# Paradox in connection with definition of limit points and order limit theorem?

I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I come across something that appears (to me) as a paradox.

Let me first write down one definition and two theorems that have been proven in the aforementioned book:

Definition 1: A point $x$ is a limit point of a set $A$ if every $\epsilon$-neighborhood $V_\epsilon (x)$ of $x$ intersects the set $A$ in some point other than $x$.

Theorem 1: A point $x$ is a limit point of a set $A$ if, and only if, $x = \lim a_n$ for some sequence $(a_n)$ contained in $A$ satisfying $a_n \neq x$ for all $n \in \mathbb{N}$.

Order limit theorem: Assume $\lim a_n = x$. Then if there exists a $c \in \mathbb{R}$ for which $a_n \leq c$ for all $n \in \mathbb{N}$, then $a \leq c$.

Now, let us consider the open interval $(1,2)$. Then, by definition $1$, $x=2$ is a limit point. This means that according to theorem $1$ we must have some sequence $(a_n) \subseteq (1,2)$ with $2 = \lim a_n$. In other words, for all $n \in \mathbb{N}$, we must have $a_n \leq 1.999999999 \ldots$. It follows by the order limit theorem that $x \leq 1.999999999 \ldots$, but we have started by saying that $x=2$ and so there appears to be a paradox.

Where am I making a mistake in the above reasoning?

$$\mbox{}1.9999999... = 2.\mbox{}$$
Also, see this related question: Is it true that $0.999999999\ldots = 1$?.