Paradox: Is $1 \in (0,1)$? Consider the set of numbers such that $x \in (0,1)$. 
Their decimal expansion is $0.b_0b_1b_2\ldots$, with $b_n \in \{0,1,2,3,4,5,6,7,8,9\}$, and they are not all zero (or else $x = 0$).
Then choose all $b_n = 9$, we have $0.999\ldots = 1$.
But $0.999\ldots = x \in (0,1)$, so $1 \in (0,1)$.
Where did we go wrong?
EDIT:
Right, the answer is that not all $0.b_0b_1b_2 \ldots$ are in $(0,1)$. Here's a follow-up:
So define $x_m = 0.b_0b_1b_2 \ldots$ where all of the $b_n$ are $9$ except $b_m = 8$. Then the limit of $x_m$ an be thought of as the largest element of $(0,1)$. But hey, wait a second... $(0,1)$ is open in the usual topology for $\mathbb{R}$! What have we done now..?
 A: You declared that $x$ had a decimal expansion consisting of the digits $1$ through $9$ without restriction, this is false.
The correct thing to say would be that  their decimal expansion is 
$$0.b_{0}b_{1}b_{2}..$$ with $b_{n} \in \{0, 1, 2 ,3,4,5,6,7,8,9 \}$,but not all $0$ and not all $9$.
A: This is a good application of the fact that, under the usual topology on $\mathbb{R}$, the interval $(0, 1)$ isn't closed.
You've found a sequence, $b_n = \frac{10^n - 1}{10^n}$ where $n = 1, 2, \ldots$ of elements of $(0, 1)$ whose limit is not in $(0, 1)$.
A: This is just a demonstration that $(0,1)$ is not closed. The sequence $\{x_n\}$ with 
$$
x_n=0.\underbrace{9\ldots9}_{n\; 9's}
$$
is such that $x_n\in (0,1)$ for each $n$, $x_n\to 1$, but $1\not\in(0,1)$.
A: Note, that, in $\mathbb{R}$ with the usual topology, $\sup((0,1)) = 1$. By definition, a closed set is a set that contains it's supremum and infimum. Since $1 \not\in (0,1)$, $(0,1)$ is open. Since it is open, for any $x \in (0,1)$, there is an $\epsilon > 0$ such that $x+\epsilon < 1$. Now, if an element has a finite number of $9$'s, such an $\epsilon$ is obvious. Symbolically, if a sequence converges to $0.999 \ldots$, we identify the symbols $0.999 \ldots$ with the symbol $1$. Thus, you reasoning is flawed, for such a number is precisely $1$. 
A: Where did we go wrong?
You forgot to give a clear definition for 
$$
0.999...
$$
What exactly do you mean by that sequence of symbols? 
A: If the assumption you're making is that 0.999... $\neq$ 1, then this question is pretty much the same as Is it true that $0.999999999\dots=1$?. The number a decimal expansion represents is the infinite sum of the part before the decimal and each digit multiplied by 10 to the power of the negative of its position after the decimal. In the case of the decimal notation 0.999..., 0.9 is $\frac{1}{10}$ away from 1; 0.99 is $\frac{1}{100}$ away from 1. 0.999... is larger than any of the numbers in that sequence so it is exactly 1.
Here, I'm using the assumption that $\mathbb{R}$ is a complete ordered field. How do we know that? How do we define $\mathbb{R}$? To avoid using the unproven assumption that $\mathbb{R}$ is a complete ordered field, we could define a real number as a decimal expansion where 0.999... $\neq$ 1 and say that's how we defined it and define addition, multiplication, and inequality in the intuitive way. Then it wouldn't be a complete ordered field because $(1 - 0.999...) + 0.999... = 0 + 0.999... = 0.999... \neq 1$. I think most mathematicians agree to define $(\mathbb{R}, 1, 0, +, \times, \leq)$ in such a way that $(\mathbb{R}, 1, 0, +, \times, \leq)$ is a complete ordered field and agree that it's unimportant which complete ordered field it's defined to be because $(\mathbb{R}, 1, 0, +, \times, \leq)$ can be explicitly constructed in ZF and shown to be a complete ordered field which is unique up to isomorphism. One way to construct it is to construct the nonnegative real numbers as a decimal expansion and define a notation will a string of trailing 9's to represent the same number as the one that increases the last digit before the trailing 9's by 1 and changes all the 9's after it to a 0 and then stick the - operation onto the beginning of any notation you already defined the meaning of to represent its additive inverse.
