A question on deficient values of entire functions Recently I come cross a question about deficient values of entire functions.
I find that  many examples in the book about functions $f$ whose deficient values are singularities of the inverse $f^{-1}$. 
I want to know whether there exist an example with the following property (in some sense it aks whether the concept of deficient value has a geometric explanation):
Is there an entire function $f$ with deficient value $a$ such that $a$ is not in the closure of the singular value set?
Another question:  For entire functions in the Eremenko-Lybuich class, is it true that the the set of finite difficiency value is also bounded?
Any comments will be appreciated. 
 A: The book of Goldberg and Ostrovskii MR2435270 contains several examples of functions whose
deficient value is not asymptotic. And in fact there are such functions
without asymptotic values at all. But deficient value must be in the closure
of the singular set. 
This follows from a theorem of E. Collingwood, which has many generalizations,
by Tumura, H. Selberg, Kakutani and H. Cartan. 
You can find one generalization as Theorem 1.2 in Chapter 7 of Goldberg - Ostrovskii. 
In particular, for a function of class $S$, only singular values can be deficient.
If you search carefully, you can find this book on the Internet.
A: I like to advocate differentiating between the singularities of the inverse function, which are denoted by $\operatorname{sing}(f^{-1})$, and the set of singular values of $f$, which I denote by $S(f)$. The latter set can be defined as the closure of $\operatorname{sing}(f^{-1})$, though it also has an intrinsic definition: it is the complement of the set of values having a neighbourhood $U$ over which $f$ is an unramified covering map. On the other hand, $\operatorname{sing}(f^{-1})$ is precisely the set of critical and asymptotic values of $f$.
Then Alex's answer says that a deficient value need not belong to $\operatorname{sing}(f^{-1})$, but must always lie in $S(f)$. The latter part should be intuitively clear, since $f$ takes any other value in a set $U$ as above the same number of times, and the set of deficient values is countable.
