Checking for NaN presence in a container

NaN is handled perfectly when I check for its presence in a list or a set. But I don't understand how. [UPDATE: no it's not; it is reported as present if the identical instance of NaN is found; if only non-identical instances of NaN are found, it is reported as absent.]

1. I thought presence in a list is tested by equality, so I expected NaN to not be found since NaN != NaN.

2. hash(NaN) and hash(0) are both 0. How do dictionaries and sets tell NaN and 0 apart?

3. Is it safe to check for NaN presence in an arbitrary container using in operator? Or is it implementation dependent?

My question is about Python 3.2.1; but if there are any changes existing/planned in future versions, I'd like to know that too.

NaN = float('nan')
print(NaN != NaN) # True
print(NaN == NaN) # False

list_ = (1, 2, NaN)
print(NaN in list_) # True; works fine but how?

set_ = {1, 2, NaN}
print(NaN in set_) # True; hash(NaN) is some fixed integer, so no surprise here
print(hash(0)) # 0
print(hash(NaN)) # 0
set_ = {1, 2, 0}
print(NaN in set_) # False; works fine, but how?

Note that if I add an instance of a user-defined class to a list, and then check for containment, the instance's __eq__ method is called (if defined) - at least in CPython. That's why I assumed that list containment is tested using operator ==.

EDIT:

Per Roman's answer, it would seem that __contains__ for list, tuple, set, dict behaves in a very strange way:

def __contains__(self, x):
  for element in self:
    if x is element:
      return True
    if x == element:
      return True
  return False

I say 'strange' because I didn't see it explained in the documentation (maybe I missed it), and I think this is something that shouldn't be left as an implement