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In proofs exhibiting bijections of $\mathbb{R}$ for the sake of proving arguments about cardinality, care is generally taken to avoid the $.999\!\ldots=1.00\!\ldots$ problem. However, it is generally accepted without comment that this is the only way to make distinct decimal expansions that represent the same real number. Although it seems obvious, how can we be sure there are there no other corner cases to be concerned with?

Formally, if $x_d$ and $y_d$ are distinct sequences of integers between $0$ and $9$ representing decimal expansions of numbers $x$ and $y$ in $[0,1)$, is it true that $x=y$ iff there is an $N\in\mathbb{N}$ such that

  • if $n<N$ then $x_n=y_n$
  • $x_N=y_N-1$, and
  • if $n>N$ then $x_n=9$ and $y_n=0$,

and if so, how does one prove this?

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up vote 4 down vote accepted

Yes (up to switching $x$ and $y$). Let $N$ be the least integer such that $x_N\ne y_N$. Assume WLOG that $y_N>$. If $y_N>x_N+1$ then since the remaining digits together contribute at most $10^{-N}$ we have $y-x\ge 2\cdot 10^{-N}-10^{-N}>0$, so $x\ne y$. If $x_{N+1}\ne 9$ or $y_{N+1}\ne 0$, then $x_{N+1}-y_{N+1}\le 8$ and since the remaining digits contribute at most $10^{-N-1}$ we have $$y-x\ge 10^{-N}-8\cdot 10^{-N-1}-10^{-N-1}>0$$ \so $x\ne y$. Thus we must have $x_{N+1}=9$ and $y_{N+1}=0$. Continuing in this manner we see that it is true for all remaining digits.

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