$0.999\cdots =1$ and $0.4999\cdots =0.5$: the second expansion of a number is exactly an eventually-all-$(b-1)$ digit sequence

Take $b=10$, so $\beta =\iota (10)$, and read a digit sequence $(d_j)$ as the series ${\sum }_{j\ge 0}\iota (d_j)\mathrm{/}{\beta }^{j+1}$ (Base-$b$ expansions: for an integer $b\ge 2$ every $x\in [0,1)$ is the sum of ${\sum }_{j\ge 0}{d}_j\mathrm{/}{b}^{j+1}$ for digits $d_j<b$, and the digit sequence is unique among those that are not eventually constantly $b-1$).

All nines. The constant sequence $d_j=9$ gives, by the geometric series,

${\sum }_{j\ge 0}\frac{9}{{10}^{j+1}}=\frac{9}{10}\cdot \frac{1}{1-1\mathrm{/}10}=1,$

which is the statement usually written $0.999\cdots =1$. Since $1\mathrm{\notin }[0,1)$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), this digit sequence is not the expansion of any point of $[0,1)$ produced by Base-$b$ expansions: for an integer $b\ge 2$ every $x\in [0,1)$ is the sum of ${\sum }_{j\ge 0}{d}_j\mathrm{/}{b}^{j+1}$ for digits $d_j<b$, and the digit sequence is unique among those that are not eventually constantly $b-1$, and no uniqueness clause is violated.

Two expansions of one number. For $x=1\mathrm{/}2$ the construction of Base-$b$ expansions: for an integer $b\ge 2$ every $x\in [0,1)$ is the sum of ${\sum }_{j\ge 0}{d}_j\mathrm{/}{b}^{j+1}$ for digits $d_j<b$, and the digit sequence is unique among those that are not eventually constantly $b-1$ returns $d_0=5$ and $d_j=0$ for $j\ge 1$: the residue $r_0=1\mathrm{/}2$ lies in $[5\mathrm{/}10,6\mathrm{/}10)$, and $r_1=10\cdot \frac{1}{2}-5=0$, after which every digit is $0$. But the sequence $d_0^{\mathrm{\prime }}=4$, $d_j^{\mathrm{\prime }}=9$ for $j\ge 1$ has

$\frac{4}{10}+{\sum }_{j\ge 1}\frac{9}{{10}^{j+1}}=\frac{4}{10}+\frac{9}{100}\cdot \frac{1}{1-1\mathrm{/}10}=\frac{4}{10}+\frac{1}{10}=\frac{1}{2},$

which is $0.4999\cdots =0.5$. So two different digit sequences have the same sum, and uniqueness in Base-$b$ expansions: for an integer $b\ge 2$ every $x\in [0,1)$ is the sum of ${\sum }_{j\ge 0}{d}_j\mathrm{/}{b}^{j+1}$ for digits $d_j<b$, and the digit sequence is unique among those that are not eventually constantly $b-1$ survives only because $(d_j^{\mathrm{\prime }})$ is terminal, being eventually constantly $b-1=9$.

This is the whole of the nonuniqueness. The uniqueness proof shows that two distinct digit sequences with the same sum must differ by one at the first index where they differ and then be all $b-1$ against all $0$. So excluding the terminal sequences excludes exactly one member of each such pair, and nothing else.

• "$0.999\cdots =1$" is an identity between a series and a number, and nothing stranger. The left side denotes the sum of ${\sum }_{j\ge 0}9\mathrm{/}{10}^{j+1}$, which step 2.1 computes to be $1$ by the geometric series. There is no approximation and no limiting process beyond the one already in Series, partial sums, convergence and the sum, divergence, and the tail series.

• Which of the two expansions the construction returns. The residue recursion of Base-$b$ expansions: for an integer $b\ge 2$ every $x\in [0,1)$ is the sum of ${\sum }_{j\ge 0}{d}_j\mathrm{/}{b}^{j+1}$ for digits $d_j<b$, and the digit sequence is unique among those that are not eventually constantly $b-1$ always takes the largest digit $d$ with $\iota (d)\mathrm{/}\beta \le r$, so it produces the expansion ending in zeros rather than the one ending in nines. That is why the constructed sequence is automatically non-terminal, as the theorem records.

• The same pair exists in every base. For $b=2$ the two expansions of $1\mathrm{/}2$ are $0.1000\dots$ and $0.0111\dots$; the argument above is the case $b=10$ of a computation that uses only ${\sum }_{j\ge 0}(b-1)\mathrm{/}{b}^{j+1}=1$.