Given a $\Sigma$ an alphabet, is $\epsilon$ in it logically?
For example, if I have a function $ f : \Sigma \to \Sigma $, can I define it for example $ f(\sigma) = \epsilon$? even if my alphabet is for example only $\Sigma = \{a,b,c\}$?
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2$\epsilon$ ususally denotes the empty string (word) of length $0$. It is not in any alphabet, but it is a string over any alphabet. – user642796 Nov 12 '13 at 11:40
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I see. So, I -can- define $f$ like I did? – TheNotMe Nov 12 '13 at 11:42
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2Your function should be denoted $f\colon\Sigma^\to\Sigma^$, but then you can define it the way you did. – Harald Hanche-Olsen Nov 12 '13 at 11:47
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As Arthur Fischer wrote in the comments: $\epsilon$ is the empty string, the word of length zero. It is not contained in any alphabet (apart from some alphabet specifically defined to contain the empty string), but it is a string over any alphabet.
You should define $f:\Sigma^* \rightarrow \Sigma^*$, which is a function from the strings over the alphabet to the strings over the alphabet. If you were to leave out the stars, it'd be just from the alphabet to the alphabet, which I don't think is what you want.
(An analogue to this is regular set notation - usually, for some set $A$, $\emptyset \notin A$ unless explicitly defined otherwise, but $\emptyset \subseteq A$ for all $A$. Similarly for $\epsilon$ and $\Sigma$.)
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