Difference between {$\lambda$}* and ø* What is the difference between {$\lambda$}* and ø*?
I know that ø* is an empty string set, but how is that different from  {$\lambda$}* ?
 A: If this is meant to be languages and Kleene star, and $\lambda$ is the empty string, it turns out that
$\begin{align}
  \emptyset^*
     &= \{\lambda\} \\
  \{\lambda\}^*
     &= \{\lambda\}
\end{align}$
so they are (suprisingly) equal. But $\emptyset \ne \{\lambda\}$, the former has no elements, the later exactly one (the empty string, of length zero).
There are lots of examples of languages $L_1 \ne L_2$ with $L_1^* = L_2^*$.
A: For any set $A$, $A^*$ is the set of all finite concatenations of elements of $A$, including $\lambda$, the empty sequence. Thus, $\lambda\in A^*$ no matter what $A$ is. In particular, $\lambda\in\varnothing^*$, and $\lambda\in\{\lambda\}^*$. The set $\varnothing$ has no elements, so it’s not possible to construct a non-empty sequence of elements of $\varnothing$ to concatenate, and $\lambda$ is therefore the only member of $\varnothing^*$: as vonbrand says, $\varnothing^*=\{\lambda\}$. It is possible to form non-empty finite sequences of elements of $\{\lambda\}$: they all have the form
$$\langle\underbrace{\lambda,\ldots,\lambda}_{n\text{ copies}}\rangle$$
for some $n\in\Bbb Z^+$, and the all concatenate to yield $\lambda$ again. Thus, $\lambda$ is also the only member of $\{\lambda\}^*$, which therefore also equals $\{\lambda\}$.
