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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?

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?

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.

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.

Vepir
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  • No, in each step your set is countable. – Mathematician 42 Jan 10 '17 at 12:59
  • No; reals in the interval $[a,b]$ are uncountable. – Mauro ALLEGRANZA Jan 10 '17 at 12:59
  • But we can never reach the last step by counting since we cannot reach infinity by counting which supposedly makes the last step uncountable? Can you explain why this assumption won't work if we try to apply it on a specific math problem? – Vepir Jan 10 '17 at 13:01
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    You have to explain how to take the limit. The most obvious thing to do is to consider $\bigcup_{i\geq 0}S_i$ where $S_i$ is the set obtained in the $i$-th step. But this union is still countable and thus the construction fails. – Mathematician 42 Jan 10 '17 at 13:07

3 Answers3

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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.

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 $|s-x|<\epsilon$.

5xum
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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. – Vepir Jan 10 '17 at 13:12
  • @Matta I'm working on it. – 5xum Jan 10 '17 at 13:13
  • @Matta I expanded my question. – 5xum Jan 10 '17 at 13:18
  • So, it conatins "almost" all real number since it contains infinently many numbers that are all infinently close to any real number? Can we then apply it as it does contain all reals, since the error would be infinently small? Like if we take a limit of $1/n$ it never reaches $0$ but the limit is $0$. So the "limit of it" does contain all reals? – Vepir Jan 10 '17 at 13:24
  • @5xum: Small remark, the $S_i$'s are not nested. For example, when $a=0$ and $b=1$, $\frac{1}{3}\in S_3$ and $\frac{1}{3}\notin S_4$. However you don't need it either. – Mathematician 42 Jan 10 '17 at 13:25
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    @Matta Actually, it contains almost no real numbers (it only contains a countable infinity of them, there's much more numbers not in $S_\infty$. And, again, there's no such thing as limits when it comes to sets. The closure of it contains all reals on $(a,b)$, yes. – 5xum Jan 10 '17 at 13:26
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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.

quid
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  • But does it still contain numbers infinently close to $\sqrt{2}-1$, making the closest number approach $\sqrt{2}-1$? – Vepir Jan 10 '17 at 13:18
  • There is no number different from $\sqrt{2}-1$ that is the closest to $\sqrt{2}-1$. What you are saying is simply noting that $S_{\infty}$ is dense in $[a,b]$. – Mathematician 42 Jan 10 '17 at 13:22
  • @Mathematician42 there is a closest number in each $S_d$ though. I think Matta meant something slightly different than density of $S_{\infty}$. See my expanded answer. – quid Jan 10 '17 at 13:25
  • Accepting this instead, because of say on the rational definition and because of my new question. – Vepir Jan 10 '17 at 14:28
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If $a$ and $b$ are rational, then all your sets $S$ from above contain no irrational number !

Your turn !

Fred
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  • But this set contains a number $(Xa+Yb)/Z$ where X,Y and Z have infinite many digits, thus meaning that that number can have infinently many decimal places, making it irrational? – Vepir Jan 10 '17 at 13:03
  • $\frac{1}{3}=0,333\dots$. Having infinitely many digits doesn't mean it's irrational. – Mathematician 42 Jan 10 '17 at 13:05
  • @Mathematician42 Not necessarily, but it can in theory since you cannot write a irrational number if those were finite. – Vepir Jan 10 '17 at 13:09
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    @Matta, you will never encounter infinite non-repeating decimal number this way. Proof is simple, rational numbers are closed under additition, subtraction, multiplication and division. Meaning that for rational $a,b$, $a+b,a-b,a\cdot b,a/b$ are also rational. So, no, it can't in theory. – Ennar Jan 10 '17 at 13:37