Hundreds of thousands of mathematicians are wrong.
Your reply says yes, but then your construction does not do it. In particular you said:
> The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation.
But the set N does not have transfinite natural numbers, so q does not map F to N, it maps F to something else.
So I ask again, is it possible to have m:N -> F, such that for every f in F, there is an n in N such that m(n)=f?
How many natural numbers are there? How many bits does it take to represent the average natural number? If you believe the natural numbers do not include transfinite numbers then how do you pick a successor when counting? There are infinite picks to be made so some of the picks must be transfinite. What I am calling a transfinite natural number must exist in N because N is an infinite set.
Assume that N has only finite numbers in it but is itself an infinite set. Would you care to tell me which number (or numbers) are listed twice? But then it is not really a set!
Infinitely many, more than any finite number.
> If you believe the natural numbers do not include transfinite numbers then how do you pick a successor when counting?
Just add one.
> There are infinite picks to be made so some of the picks must be transfinite.
No, adding one to a finite number does not result in a transfinite number.
> What I am calling a transfinite natural number must exist in N because N is an infinite set.
The definition is that N is the smallest set that contains 0, and every successor of something already in N. Every finite non-negative integer is there, and nothing else - they are all finite.
> Assume that N has only finite numbers in it but is itself an infinite set. Would you care to tell me which number (or numbers) are listed twice?
No number has to be listed twice - just because each thing in the the set is finite, that does not mean that the set has to be finite. There is no contradiction in N being infinite, but all its elements being finite.
You appear to be using words in a non-standard manner. As such, quite simply, you need to be amazingly careful, or you will not be understood.
Certainly I don't understand you.
The leading/trailing zeros are confusing. So many digits are confusing. Let's pick an easy model: The "naïve roman numbers" (This is a made up name, not a technical name.)
Let's consider the set that has
I
II
III
IIII
IIIII
IIIIII
...
IIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIIIIIIIIIIIIII
IIIIIIIIIIIIIIIIIIIIIIIIII
...
...
i.e. all the strings that have a bunch of "I" characters.
Each element has a finite amount of "I". If you delete one of them you get a shorter string/number.
The set has strings that are as long as you want. If you print them in a 8pt times new roman font, you can pick one of them that will be long enough to wrap around the Earth, one of the is long enough to reach the Moon, ...
This is an infinite set, where each element has a finite amount of "I".
No. No natural number has an infinite number of digits, and you are wrong.
If you weren't, of course, it would be easy to prove me wrong -- simply name the natural number to which the successor function is applied that results in a "transfinite" number, whatever that is.
But transfinite numbers are not "natural numbers", they are not in the set N, they don't have infinitely many digits, and in the context of this thread, they are a red herring.
Perhaps you simply don't understand the argument in detail, and are relying on your intuition. And perhaps your intuition is faulty.
Which seems more likely?
So let me try to provide a better insight for you. Consider the collection of natural numbers, including 0. Call it N. We all agree that N, also described as the set of non-negative integers, is infinite.
Now imagine flipping a coin at time t_0, t_1, t_2, etc. If you're worried that this will take infinite amounts of time, we can suppose that each flip - because we are practised - takes half the time of the previous flip, so all the flips can be done in finite time. However, we're in the realm of Pure Mathematics now, chasing the puzzle for its own sake, and not worrying about practicalities.
So what might the result be? Well, you might get all heads, you might get all tails, you might get alternating heads and tail, in practice, of course, you'll get something that looks random.
Let's think about all the possible results of flipping the coin. All possible results. Let's let F be the set of all possible results obtained from flipping the coin are each of t_0, t_1, t_2, and so on. We can think of F as functions from N to {H,T}.
So we have F and N. Let's wonder if it's possible to have a function from N to F that hits every element of F. Suppose we can.
So we have m:N -> F, and for every f in F, there is an n in N such that m(n)=f.
Do you think that's possible? Because hundreds of thousands of mathematicians say that it's not possible.