>>69
It depends you idiot. What happens if I have some function called f defined over a set of natural numbers that doesn't satisfy either the onto on one-to-one condition that is necessary for countability?
For your statement to make even a shred of sense it needs to be a function defined over some other set to the natural numbers, not over some subset of the natural numbers. And even then you have proved nothing, you may always construct such a function from any set to another set, consider a set X now let f(x) = 0 for each x in X, this is a function which is neither bijective or surjective while X might be countable or uncountable.
If I'm generous I'm guessing what you really mean is that if there does not exist any function f that is an isomorphism from the set to the natural numbers then the set isn't countable.
>>71
Yeah the metric you're looking for is d(n, m) = if n != m then 1 else 0, the naturals are bounded with respect to this metric.
>>75
Is correct and the proof is even in the post.
>>83
You seem a bit confused here, [0, 1] is a real interval, it is defined as the set { x in the Reals : 0 <= x <= 1}, and that is not a subset of the natural numbers.
If I'm being generous I suppose what you're trying to state is that {0, 1} is a subset of the natural numbers, which might or might not be true, most definitions of the natural numbers don't contain 0 but some do, let's be generous and assume that 0 is a member of the natural numbers.