# What is the language of Sigma^n? Confused about meanining

I am learning the Theory of Computation, and I came across the language $$\Sigma^n$$. Could someone please explain what that could mean if $$\Sigma$$ is the alphabet?

Thank you so much!

• We identify $\Sigma$ with the set of all words over $\Sigma$ of length $1$. Then $\Sigma^n$ is just the $n$-th power of the language $\Sigma$, a definition you must have seen in class. Dec 13 '20 at 16:31
• For example, if $\Sigma = \{a,b\}$ then we identify $\Sigma$ with the language consisting of the words $a$ and $b$. Dec 13 '20 at 16:31

By definition $$\Sigma^n$$ is simply $$\underbrace{\Sigma\times \Sigma \times \cdots \times\Sigma}_{n \ \text{times}} = \{(\sigma_1, \sigma_2, \ldots, \sigma_n): \forall \ i\in [n], \sigma_i \in \Sigma \}$$ which is the set of ordered tuples of length $$n$$ over $$\Sigma$$. Usually, when we refer to a tuple $$w = (\sigma_1, \sigma_2, \ldots, \sigma_n)$$, we omit the commas "," and the parenthesis "(" and ")", so the tuple $$w$$ is written as $$\sigma_1\cdot \sigma_2\cdots \sigma_n$$ to which we refer as a word of length $$n$$ over $$\Sigma$$. In other words, $$\Sigma^n$$ is the set of all words of length $$n$$ over $$\Sigma$$.
Formally, the map that maps a tuple in $$\Sigma^n$$ to the word that we get by omitting the commas and the parenthesis from the tuples, is a bijection from $$\Sigma^n$$ to the words of length $$n$$ over $$\Sigma$$.
• Thank you for the answers! If $n = 5$, could I have any words of length 2? Dec 13 '20 at 16:47
• No. $\Sigma^5$ contains all words of length $5$, and only them. Dec 13 '20 at 16:50