During a test in Automata I had to prove that $L[(\epsilon)^*]=\{\epsilon\}$ for REs, where $\epsilon$ is the empty word. I didn't prove the following last step because I thought it was trivial:
$$ \bigcup_{i=0}^{\infty}\{\epsilon\}=\{\epsilon\} $$ And lost two points. Now that I'm trying to think of a proof for that simple theorem, I can't think of one. I can't use induction because the amount of $i$'s is not finite. How should I prove it?
On
You didn't really prove the equality. You need to start with the formal definition of $L^*$: $$ L^* = \bigcup_{n \geqslant 0} L^n $$ with the convention that $L^0 = \{\epsilon\}$. Now, you have to apply this definition with $L = \{\epsilon\}$ and show that, for all $n \geqslant 0$, $\{\epsilon\}^n = \{\epsilon\}$. This can easily be proved by induction, but this is really the missing part in your proof. Once you have proved this, you can use your infinite union.
Two sets $A$ and $B$ are equal if for every $x$ we have that $x \in A$ if and only if $x \in B$.
The countable union $\bigcup^{\infty}_{i=1} A_i$ is the set of elements $\{ x \mid \exists 1 \leq i < \infty. x \in A_i \}$. From this we see, that if $A_i = A_j = A$ for every $i,j$ and some $A$, we then have that $\bigcup^{\infty}_{i=1} A_i = A$.