Does this set contain all real numbers between the $a$ and $b$?

Question

If we define a set with $2$ elements in it $S=\{a,b\}$ and a variable "density" $d = 1$ here.

Then if we continue to expand the set with more elements relative to variable $d$ arithmetically, in such a way that:

$$(d=2) \to S= \{a, \frac{a+b}{2},b\}$$ $$(d=3) \to S= \{a, \frac{2a+b}{3}, \frac{a+2b}{3}, b\}$$ $$(d=4)\to S= \{a, \frac{3a+b}{4}, \frac{a+b}{2}, \frac{a+3b}{4}, b\}$$ $$(d=5)\to S= \{a, \frac{4a+b}{5}, \frac{3a+2b}{5}, \frac{2a+3b}{5}, \frac{a+4b}{5}, b\}$$

$$\dots$$

Is it true that $$(d\to\infty )\to S = [a,b]$$

When $d$ approaches infinity, the set $S$ contains all real numbers between the $a$ and $b$?

How can I prove or disprove it?

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I'm trying to visualize this as a infinently dense segment of a line where each element in the set is a point $(S_n,0)$ on that segment, and by that claim, the set supposedly assigns a point to each point on the line as $d$ goes to infinity, thus it contains all reals in that range?

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Also, this set contains infinently many numbers $$\frac{Xa+Yb}{Z}$$

Where $X,Y,Z$ each have infinently many digits, giving the number a possibility to contain infinently many decimal places, meaning that it can contain irrational, transcendental numbers?

If not, can someone explain why because $\infty$ boggles me.

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If I haven't properly defined my statement, can someone help me define it in both a way it could work and a way it can't work, if either is possible? To help me understand more how to construct such a statement.

Answer 1

First, your definition of $S$ is not quite precise. Let us sidestep this and take for the final set the union of all the sets you write down. That is,

$$(d=2) \to S_2= \{a, \frac{a+b}{2},b\}$$ $$(d=3) \to S_3= \{a, \frac{2a+b}{3}, \frac{a+2b}{3}, b\}$$ $$(d=4)\to S_4= \{a, \frac{3a+b}{4}, \frac{a+b}{2}, \frac{a+3b}{4}, b\}$$ $$(d=5)\to S_5= \{a, \frac{4a+b}{5}, \frac{3a+2b}{5}, \frac{2a+3b}{5}, \frac{a+4b}{5}, b\}$$

$$\dots$$

And then $S_{\infty} = \bigcup_{d \ge 1} S_d$.

This set $S_\infty$ is then equal to $\{a+ q (b-a)\colon q \in \mathbb{Q}, \, 0\le q \le 1\}$.

In particular, if you chose $a=0$ and $b=1$ then your set is the set of all rational numbers from $0$ to $1$. And generally for rational $a,b$ it is the set of all rational numbers from $a$ to $b$.

But this is not what one usually means by an interval; your set for $a=0$ and $b=1$ does not contain $\sqrt{2}-1$ for example.

You could say though that for rational $a,b$ the set is the (closed) rational interval from $a$ to $b$. This is not very common terminology, but not completely unheard of either.

To answer the comment: yes, the set $S_{\infty}$ is dense in $[a,b]$. That is every number in $[a,b]$ can be approximated to any precision, by elements from $S_{\infty}$. But differently $\overline{S_{\infty}}$, the topological closure, is $[a,b]$.

Yet, it is even true that, and something slightly different, that all the elements that you can write as a limit of a sequence $a_d$ with $a_d \in S_d$ for each $d$, is the interval $[a,b]$.

If you meant this by taking the limit then it is indeed the interval.

Answer 2

If $a$ and $b$ are rational, then all your sets $S$ from above contain no irrational number !

Your turn !

Answer 3

You ask "How can I disprove it", but you didn't really define a strict mathematical statement. Your statement

As $d\to\infty$, $S=[a,b]$

lacks definitions. You seem to imply that for a sequence of sets $A_1,A_2,\dots $, there exists a limit $$\lim_{n\to\infty} A_n$$ but limits are really only defined for real numbers, so unless you define what you mean, there is no point in asking your question.

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Now, there are cases where a "sort of" limit makes sense. For example, if $A_1\subseteq A_2 \subseteq A_3\cdots$, then at each step, $A_i$ grows to $A_{i+1}$ and you could say that $$\bigcup_{n=1}^\infty A_n$$ is what it is growing towards "at infinity".

In your case, the condition is sort of met since for each $n$, there exist infinitely many such $m$ that $S_n\subseteq S_m$, and your question could then be

Is $$S_\infty:=\bigcup_{d=1}^\infty S_d$$ equal to $(a,b)$?

In which case the answer is no because, if $a$ and $b$ are rational, the set $S_\infty$ contains only rational numbers.

However, the set $S_\infty$ is dense in $(a,b)$, meaning that the closure of $S_\infty$ is, indeed, $(a,b)$.

In other words, this means you can get arbitrarily close to any number in $(a,b)$, even though you maybe can't reach the number itself. More strictly, for every number $x\in(a,b)$, and any $\epsilon > 0$, you can find some $s\in S_\infty$ such that $|s-x|<\epsilon$.