====== Languages and Grammar ======
===== Languages =====
==== Powers of languages ====

If $L$ is a language, then the product $L \cdot L$ is denoted by $L^2$.
The language product $L^n$ for every $n \in \{0,1,2,...\} is defined as follows:

$$
  \begin{aligned}
    L^0 &= \{\Lambda\} \\
    L^n &= L \cdot L^{n-1}, \text{if} \ n > 0
  \end{aligned}
$$

**Example.** If $L = \{a,bb\}$ then the first few powers of $L$ are...
$$
  \begin{aligned}
    L^0 &= \{\Lambda\} \\
    L^1 &= L = \{a,bb\} \\
    L^2 &= L \cdot L = \{aa,abb,bba,bbbb\} \\
    L^3 &= L \cdot L^2 = \{aaa,aabb,abba,abbbb,bbaa,bbabb,bbbba,bbbbbb\}
  \end{aligned}
$$

==== Closures of a Language ====

If L is a language over $\Sigma$ (i.e. $L \subset \Sigma^{\ast}$) then **the closure** of $L$ is denoted $L^{\ast}$, and the **positive closure** of $L$ is denoted $L^+$.

$$
  \begin{aligned}
    L^{\ast} &= L^0 \cup L^1 \cup L^2 \cup ... \\
    L^+ &= L^1 \cup L^2 \cup L^3 \cup ...
  \end{aligned}
$$

==== Closure of an alphabet ====

The closure of $\Sigma$ coincides with out definition of $\Sigma^{\ast}$ as the set f all string over $\Sigma$. In other words, we have a nice representation of $\Sigma^{\ast}$ as follows:

$$
  \Sigma^{\ast} = \Sigma^0 \cup \Sigma^1 \cup \Sigma^1 \cup ...
$$

Where $\Sigma^k$ denotes the set of strings of length $k$, each of whose symmbols in $\Sigma$.
===== Grammar =====