Maximum vs Least Upper Bound - Confused in infinity I know I am missing something and I just found this quite counter-intuitive:

$x = 1.9999999... \implies x = 2$

Does this mean $x$ is NOT an element of $(0, 2) = \{x ∈ \mathbb{R}: 0 < x < 2\}$?
Otherwise would the set $(0, 2)$ have a maximum instead of just a least upper bound?
 A: Indeed, $x = 1.9\dots = 2$ is not an element of $(0,2)$. To see why, we have to know what $1.9\dots$ really is.
To answer the rest of your question: 2 does not lie in your set, but it is larger than every element in you set so it is an upper bound. If 2 were to lie in you set and still be strictly larger than every other element, it would be a maximum. This is the case with e.g. $(0,2]$.
Let's look at what $1.9\dots$ really means.
It is not a one with a dot and infinite nines behind it, although that captures the spirit of what $x$ is.
Analysis manages to formalize this idea using limits.
We start by defining a sequence of numbers:

*

*$x_0 = 1$

*$x_1 = 1.9$

*$x_2 = 1.99$

*$x_3 = 1.999$
and so on.
Here, $x_n$ is a one and a dot and $n$ nines at the end, which makes it somehow connected to $1.9\dots$.
We could write a term in the sequence as the sum of its digits:
$$x_n=1 + 0.9 + 0.09 + \dots + \underbrace{0.0\dots0}_{n\text{ zeros}}9=1+\sum_{i=1}^n9\cdot10^{-i}.$$
One could say that as $n$ gets larger, $x_n$ gets closer to 2.
One could even say that we can get $x_n$ and subsequent terms as close to 2 as we want, just by choosing $n$ large enough.
By definition, 2 would then be the limit of the sequence $(x_n)_{n\in\mathbb N}$.
At the end of this post I show how we can prove this.
So now we know that the limit of $x_n$ is equal to 2, we can look at an alternative way to write this limit:
$$\lim_{n\to\infty}x_n=1 + 0.9 + 0.09 + \dots =1+\sum_{i=1}^\infty9\cdot10^{-i}.$$
That last sum can of course be written in short as $1.9\dots$, which means $1.9\dots = 2$.
The very reason we can speak and reason about numbers as $1.9\dots$ and e.g. $\pi=3.141592\dots$, is because we are able to define them.
One of the ways to do that, is using limits.

The limit of the sequence $(x_n)_{n\in\mathbb N}$ is 2. Proof:
Let's say we want to find an $x_n$ such that the distance between it and 2 is smaller than some real, strictly positive number $\varepsilon$.
Note first that we can write $x_n = 2-10^{-n}$.
Using some inequality juggling, we can get an $n$ that fulfills this property.
$$
\begin{aligned}
&\lvert 2 - x_n\rvert = \lvert 2 - (2 - 10^{-n})\rvert < \varepsilon \\
\iff &10^{-n} < \varepsilon \\
\iff & n < -\log_{10}\epsilon \\
\iff & n < \log_{10}\frac1\epsilon
\end{aligned}
$$
Additionally, if $m$ is an integer strictly larger than $n$, then $x_m$ lies closer to 2 than $x_n$, because
$$
\begin{aligned}
\lvert 2 - x_n\rvert &= \lvert 2 - (2 - 10^{-n})\rvert \\
&= 10^{-n} \\
&> 10^{-m} \\
&= \lvert 2 - (2 - 10^{-m})\rvert \\
&= \lvert 2 - x_m\rvert.
\end{aligned}
$$
This means that for any given strictly positive number $\varepsilon$, at some point in the sequence suddenly the distance between every element and 2 is less than $\varepsilon$.
A: It looks weird, but $x=1.9999 \ldots=2=sup((0,2))\neq max((0,2))$; because the open interval does not have a maximum. So, even though it looks weird, $x\notin (0,2)$, because $x$ actually is $2$.
