Example of a discrete bounded subset of $\mathbb{R}$ such that for every two points of that set there is another point between them of that set Can anyone give an example of a subset say $A$ of $\mathbb{R}$ such that every point of that set is isolated i.e. for every $x \in A$  there exist a neighborhood say $N(x)$ containing x such that $N(x) \cap A = \{x\} $ and for any two points say $x,y\in A$ there exist $z\in A$ such that $x<z<y$.
Thanks in advance. note that A should have at least two elements
 A: Same idea as Dunston's answer.  Take the set $A$ of all the midpoints of removed intervals in the construction of the Cantor middle-thirds set.
Each point $a\in A$ is isolated, since the removed interval interval $J$ of which it is the center has $J \cap A = \{a\}$.


So $A$ consists of:
$1/2$, the center if $(1/3,2/3)$,
$1/6$, the center of $(1/9,2/9)$,
$5/6$, the center of $(7/9,8/9)$,
and so on.
The infinite set $A$ is countable.  The set of accumulation points of $A$ is the Cantor set, with cardinality $2^{\aleph_0}$.
A: Such a set absolutely does exist. I don't know if there's a brief description of such a set, but it's not too difficult to construct one recursively.
Below is just annoying notation required to make an intuitively clear concept somewhat precise.
Let $S = [0, 1] \cap D$ where $D$ is the set of dyadic rational numbers, and let $d_0=0, d_1=1, d_2=1/2, \ldots$ be an enumeration of $S$ (the exact enumeration is arbitrary, but sorting lexicographically first by reduced fraction denominator and then by increasing numerator is what I have in mind). We will construct, recursively, a sequence of positive real numbers $\epsilon_0, \epsilon_1, \ldots$ and an increasing function $f: S \to [0, 1]$ with the property that the closed intervals $\big[f(d_n)-\epsilon_n, f(d_n)+\epsilon_n\big]$ are all disjoint from each other.
With this setup, it should be obvious that this is possible. Define $f(0)=0$, $f(1)=1$, and then $\epsilon_0, \epsilon_1$ are somewhat arbitrary as long as they add up to less than $1$. Let's define them to be both $1/4$.
Having defined $f(d_0), \ldots, f(d_{n-1})$ and $\epsilon_0, \ldots, \epsilon_{n-1}$, let $\{i_0, \ldots, i_n\} = \{0, 1, \ldots, n\}$ be the indices so that $d_{i_0} < d_{i_1} < \cdots < d_{i_n}$, and some $i_k$ is equal to $n$. Select $f(d_n) \in \big(f(d_{i_{k-1}})+\epsilon_{i_{k-1}}, f(d_{i_{k+1}})-\epsilon_{i_{k+1}}\big)$, which exists by construction, and select $\epsilon_n$ to delineate a neighborhood of $f(d_n)$ which misses both $f(d_{i_{k-1}})+\epsilon_{i_{k-1}}$ and $f(d_{i_{k+1}})-\epsilon_{i_{k+1}}$ (this again exists by construction).
After infinitely many steps, $f$ is constructed, and $f(S)$ consists entirely of isolated points in $\mathbb R$ and is order-theoretically dense.
