0.999.../Proof by equality of Dedekind cuts

Assumptions

 * Construction of the real numbers from Dedekind cuts
 * Definition from Dedekind cuts

Proof
In the Dedekind cut approach, the real number 1 is the set of all rational numbers that are less than 1.
 * $$1 = \left\{ r | r \isin \textbf{Q} \land r < \tfrac{1}{1}\right\}$$

Meanwhile, the real number 0.999... is the set of rational numbers r such that r < 0, or r < 0.9, or r < 0.99, or r is less than some other number of the form
 * $$1-\tfrac{1}{10^n}$$

In full
 * $$0.999... = \left\{ r | r \isin \textbf{Q} \land \exists n \isin \textbf{N} :r < 1-\tfrac{1}{10^n} \right\} $$

The proof that these two Dedekind cuts are equal then relies on proving that these two set conditions are equivalent.

It can be shown that any number rational number smaller than 0.999... is also smaller than 1, since any non-negative n will give a value of 1-(1/10)^n which is smaller than 1.

All that remains is to prove that if a rational is smaller than 1, that it is always in 0.999... that:
 * $$r < 1 \rightarrow \exists n \isin \textbf{N} : r < 1-\tfrac{1}{10^n} $$

for any r in Q

Substituting r for a natural fraction a/b


 * $$\tfrac{a}{b} < 1 \rightarrow \exists n \isin \textbf{N} : \tfrac{a}{b} < 1-\tfrac{1}{10^n} $$

And letting n in the Quantification equal the denominator b gives


 * $$\tfrac{a}{b} < 1-\tfrac{1}{10^b} $$

cross-multiplication by 10^b gives
 * $$a \times 10^b < b \times 10^b - b $$

and since we know that a is smaller than b, and 10 is positive, this will always be true.

Thus the two sets are equivalent, and 0.999... = 1.