Epsilon and phi in automata

Question

If $L_1 = Ø$, I read that $L_1^\ast$ is $\{\varepsilon\}$. Why is this so? I am unable to understand this. Can someone please explain? I understand that $Ø$ is an empty string while $\varepsilon$ allows us to make transitions without symbols. But, why is $L_1^\ast$ not $Ø$ and why is it $\{\varepsilon\}$?

Answer 1

The definition of Kleene star: $$ A^\ast:=\bigcup_{i=0}^n A^n $$ where $A^0=\{\varepsilon\}$ (see, for example, wiki: http://en.wikipedia.org/wiki/Kleene_star). If $A = \emptyset$ then $A^i = \emptyset$ for $i \geq 1$, but $A^0$ is still there.

Answer 2

You say

• Ø is an empty string - wrong.
• ε allows us to make transitions without symbols - so what is it?

In fact

• $\emptyset$ is the empty set.
• $\epsilon$ is the empty string.

All the strings that can be made with no things (letters or whatever you want to call them) are just one: the empty string.

Answer 3

Going back to Kleene star's definition

If $L$ is a set of symbols or characters then $L^*$ is the set of all strings over symbols in $L$, including the empty string $\epsilon$.

In the case $L=\varnothing$, following the definition we get $L^{*} = \{\epsilon\}$