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I was wondering how to correctly mathematically describe the following observation/fact.

Let us consider the set of points on the real line defined as $\frac{K}{n}$, where $K$ is a chosen real constant and $n \in \mathbb{N}$.

Let us take $K=1$: the corresponding set, named $\mathcal{P}_1$, contains ${1,\frac{1}{2},\frac{1}{3}, \frac{1}{4}, \dots}$.

Doubling $K$, the corresponding set $\mathcal{P}_2$ contains all the points in $\mathcal{P}_1$, as $ \frac{1}{2} =\frac{2}{4}, \frac{1}{3} =\frac{2}{6} $, and more are added to the set. e.g. $ \frac{2}{3}, \frac{2}{5}$.

Intuitively, as $K \to \infty$, the set $\mathcal{P}_K$ will contain more and more points in any fived interval $[0,a)$, say $[0,1)$.

Is there a sense/definition whereby one can state that $[0,1) \subset \mathcal{P}_K $ as $k \to \infty$?

EDIT

Following Gerry Myerson comment, please consider the following re-phrase of the last question:

Is there a sense/definition whereby one can state that $\mathbb{Q}[0,1) \subset \mathcal{P}_K $ as $k \to \infty$, where $\mathbb{Q}[0,1)$ stands for all the rationals in the interval $[0,1)$?

Or anything alternative of course, expressing the fact that, as $K$ grows, the distance between any number in $[0,1)$ and a suitably chosen element of $\mathbb{P} _K$ goes to $0$ ?

In other words, how does the mathematician express the fact that, as $K$ grows, $\mathcal{P} _K$ "almost fills" the interval $[0,1)$ ?

Andrés E. Caicedo
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An aedonist
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  • No. Have you not heard of irrational numbers? – Gerry Myerson Mar 19 '19 at 11:10
  • I have indeed. That is why I asked if there is such a definition, implying that I am not aware of any. Should have been clearer, my fault of course. – An aedonist Mar 19 '19 at 11:16
  • As you have heard of irrational numbers, then you can tell whether there are any in $[0,1)$, and you can tell whether there are any in $P_K$ (or $P_k$, you use both notations), and so you can tell whether $[0,1)$ is contained in $P_k$ as $k\to\infty$. – Gerry Myerson Mar 19 '19 at 11:18

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For $K$ rational, you can say that $\mathcal{P}_k$ "in the limit" contains every rational number in the interval $[0, K]$ by $$ \liminf_{k \to \infty} \mathcal{P}_k = \bigcup_{n = 1}^\infty \bigcap_{k = n}^\infty \mathcal{P}_k = [0, K] \cap \mathbb{Q}. $$ In the general case of arbitrary $K \in \mathbb{R}$, you will find $\liminf_{k \to \infty} \mathcal{P}_k = [0, K] \cap K \mathbb{Q}$, where $K \mathbb{Q}$ means the set formed by multiplying all of the rationals by $K$. (In this particular case the $\liminf$ above ends up just being a union.)

However, for some fixed $K$ you will only be able to "reach" the rational multiples of $K$ via this method (and only those rational multiples in the interval $[0, K]$, even). For example, if you take $K$ to be rational, none of the $\mathcal{P}_k$s can ever contain any irrational number in the interval $[0, K]$ (of which there are a lot). So your sets $\mathcal{P}_k$ will always have lots of "holes", and will never contain all of $[0, K]$ in this sense.

To address your edit: we could say that the closure $\overline{[0, K] \cap K\mathbb{Q}}$ of the set $[0, K] \cap K\mathbb{Q}$ is all of $[0, K]$, meaning that every element of $[0, K]$ can be approximated arbitrarily closely by a sequence in $[0, K] \cap K\mathbb{Q}$. A mathematician then wouldn't say that that the union of all of the $\mathcal{P}_k$s "fills up" $[0, K]$, but they might say that $\liminf_{k \to \infty} \mathcal{P}_k$ is dense in $[0, K]$.

Keeley Hoek
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