Consider the interval $[0, 1)$. Imagine joining the two ends of the interval, creating a circle. Start at the point $0$, then move along the circle a total length of $\frac 1 {2\pi}$, labeling the point $1$. then, move along the circle a distance of $\frac 1 {2\pi}$ again and label the next point $2$. you will pass by point $0$, but you will never come back to it, marking the points of the circle and marking a total of $\aleph_0$ points. a bijective function can represent this as:
$f:\Bbb N\to\Bbb R$ such that $f(n) = \frac n {2\pi}\bmod1$

Next, do the same for the intervals [1, 2), [-1, 0), [2, 3), and so on, totaling to $\aleph_0$ intervals. this will count all the intervals bounded by integers and the sequence of the lower bounds can be written as:
$g:\Bbb N\to\Bbb Z$ such that $g(0) = 0$, $g(1) = 1$, $g(n) = -\frac n 2$ for when $n$ is even and is not $0$, and $g(n) = \frac {n+1} 2$ for when $n$ is odd and is not $1$. Each of the intervals would be $[n, n+1)$.

Let $p(n)$ be a function that maps the natural numbers to the primes. It maps $n$ to the $(n+1)^{st}$ prime number, meaning $0$ maps to $2$, $1$ maps to $3$, and $2$ maps to $5$.

Let $M(x) = p(g^{-1}(\lfloor x\rfloor))^{f^{-1}(x) + 1}$.

Since primes can't be divided into smaller natural numbers, nonzero natural number powers of primes will never create other primes or multiples of other primes. Every real number is mapped to a distinct natural number, meaning that the function $M(x)$ is bijective.

Which leaves the question, if there is a bijective function that maps the real numbers to the natural numbers, then is the set of real numbers countable?

  • $\begingroup$ You claim $f$ is a bijective function; what is it in bijection with? The set $[0, 1)$ is already uncountable, so proving this is a surjection is the difficult part. (That is, the impossible part, unless a lot of set theory is wrong.) $\endgroup$ Nov 5, 2019 at 23:29
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    $\begingroup$ Start at zero, check. Move a distance 1 and label the point (which is the number 0) 1, check. Then move along the circle a distance of 1 again (landing on the number 0, which we labeled 1) and label the NEXT point - there is no such number. There is no smallest positive real number. $\endgroup$
    – Joe
    Nov 5, 2019 at 23:30
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    $\begingroup$ Actually, when I was a freshman in college I debated the cardinality of the reals for the whole year with my professor. When I finally “got it”, it felt like an epiphany. So I applaud your enthusiasm. Good luck on your journey. I enjoyed mine. $\endgroup$
    – Joe
    Nov 5, 2019 at 23:47
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    $\begingroup$ You seem to be making the classic error of assume that all points are can be found to be close to some $f(n);n\in \mathbb N$ as precisely as we like that, that would mean there are countably many such bounds. That is false as the diagonalization arguments show. You method of finding this be asynchronous circlings is in essence no better or worse way of generation bounds as finding the decimal expansion. It is is fallacious for the exact same reasons. $\endgroup$
    – fleablood
    Nov 6, 2019 at 0:31
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    $\begingroup$ I don't know what $f^{-1}(e)$ maps to $\endgroup$
    – Nick
    Nov 6, 2019 at 0:42

2 Answers 2


a bijective function can represent this as:
$f:\Bbb N\to \left(\Bbb R \cap [0,1) \right)$ such that $f(n) = \frac n {2\pi}\bmod1$

The claim that $f$ is bijective implies that $f$ maps onto $\left(\Bbb R \cap [0,1) \right)$, and that implies the following: $$ \forall r \in \left(\Bbb R \cap [0,1) \right) \exists n \in \Bbb N \exists m \in \Bbb N \left( \frac{n}{2\pi} - m \right) = r$$

Let $r = \frac{3}{\pi}$, so $r \in \left(\Bbb R \cap [0,1) \right)$

Therefore, $$\exists n \in \Bbb N \exists m \in \Bbb N \left( \frac{n}{2\pi} - m \right) = \frac{3}{\pi}$$

Simplifying, we obtain $\pi = \frac{n - 6}{2 \cdot m}$, but that contradicts the well-known theorem that $\pi$ is irrational.

  • $\begingroup$ my apologies, I confused the radius, which was definitely not 1, with the circumference, which was one. I meant to say move along the circle by an arc length of 1/2pi. I also corrected the function since the cosine function was incorrect. $\endgroup$
    – Nick
    Nov 6, 2019 at 0:22
  • $\begingroup$ It's okay. I revised my answer to attempt to handle the revised question. $\endgroup$ Nov 6, 2019 at 0:50
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    $\begingroup$ Ah, thank you! Your response is very helpful. $\endgroup$
    – Nick
    Nov 6, 2019 at 1:03

if there is a bijective function that maps the real numbers to the natural numbers, then is the set of real numbers countable?

Yes. This is the definition of being countably infinite.

Unfortunately, no such bijection exists, for if it did, we could find a contradiction by constructing a real number that is not in the list, with the usual argument. This might not be a satisfactory answer in context of the OP but it’s true. Give us a list of all reals in $(0,1]$ and expand them in decimal notation as $x_n=0.a_{n1}a_{n2}\dotso$ for each $n\in \mathbb{N}$. Define a real $x=0.b_1b_2\dotso$ by $b_n=2$ if $a_{nn}\neq 2$ and $b_n=3$ else. Then $x$ differs from $x_n$ at the $n$-th decimal place for each natural $n$ and hence cannot be in the list $x_1,x_2,\dotsc$, a contradiction.


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