I would like to discuss the following proof about the uncontability of $\Bbb R$.

So first of all we remember the following two important results

Theorem -of density

Between any two distinct real numbers there is a rational real number.

Theorem -of supremum/infimum

Any subset $X$ of $\Bbb R$ which has an upper/lower bound it has a supremum/infimum

Now let we prove that for any sequence $(x_n)_{n\in\Bbb N}$ of real numbers there exist a real number $x$ not belonging to it. So given the interval $[x_0-1,x_0]$ by the density theorem there exist two different real numbers $a_0$ and $b_0$ such that $$ x_0-1<a_0<b_0<x_0 $$ Now if $x_1$ lies to $[a_0,b_0]$ then with the density theorem we can find two different real number $a_1$ and $b_1$ such that $$ a_0<a_1<b_1<x_1<b_0 $$ Instead, if $x_1$ not lies to $[a_0,b_0]$ then we put $a_1=a_0$ and $b_1=b_0$.

So in this way we can make an increasing sequence $(a_n)_{n\in\Bbb N}$ and a decreasing sequence $(b_n)_{n\in\Bbb N}$ such that $$ x_n\notin[a_n,b_n]\subset\dots\subset[a_0,b_0] $$ Now the above chain say us that two sequences are bounded so that they have a supremum $M$ and an infimum $m$.

So unfortunately I did not able to complete the proof: I tried to show that $M\le m$ because in this case it would be $$ x_n\notin [M,m] $$ for any $n\in\Bbb N$ since otherwise $x_n\in [a_n,b_n]$.

I point out I found the proof here but unfortunately is not well expalined so that I tried to rewrite them but I faild and so I tried to elaborate a similar proof but unfortunately I filed again.

So could someone help me to complete or to adjust the proof, please?

  • 2
    $\begingroup$ If you replace "real numbers" everywhere in your proof with "rational numbers", would you get the same result? Remember that rational numbers are also dense but despite this the rationals are countable. If you want to prove something about real numbers which is not true about rational numbers, you must use something unique about real numbers in the proof. $\endgroup$
    – JMoravitz
    Commented Mar 10, 2022 at 15:30
  • $\begingroup$ @JMoravitz I used the sumpremum/infimum theorem which is not true for rational number, right? $\endgroup$ Commented Mar 10, 2022 at 15:47
  • $\begingroup$ @JMoravitz The unique thing used is that an increasing bounded sequence converges. $\endgroup$ Commented Mar 10, 2022 at 16:02
  • $\begingroup$ @JMoravitz Anyway I do not understand why I have to use only results about real numbers: indeed I think that I have to use at least a result that is true only for real numbers; is perhaps this incorrect? $\endgroup$ Commented Mar 10, 2022 at 17:07

1 Answer 1


Fix $m$. For every $n$ you have $a_n<b_m$, so $b_m$ is an upper bound of the set $\{a_n:n\in\mathbb{N}\}$. As $M$ is the least upper bound we have $M\le b_m$.

This holds for all $m$, so $M$ is a lower bound for $\{b_m:m\in\mathbb{N}\}$. As $m$ is the largest lower bound we get $M\le m$.

  • $\begingroup$ So if $b_n<a_m$ for any $m,n\in\Bbb N$ then $b_n<a_m<b_m$ and so $n>m$ because $(b_n)_{n\in\Bbb N}$ is decreasing; however $a_n<b_n<a_m$ and so $n<m$ because $(a_n)_{n\in\Bbb N}$ is increasing. So we conclude that it must be $a_n<b_m$ for any $m,n\in\Bbb N$ so that $M\le b_n$ for any $n\in\Bbb N$ and finally $M\le m$, right? $\endgroup$ Commented Mar 10, 2022 at 16:46
  • $\begingroup$ Moreover, is true that $x_n\notin[M,m]$ so that $M$ and $m$ are not in the range of the sequence $(x_n)_{n\in\Bbb n}$? So is the proof correct? $\endgroup$ Commented Mar 10, 2022 at 16:47
  • $\begingroup$ Yes, the proof is correct. $\endgroup$
    – hartkp
    Commented Mar 18, 2022 at 20:47
  • $\begingroup$ Okay, perfect. Thanks for your assistance! :) $\endgroup$ Commented Mar 19, 2022 at 7:28

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