[proofplan] We prove the contrapositive obstruction to countability by applying Cantor's diagonal argument to the interval $(0,1)$. Assuming $(0,1)$ has an enumeration, we choose one decimal expansion for each listed [real number](/page/Real%20Numbers) and then build a new real number whose $n$-th decimal digit disagrees with the $n$-th digit of the $n$-th listed number. The construction uses only the digits $1$ and $2$, which prevents decimal-expansion ambiguity and lets the digit disagreement imply genuine numerical inequality. This produces an element of $(0,1)$ missing from the proposed enumeration, contradicting countability. [/proofplan] [step:Reduce uncountability of $\mathbb{R}$ to uncountability of $(0,1)$] Suppose, toward a contradiction, that $\mathbb{R}$ is countable. Since $(0,1) \subseteq \mathbb{R}$, the inclusion [map](/page/Function) \begin{align*} \iota: (0,1) &\to \mathbb{R} \\ x &\mapsto x \end{align*} is injective. Therefore $(0,1)$ would be countable by the standard result that every subset of a [countable set](/page/Countable%20Set) is countable, equivalently by the [Subsets of Natural Numbers are Countable](/theorems/752) argument after composing with an enumeration of $\mathbb{R}$. It is enough to contradict this conclusion. Hence assume that there is a surjective map \begin{align*} a: \mathbb{N} &\to (0,1) \\ n &\mapsto r_n. \end{align*} [guided] We begin by reducing the theorem to the interval $(0,1)$. Suppose, toward a contradiction, that $\mathbb{R}$ is countable. The inclusion \begin{align*} \iota: (0,1) &\to \mathbb{R} \\ x &\mapsto x \end{align*} is an injective map because distinct real numbers in $(0,1)$ remain distinct as real numbers. A subset of a countable set is countable; in Androma this is the same counting principle used in [Subsets of Natural Numbers are Countable](/theorems/752), applied after identifying a countable set with the image of an enumeration. Therefore, if $\mathbb{R}$ were countable, then its subset $(0,1)$ would also be countable. Thus it suffices to prove that $(0,1)$ is not countable. We assume the opposite: there exists a surjective map \begin{align*} a: \mathbb{N} &\to (0,1) \\ n &\mapsto r_n. \end{align*} Surjectivity means that every real number in $(0,1)$ appears somewhere in the list $r_1,r_2,r_3,\dots$. The rest of the proof constructs an element of $(0,1)$ that is not equal to any $r_n$, contradicting this surjectivity. [/guided] [/step] [step:Assign decimal digits to every listed real number] For each $n \in \mathbb{N}$, choose a decimal expansion of $r_n$. Equivalently, define a digit function \begin{align*} d: \mathbb{N} \times \mathbb{N} &\to \{0,1,2,3,4,5,6,7,8,9\} \\ (n,k) &\mapsto d_{nk} \end{align*} such that, for every $n \in \mathbb{N}$, \begin{align*} r_n = \sum_{k=1}^{\infty} d_{nk}10^{-k}. \end{align*} In decimal notation, this says $r_n = 0.d_{n1}d_{n2}d_{n3}\cdots$. [/step] [step:Build a real number by changing the diagonal digits] Define the anti-diagonal digit function \begin{align*} e: \mathbb{N} &\to \{1,2\} \\ n &\mapsto \begin{cases} 1, & \text{if } d_{nn} \ne 1, \\ 2, & \text{if } d_{nn} = 1. \end{cases} \end{align*} Now define \begin{align*} r := \sum_{n=1}^{\infty} e_n10^{-n}. \end{align*} Since $1 \le e_n \le 2$ for every $n \in \mathbb{N}$, comparison with the [geometric series](/page/Series) gives \begin{align*} 0 < r \le \sum_{n=1}^{\infty} 2 \cdot 10^{-n} = 2\sum_{n=1}^{\infty}10^{-n} = 2 \cdot \frac{1/10}{1-1/10} = \frac{2}{9} < 1. \end{align*} Thus $r \in (0,1)$. Also, by construction, $e_n \ne d_{nn}$ for every $n \in \mathbb{N}$. [guided] The goal is to manufacture a number whose decimal expansion disagrees with the list in a controlled way. We define the anti-diagonal