The Reals in (0,1) Are Not Countably Infinite

See also – Cardinality of the Rationals – Positive and Negative included

Claim

The interval (0,1)\subset \mathbb{R} is uncountable. Equivalently, there is no bijection
f:\mathbb{N}\to(0,1).

Proof (Cantor’s Diagonal Argument)

Suppose for contradiction that (0,1) is countable. Then its elements can be listed as a sequence:

    \[ x_1,\, x_2,\, x_3,\, \dots \]

Write each x_n in (a chosen) decimal expansion:

    \[ x_n \;=\; 0.d_{n1}d_{n2}d_{n3}\dots,\qquad d_{nk}\in\{0,1,\dots,9\}. \]

Convention: If a real admits two decimal expansions (e.g., 0.4999\ldots=0.5), choose the one
not ending with an infinite tail of 9s. This removes ambiguity.

Construct a new number y\in(0,1) by defining its digits c_1,c_2,c_3,\dots along the diagonal:

    \[ c_n \;=\; \begin{cases} 1, & \text{if } d_{nn}\neq 1,\\ 2, & \text{if } d_{nn}=1, \end{cases} \qquad\text{and}\qquad y \;=\; 0.c_1c_2c_3\ldots \]

By construction, y differs from x_n in the n-th digit, so y\neq x_n for all n. This contradicts the assumption that \{x_n\} lists all elements of (0,1).

Therefore, (0,1) is uncountable. \square

Consequences

  • (0,1) has the same cardinality as \mathbb{R} (the continuum):

    \[ |(0,1)| \;=\; |\mathbb{R}|. \]

  • The natural numbers are strictly “smaller” in cardinality:

    \[ |\mathbb{N}| \;<\; |(0,1)| \;=\; |\mathbb{R}|. \]