Proposed $1-1$ map $\mathbb{R} \to \mathbb{N}$ [closed]

Let $n=0.$ Every time you think of a particular real number $r$, let $n=n+1$ and map $r$ to $n$.

Edit: is this even a "map" according to the definition?

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What is the question? –  Did May 23 '12 at 18:15
what makes you assume you can 'think of a particular real number'? Can you hold all the digits in your head? –  Robert Mastragostino May 23 '12 at 18:16
I can't hold all the digits in my head but I can picture a length of 1 and a real number as another length. Or I can picture various other structures which contain representations of real numbers such as the circumference of a circle. –  AbstractionOfMe May 23 '12 at 18:19
Never in the context of the real numbers $n=n+1$. –  Asaf Karagila May 23 '12 at 18:27
In you sense, indeed the number of reals is finite, since in my finite life I will only think of finitely many explicit reals. That is a point of view that has in fact been argued by a (very) small number of mathematicians. I do not consider it a mathematically useful point of view. –  André Nicolas May 23 '12 at 19:27

closed as not a real question by Asaf Karagila, Davide Giraudo, Benjamin Lim, Pete L. Clark, t.b.May 24 '12 at 16:33

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I think we have a bit of confusion. Since you're mapping $r$ to an integer $n$, it seems to me that you are actually trying to come up with a $1-1$ map $\mathbb{R} \to \mathbb{N}$.
The problem here is that you are "thinking of a real number", saying where to put it, and moving on to the next real number. What this means is that you are making an ordered list of the real numbers. However, a fundamental fact about the real numbers is that there are 'too many' to be put on any ordered list, even one that's infinitely long. So this would only be a map from some countable subset of $\mathbb{R}$ to $\mathbb{N}$.
If it turns out you just want a $1-1$ map $\mathbb{N} \to \mathbb{R}$, why not do the inclusion map $n \mapsto n$?