Prove $a_t \rightarrow x$ using the Betweenness Property Prove that for any $x \in \Bbb R$ there is a strictly increasing sequence ($a_t$) in $\Bbb Q$ such that ($a_t$) converges to $x$, (i.e. $a_t \rightarrow x$)
I want to prove this using the Betweenness property of $\Bbb R$, that is, $\forall x,y \in \Bbb R$, if $x\lt y$ then $\exists \omega \in \Bbb Q$ such that $x\lt \omega\lt y$. Also, $a_t \rightarrow x$ iff $\forall \epsilon \gt 0$, $\exists T$ such that $\forall t\gt T$ we have $a_t \in N_{\epsilon} (x)$. Could someone give a proof using the Betweenness property? Thanks.
 A: Hint Use the betweeness property to pick some $a_n$ between $x$ and $x-\frac{1}{n}$.
A: For all $n \in \mathbb{N}$ there is by the "Betweenness Property" a rational number $x + \frac{1}{n} > a_n > x$. Can you do the rest?
A: Basically, you can use the decimal expansion for $x$ for this. Given a positive integer $n$ you let $k_n$ be the largest integer such that ${\displaystyle {k_n \over 10^n}
\leq x}$; such a $k_n$ must exist since otherwise $10^n x $ would be greater than every integer and the basic properties of ${\mathbb N}$ ensure this is not possible. So you have
$${k_n \over 10^n} \leq x < {k_n + 1\over 10^n}$$
Thus the ${\displaystyle {k_n \over 10^n}}$ are an increasing sequence with
$$\bigg|{k_n \over 10^n} - x\bigg| \leq {1 \over 10^n}$$
So
$$\lim_{n \rightarrow \infty} {k_n \over 10^n} = x$$
A: Sure. By the betweenness property, there exists $a_0\in \Bbb Q$ with $x-1<a_0<x$.
Assume we have founs $a_t\in\Bbb Q$ with $a_t<x$. Then also $\frac{a_t+x}2<x$ and by betweenness we find $a_{t+1}\in\Bbb Q$ with $\frac{a_t+x}2<a_{t+1}<x$.
This defines a sequence of rational numbers. From $a_t<\frac{a_t+x}{2}<a_{t+1}$ it is clear that $a_t$ is strictly increasing. We also have $2^t\cdot |x-a_t|<1$ for all $t\in\Bbb N_0$. As $2^t>\frac 1\epsilon$  for $t$ large enough, we have $a_t\to x$.
A: Start off by choosing $a_0\in\mathbb{Q}$ with $x-1<a_0<x$. Suppose you have built
$$
a_0<a_1<\dots<a_n<x
$$
with the property that $a_k\in\mathbb{Q}$ and $x-\frac{1}{k}<a_k<x$ (for $k=1,2,\dots,n$).
If $x-\frac{1}{n+1}>a_n$, select $a_{n+1}$ such that
$$
x-\frac{1}{n+1}<a_{n+1}<x
$$
Otherwise, select $a_{n+1}$ such that
$$
a_n<a_{n+1}<x
$$
In both cases we have
$$
a_n<a_{n+1}
\qquad\text{and}\qquad
x-\frac{1}{n+1}<a_{n+1}<x
$$
This recursively builds the sequence you were looking for.
