Why are Swift ranges are not comparable?

Why are Swift's range types are not Comparable, if their element type T is comparable? (such as Int, String, Double, etc.).

For example:

let rangeFive = 1...10
let rangeSix = 1...10
rangeFive < rangeSix

OR

let rangeFive = 1...10
let rangeSix = 1...10
rangeFive <= rangeSix

OR

let rangeFive = 1...10
let rangeSix = 1...10
rangeFive == rangeSix


let minRangeFive = rangeFive.min()
let maxRangeSix = rangeSix.max()
minRangeFive < maxRangeSix


// Error: Binary operator '<' cannot be applied to two 'ClosedRange<Int>.Element?' (aka 'Optional<Int>') operands

I've tried rangeFive == rangeSix - and it works, but not with other comparison operators. Why is it that when range types contain Comparable data, all range types are not Comparable and do not conform to the Comparable protocol? comparable swift ranges are not comparable, i.e. ‘all range types are not comparable, i.e. do not meet the requirements of the Comparable protocol’, while the range data type partialrangefrom, partialrangeupto, partialrangethrought<T>, where T is a comparable data type(Int, String, Double), is comparable.

Making Ranges comparable would lead to more confusion than it's worth, because there's no obviously-correct answer for what it means for one range to be greater than another.

We can agree that 1...2 is definitely less than 3...4, and that 1... should probably be considered less than 2....

But what about overlapping ranges?

Should 10...20 be less than 10...100?

There are several ways we could define it, and each of them has issues:

1. Range a is less than range b if both its lower and upper bounds are both less than b's upper bounds.
   • This doesn't work for overlapping range, where neither is less or greater than the other. This violates Comparable's requirement to be a total order.
2. Range a is less than range b only if its lower bound is lower than the lower bound of the other
   • This leads to other strange situations, where 1...10 < 2...9 would be true (because 1 < 2), but simultanously 1...10 > 2...9 would be true (because 10 > 9).
3. Perhaps you could define it in terms of the "center" of the range, but:
   • That doesn't work for infinite ranges
   • Doesn't work for elements whose average can't be computed. E.g. what's the mid-point of the range "foo"..."bar"?

Equality doesn't have this issue, because you can just compare the bounds, and have an unambiguously correct answer. That's why ranges are conditionally Equatable, but never Comparable.