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 $b \neq 1$.

$\therefore 0.999... \neq 1$.

█

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.)*