[zelah]

It is possible. The inverse of m is called q. The function q takes an infinite sequence of coin flips and one by one changes every heads to 1 and every tails to 0. The infinite string of zeros and ones is then prepended with a 1 and interpreted as a transfinite natural number in binary notation. This will not take forever because each change will only take half as long as the previous one. A transfinite natural number can exist since there are an infinite number of natural numbers. If we stop at 1-bit numbers then the natural numbers are not an infinite set. Likewise, if we stop at 2-bit numbers then the natural numbers are not an infinite set. Our only option is to concede that transfinite natural numbers do actually exist and that they can be put into correspondence with all sequences of coin flips.

Hundreds of thousands of mathematicians are wrong.

[ColinWright]

So I asked: 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?

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?

[zelah]

>But the set N does not have transfinite natural numbers, so q does not map F to N, it maps F to something else.

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!

[ColinWright]

> How many natural numbers are there?

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.

[gus_massa]

I think you are confusing "arbitrary long" with "infinite".

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".

[dogecoinbase]

Hundreds of thousands of mathematicians are wrong.

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.

[ColinWright]

Be careful, there are such things as transfinite numbers, and they are well-founded. They let us do wonderful things like transfinite induction, and to prove, for example, that Goodstein's Theorem is true, even though it's unproveable in Peano Arithmetic.

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.

[dogecoinbase]

I'm fine with nonstandard models; my scare quotes on "transfinite" were more to emphasize that the term was being used without basis or definition.