digit function \begin{align*} e: \mathbb{N} &\to \{1,2\} \\ n &\mapsto \begin{cases} 1, & \text{if } d_{nn} \ne 1, \\ 2, & \text{if } d_{nn} = 1. \end{cases} \end{align*} This definition compares only with the diagonal digit $d_{nn}$, the $n$-th digit of the $n$-th listed number. In both cases, the new digit differs from the diagonal digit: if $d_{nn} \ne 1$, then $e_n = 1 \ne d_{nn}$; if $d_{nn}=1$, then $e_n = 2 \ne d_{nn}$. Now define the real number \begin{align*} r := \sum_{n=1}^{\infty} e_n10^{-n}. \end{align*} This is a convergent series because $0 \le e_n10^{-n} \le 2\cdot 10^{-n}$ and the geometric series $\sum_{n=1}^{\infty}10^{-n}$ converges. Moreover, \begin{align*} 0 < r \le \sum_{n=1}^{\infty} 2 \cdot 10^{-n} = 2\sum_{n=1}^{\infty}10^{-n} = 2 \cdot \frac{1/10}{1-1/10} = \frac{2}{9} < 1. \end{align*} Therefore $r \in (0,1)$. The choice of digits $1$ and $2$ is important: it ensures that the decimal expansion of $r$ is not eventually all $0$ and not eventually all $9$, so the usual decimal ambiguity cannot make $r$ equal to a listed number despite a diagonal digit disagreement. [/guided] [/step] [step:Show the anti-diagonal number differs from every listed number] Fix $k \in \mathbb{N}$. The number $r_k$ has the chosen decimal expansion \begin{align*} r_k = \sum_{n=1}^{\infty} d_{kn}10^{-n}, \end{align*} while $r$ has the decimal expansion \begin{align*} r = \sum_{n=1}^{\infty} e_n10^{-n}. \end{align*} At the $k$-th digit, the two expansions disagree because $e_k \ne d_{kk}$. Since the expansion of $r$ uses only the digits $1$ and $2$, it is not eventually all $0$ and not eventually all $9$; hence decimal-expansion ambiguity cannot identify it with a different expansion of the same real number. Therefore $r \ne r_k$. Since $k \in \mathbb{N}$ was arbitrary, $r \ne r_k$ for every $k \in \mathbb{N}$. [guided] Fix $k \in \mathbb{N}$. We compare $r$ with the $k$-th number in the alleged enumeration. The chosen decimal expansion of $r_k$ is \begin{align*} r_k = \sum_{n=1}^{\infty} d_{kn}10^{-n}. \end{align*} The decimal expansion of the constructed number is \begin{align*} r = \sum_{n=1}^{\infty} e_n10^{-n}. \end{align*} The decisive comparison is at the $k$-th digit. By the definition of $e_k$, we have $e_k \ne d_{kk}$. Thus the $k$-th digit of $r$ differs from the $k$-th digit of $r_k$ in the chosen expansion of $r_k$. There is one standard issue to address: decimal expansions are not always unique, because a terminating decimal can also be written with an infinite tail of $9$s. Our construction avoids this ambiguity for $r$. Every digit $e_n$ is either $1$ or $2$, so the decimal expansion of $r$ is neither eventually all $0$ nor eventually all $9$. Therefore it is the non-ambiguous decimal expansion of $r$. If $r$ were equal to $r_k$, then a decimal expansion of $r_k$ could represent $r$ only through the standard decimal ambiguity; but that ambiguity requires one of the two expansions to be eventually all $0$ and the other eventually all $9$. The expansion of $r$ has neither behaviour. Hence the digit disagreement at position $k$ forces $r \ne r_k$. Since the index $k$ was arbitrary, the same argument proves \begin{align*} r \ne r_k \end{align*} for every $k \in \mathbb{N}$. [/guided] [/step] [step:Contradict the enumeration and conclude $\mathbb{R}$ is uncountable] We have constructed $r \in (0,1)$ such that $r \ne r_k$ for every $k \in \mathbb{N}$. This contradicts the surjectivity of \begin{align*} a: \mathbb{N} &\to (0,1) \\ k &\mapsto r_k. \end{align*} Therefore $(0,1)$ is uncountable. If $\mathbb{R}$ were countable, then its subset $(0,1)$ would be countable, contrary to what we have just proved. Hence the [set](/page/Set) $\mathbb{R}$ is uncountable. [/step]