The range objects are special:

Python will compare range objects as Sequences. What that essentially means is that the comparison doesn't evaluate how they represent a given sequence but rather what they represent.

The fact that the start, stop and step parameters are completely different plays no difference here because they all represent an empty list when expanded:

For example, the first range object:

list(range(0))  # []

and the second range object:

list(range(2, 2, 2)) # []

Both represent an empty list and since two empty lists compare equal (True) so will the range objects that represent them.

As a result, you can have completely different looking range objects; if they represent the same sequence they will compare equal:

range(1, 5, 100) == range(1, 30, 100) 

Both represent a list with a single element [1] so these two will also compare equal.

No, range objects are really special:

Do note, though, that even though the comparison doesn't evaluate how they represent a sequence the result of comparing can be achieved using solely the values of start, step along with the len of the range objects; this has very interesting implications with the speed of comparisons:

r0 = range(1, 1000000)    
r1 = range(1, 1000000)

l0 = list(r0)    
l1 = list(r1)

Ranges compares super fast:

%timeit r0 == r1
The slowest run took 28.82 times longer than the fastest. This could mean that an intermediate result is being cached 
10000000 loops, best of 3: 160 ns per loop

on the other hand, the lists..

%timeit l0 == l1
10 loops, best of 3: 27.8 ms per loop

Yeah..

As @SuperBiasedMan noted, this only applies to the range objects in Python 3. Python 2 range() is a plain ol' function that returns a list while the 2.x xrange object doesn't have the comparing capabilies (and not only these..) that range objects have in Python 3.

Look at @ajcr's answer for quotes directly from the source code on Python 3 range objects. It's documented in there what the comparison between two different ranges actually entails: Simple quick operations. The range_equals function is utilized in the range_richcompare function for EQ and NE cases and assigned to the tp_richcompare slot for PyRange_Type types.

I believe the implementation of range_equals is pretty readable (because it is nice as simple) to add here:

/* r0 and r1 are pointers to rangeobjects */

/* Check if pointers point to same object, example:    
       >>> r1 = r2 = range(0, 10)
       >>> r1 == r2
   obviously returns True. */
if (r0 == r1)
    return 1;

/* Compare the length of the ranges, if they are equal 
   the checks continue. If they are not, False is returned. */
cmp_result = PyObject_RichCompareBool(r0->length, r1->length, Py_EQ);
/* Return False or error to the caller
       >>> range(0, 10) == range(0, 10, 2)  
   fails here */
if (cmp_result != 1)
    return cmp_result;

/* See if the range has a lenght (non-empty). If the length is 0
   then due to to previous check, the length of the other range is 
   equal to 0. They are equal. */
cmp_result = PyObject_Not(r0->length);
/* Return True or error to the caller. 
       >>> range(0) == range(2, 2, 2)  # True
   (True) gets caught here. Lengths are both zero. */
if (cmp_result != 0)
    return cmp_result;

/* Compare the start values for the ranges, if they don't match
   then we're not dealing with equal ranges. */
cmp_result = PyObject_RichCompareBool(r0->start, r1->start, Py_EQ);
/* Return False or error to the caller. 
   lens are equal, this checks their starting values
       >>> range(0, 10) == range(10, 20)  # False
   Lengths are equal and non-zero, steps don't match.*/
if (cmp_result != 1)
    return cmp_result;

/* Check if the length is equal to 1. 
   If start is the same and length is 1, they represent the same sequence:
       >>> range(0, 10, 10) == range(0, 20, 20)  # True */
one = PyLong_FromLong(1);
if (!one)
    return -1;
cmp_result = PyObject_RichCompareBool(r0->length, one, Py_EQ);
Py_DECREF(one);
/* Return True or error to the caller. */
if (cmp_result != 0)
    return cmp_result;

/* Finally, just compare their steps */
return PyObject_RichCompareBool(r0->step, r1->step, Py_EQ);

I've also scattered some of my own comments here; look at @ajcr's answer for the Python equivalent.

‍range(a, b, s) is [a, a+s, a+2*s..., x] where x < b.

So range(0) -> range(0,0,1) generates an empty list. This means the inside of the for loop is skipped which returns True.

In Python 3.x ,

range(0,3) returns a class of immutable iterable objects that lets you iterate over them, it does not produce lists, and they do not store all the elements in the range in memory, instead they produce the elements on the fly (as you are iterating over them) , whereas list(range(0,3)) produces a list (by iterating over all the elements and appending to the list internally) .

Example -

>>> range(0,3)
range(0, 3)
>>> list(range(0,3))
[0, 1, 2]

Ideally, if you only want to iterate over that range of values , range(0,3) would be faster than (list(range(0,3)) because the latter has the overhead of producing a list before you start iterating over it.

In Python 2.x , range(0,3) produces an list, instead we also had an xrange() function that has similar behavior of range() function from Python 3.x (xrange was renamed to range in Python 3.x)

For Python 3.5, From the documentation -

Range objects implement the collections.abc.Sequence ABC, and provide features such as containment tests, element index lookup, slicing and support for negative indices

So you can do things like -

>>> range(0,10)[5]
5
>>> range(0,10)[3:7]
range(3, 7)
>>> 5 in range(6,10)
False
>>> 7 in range(1,8)
True

And all of these are constant time operations , as can be seen from this test -

In [11]: %timeit a = xrange(0,1000000)[1000]
1000000 loops, best of 3: 342 ns per loop

In [12]: %timeit a = xrange(0,1000000)[10000]
1000000 loops, best of 3: 342 ns per loop

In [13]: %timeit a = xrange(0,1000000)[100000]
1000000 loops, best of 3: 342 ns per loop

In [14]: %timeit a = xrange(0,1000000)[999999]
1000000 loops, best of 3: 342 ns per loop

In [15]: %timeit a = xrange(0,10000000)[9999999]
1000000 loops, best of 3: 339 ns per loop

In [16]: %timeit a = xrange(0,1000000000000)[9999999999]
1000000 loops, best of 3: 341 ns per loop