Paradox: Is $1 \in (0,1)$?

Question

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..?

Answer 1

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 $9$.

Answer 2

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 $(0, 1)$.

Answer 3

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 $1\not\in(0,1)$.

Answer 4

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 $1$.

Answer 5

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?