rereading that makes me feel like an idiot. Thanks so much! – 2017-01-27
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    For a general solution, you can use a proof similar to Arden's lemma : assume $LL= L$. Then let $m \in L^*$, e.g. $m \in L^k$, then $m\in L$ (easy induction). Since $L\subset L^*$ is obvious, we have $L=L^*$. The converse shouldn't be troublesome if you understand what I've done there ;) – 2017-01-27
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    I actually lost you at $L \subset L^*$ is obvious. I hate saying this phrase as it makes me sound dumb, but can you explain in plain english? My intuition so far is: If $L$ is the same as $LL$ then $L$ is this magic thing that is an infinite repetition of the same thing, which is in essence $L^*$. – 2017-01-29
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    Ok so up to $L\subset L^*$ you understand ? Remember that $L^* = \displaystyle \bigcup_{k\in \mathbb{N} } L^k$ by definition, where $L^k$ is the language consisting of the words that are made up of $k$ words of $L$, glued together. Now what is $L^1$ ? That should answer this. Now since we have both $L\subset L^*$ and $L^* \subset L$, what do we have ? Now for the converse, assume $L= L^*$. Then what is $LL$ ? It's the language consisting of words that are two words of $L$ glued together, in other words $L^2$. But look at the definition of $L^*$ : that means $L^2 \subset L^* = L$. – 2017-01-29
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    So we have $LL \subset L$. Can you prove the converse inclusion ? (Knowing that we assumed $L^* = L$) – 2017-01-29
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    Ohh yes I can, isn't this pretty straightforward? $L \subset LL$ because if $L = L^*$ then $\epsilon \in L$ which means that when $LL = L\epsilon$ then $LL = L$? Is that acceptable as a proof? (I already submitted my hw, just asking in general) – 2017-01-29
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    Well more precisely, if $m$ is in $L$, then since $\epsilon \in L$, $m=m\epsilon \in LL$ but I think you understood that – 2017-01-29
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    Oh ok. Thanks for that last part. I always have trouble putting my thoughts into a presentable format. and thanks for everything!! – 2017-01-29