Is there a set without a predicate? Is there a set that has no predicate that defines it? I limit this question to the pure set theory.
It seems there are sets whose members have no common exclusive property and so the only way to define such sets is by listing their members. For example a set whose members are number 2, the equilateral triangle with area equal to 10, and the function f(x)=5x+1.
Maybe it is just difficult to find the predicate for some sets but shouldn't there be a predicate for every set, in principle?
EDIT: I mean to ask if every set can be defined by a common property/condition as opposed to listing individually all its members. 
 A: If I've understood your question correctly, you're not interested in predicates per se, since a predicate on a set $X$ is most commonly defined as a function $X \rightarrow \{\mathrm{True},\mathrm{False}\}$. Rather, I think you're interested in definable elements and definable sets. Lets start by clearing something up.
Let $X=\{x_0,\ldots,x_{n-1}\}.$ Then: $$\forall a(a \in X \iff a = x_0 \vee \ldots \vee a = x_{n-1})$$
So if each of the elements $x_0,\ldots,x_{n-1}$ is definable, then the set $\{x_0,\ldots,x_{n-1}\}$ is definable.
But there's a lot more to be said here.
Firstly, make sure you know about Skolem's paradox.
Now. As I understand it:


*

*Every model of ZFC has undefinable sets.

*Every uncountable model of ZFC has undefinable elements.

*Nonetheless, there exist (necessarily countable) models of ZFC whose every element is definable.


A few more comments:


*

*By "undefinable", I mean "not first-order definable without parameters."

*The whole concept of definability is heavily language dependent. If you change first-order logic to something else, don't expect the undefinable elements and/or undefinable sets to remain unchanged.

*Points 1. and 2. hold more generally if we replace ZFC with any countable first-order language. I think Point 3. is specific to ZFC.

