Given the following constraints, can this set be infinitely large? I am studying a set $X$ with the following properties. Let $x_{i}$ denote the $i^{th}$ element in $X$. Let $n$ denote the cardinality of $X$ ($ = |X|$). The following properties apply:
a) $X$ is a set of positive integers, the values of which must not exceed $2n$.
b) No element $x_{i}$ in $X$ is allowed to divide any other element $x_{j}$ in $X$.
I am curious as to the maximum number of integers (or maximum cardinality) that set $X$ can possess.
If we force $x_{1} = n$, set $n \ge 3$ and ensure that $x_1 \lt x_2 \lt ... \lt x_{n}$. Then provided that each successive element is one greater than the last, can't $X$ contain all elements from $x_1$ up to but not including the element whose value is $2 \cdot x_1$? In turn, would this mean that $X$ can be infinitely large given $n$ is arbitrary? [You can check easily why it is best to start at $x_1 = n = 3$; If $x_1=1$ or $x_1=2$ you can't add many further integers to the set without violating a) or b)]
For example, let's take $n = 3$. Then set $X$ contains three successive integers starting at  $x_1 = n = 3$. So
$X = \{3, 4, 5\}$. Here, the conditions are met that all elements are less than $2n$. If we add the next integer ($6$) then $x_n = 2\cdot x_1$ and hence $x_1$ divides $x_n$ which violates b).
For any $n \ge 3$, $x_1 = n$, we can let S contain all the integers whose values do not exceed $2\cdot x_1$. As another example, take $n=20$, then the rule is to start with $20$ and include all elements less than $2(20) = 40$. So $X = \{20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39\}$.
And indeed $X$ contains $n$ elements with conditions a) and b) satisfied.
Hopefully now you understand my logic and thought process to this problem. As it seems, the cardinality of $X$ can be arbitrarily large provided we make $n$ arbitrarily large, set $x_1 = n$ and count up one at a time until we reach the element $x_n$ which has a value that is 1 less than $2 \cdot x_1$. My reasoning here feels a bit messy though. If indeed my logic is correct, is there a more formal way to prove that $X$ can be infinitely large in this scenario since we can make $n$ arbitrarily large?
Hope I have been clear. Any help appreciated!
 A: As Daniel Wainfleet has said, "Arbitrarily large and infinitely large are different things."
If $n$ is a natural number, then it is finite. So if $|X| = n$, then $X$ is a finite set. Alternatively, the set of all positive integers $\le 2n$ is finite, and $X$ is a subset of that. A subset of a finite set is also finite. So again, $X$ is finite.
It doesn't matter that $n$ can be arbitrarily large. $X$ is not a single set. What set it is depends on $n$ (and even for a fixed $n$, there are many possible sets $X$). By choosing a larger $n$, you get a larger $X$, but that $X$ is still a finite set. As you've already discovered, for each positive $n$, the set of all numbers between $n$ and $2n - 1$ meets all your conditions to be $X$. Thus there is at least one such $X$ for any $n$.
Can $n$ be an infinite cardinal, instead of a natural number? Yes. Though $\{ k\in \Bbb N\mid n \le k < 2n\}$ no longer works as an example, since $n$ and $2n$ are not integers. Instead, you can let $X$ be the set of prime numbers.
