Assumption: $0.999...\neq 1$.
Proof:
Let set $S = \{x | x \in \mathbb{R}$ and $0 < x < 1\}$.
Let $S$ converge to $1$ by definition of convergence to a point.
By construction and the definition of an open point set, $1 \notin S$.
By the definition of a limit point, $1$ is a limit point of $S$.
Let there be a point $p \in S$ such that there is also a point $b$ where $p < b < 1$.
(an example of such a set of points is where $p = 0.8$ and $b=0.9$, $0.8 < 0.9 < 1$).
By definition of a limit point, $b \in S$ and $b$ is between $p$ and $1$.
By the axiom, for any two points $p$ and $q$ there is a point between them.
By definition of first point to the right, $1$ is the first point to the right of $S$ and greater then every point in $S$.
By the definition of infinite, $S$ is infinite meaning not finite and there are always $n + 1$ points in $S$ where $n \in \mathbb{N}$ showing that $S$ does not meet the definition of finite.
Let $M$ be an open interval $(a, b)$ containing $1$ and $p$ by the definition of a limit point.
For all values of $p < 1$, $b \in M$.
For any value of $p < 0.999...$, $b > p$.
For some $p \leq 0.999...$, $b > p \in S$.
By the definition of infinite there is a value of $b = 0.999...$ for some $M$ where $b \in S$.
Since $b \in S$ and $1 \notin S$, then $b \neq 1$.
$\therefore 0.999... \neq 1$.
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Q.E.F. (Latin: Quod Erat Faciendum, Greek: oper edei poiesai OR $o\pi\epsilon\rho$ $\epsilon\delta\epsilon\iota$ $\pi o\iota\eta\sigma\alpha\iota$, English: Which was to be done. This is appearantly how Euclid said it in Greek.)
