1. kleene-closure

Hello, I stuck with this exercise

The kleene-closure of an Language L ist the smallest language, which contines L as subset and the empty word as an element and which is closend under concatenation. I have to profe by solving a-e

a) L $\subseteq$ L*
b) L* $\cdot$ L* $\subseteq$ L*
c) For every L' (over same Alphabet like L) with L $\subseteq$ L' , $\epsilon$ $\in$ L' and L' $\cdot$ L' $\subseteq$ L' obtains L* $\subseteq$ L'.

I think I solved a) by showing

L = {w1 | w1 $\in$ L} L*={w1,.....,wn | n $\in$ N, $w_i$ $\in$ L for all i = 1....n}

=> w1 $\in$ L and w1 $\in$L* => L $\subseteq$L*

b) and c) I stuck and any help will be appreciated

2. Re: kleene-closure

Originally Posted by suslik
b) L* $\cdot$ L* $\subseteq$ L*
Suppose $w\in L^*\cdot L^*$. Then $w=uv$ for some $u, v\in L^*$, i.e., u and v are concatenations of words from L. What can you way about uv?

Originally Posted by suslik
c) For every L' (over same Alphabet like L) with L $\subseteq$ L' , $\epsilon$ $\in$ L' and L' $\cdot$ L' $\subseteq$ L' obtains L* $\subseteq$ L'.
The definition of $L^*$ says that it is the smallest language satisfying these properties.

3. Re: kleene-closure

Perhaps that uv is already in L* because every concatenation of L is already in L* ?

4. Re: kleene-closure

Yes, if u and v are concatenations of words in L, then so is uv, and therefore it is in L*.