Basic Topology of R (Just want hints)

Question

I just want to know if I'm on to something for the following problem from intro to analysis. I laid out the problem, my scratchwork follows and finally, my questions are found under "Thoughts" in boldface. The topic is, as in the title, the basic topology of the reals under a section called "Perfect Sets." There's a subsection called "Cannected Sets." I already covered compact sets. Thanks a bunch in advance for any help I can get (again, not the solution please; I'm one of those).

$\textbf{Problem}$

Let $r_{1},r_{2},\ldots$ be an enumeration for the rational numbers (I guess the author means that these are all the rational numbers) and for each $n\in \mathbb{N}$, let $\epsilon_{n} = 1/2^{n}$. Set $O=\bigcup_{n=1}^{\infty}V_{\epsilon_{n}}(r_{n})$ and $F=O^{c}$ where $V_{\epsilon}(x)=(x-\epsilon,x+\epsilon)$ centered at $x$. Then $F\neq \emptyset$.

$\textbf{Scratchwork}$

I'm assuming that $[0,3]\subseteq O\Rightarrow \exists r_{n_{1}},\ldots r_{n_{k}}$ such that $[0,3]\subseteq \bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}})$. But the length of $V_{\epsilon_{n}}(r_{n})$ is $1/2^{n-1}$ for each $n\Rightarrow$ the length of $O$ is $\sum_{n=1}^{\infty}1/2^{n-1}=\displaystyle \frac{1}{1-1/2}=2\Rightarrow $the length of

$\bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}}) $

is at most $2$. But the length of $[0,3]$ is $3\Rightarrow \bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}})$ does not cover $[0,3]$

Remark: The problem does not mention $[0,3]$. My strategy to proving that $F\neq \emptyset$ is to consider $[0,3]$ and show that it cannot be covered by $O$.

$\textbf{Thoughts}$

Am I right with what I have ?(don't want the solution please). If so, can I just say it like that? (then the "scratchwork" is actually a proof). Is there a better strategy? And on a final note, I did not use any of the ideas provided in this section but This problem is just an indexed question so I'm not too worried about that.

Answer 1

So the essence of your proof is that $\mu(O)\le M$ for some constant $M$ (3 maybe) and so $O$ cannot be $\mathbb{R}$ which has infinite Lebesgue measure. The idea seems OK to me.

Answer 2

I came up with this:

If $|x|<\epsilon \wedge |y|<\delta \Rightarrow |x-y|\leq |x|+|y|<\epsilon +\delta$. Suppose $[0,3]\subseteq \bigcup_{n=1}^{\infty}V_{\epsilon_{n}}(r_{n})\Rightarrow \exists r_{1},\ldots r_{k}$ such that $[0,3]\subseteq \bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}})$. Let $x,y\in \bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}})\Rightarrow \exists p,q\in \mathbb{Z}^{+}$ such that $x\in V_{\epsilon_{p}}(r_{n_{p}})$ and $y\in V_{\epsilon_{q}}(r_{n_{q}})$. We have $r_{n_{p}}\in (-1/2^{n_{p}}, 1/2^{n_{p}})$ and $r_{n_{q}}\in (-1/2^{n_{q}},1/2^{n_{q}})\Rightarrow |r_{n_{p}}-r_{n_{q}}|\leq 1/2^{n_{p}}+1/2^{n_{q}}\Rightarrow |x-y|=|x-r_{n_{p}}+r_{n_{q}}-y+r_{n_{p}}-r_{n_{q}}|\leq |x-r_{n_{p}}|+|y-r_{n_{q}}|+|r_{n_{p}}-r_{n_{q}}|<1/2^{n_{p}}+1/2^{n_{q}}+1/2^{n_{p}}+1/2^{n_{q}}=\displaystyle \frac{1}{2^{n_{p}-1}}+\frac{1}{2^{n_{q}-1}}\leq \sum_{i=1}^{\infty}\frac{1}{2^{n_{i}-1}}=2 $. But if $x=0, y=3\Rightarrow x,y\in [03]\wedge |x-y|=3$ contrary to $x,y\in \bigcup_{i=1}^{k}V_{\epsilon_{n_{i}}}(r_{n_{i}}) \Rightarrow F\neq \emptyset$.

Answer 3

An enumeration $r_1,r_2,r_3,...,$ or $\{r_n:n\in \Bbb Z^+\},$ or (abbreviated) $\{r_n\}_n$ of a set $S$ customarily means that $S=\{r_n: n\in \Bbb Z^+\},$ and many authors also assume that $r_i\ne r_j$ when $i\ne j$, when talking of an enumeration.

The measure $m(O)$ (not the length) of $O$ is less than $2$ because $m(O)=$ $m(\cup_{n\in \Bbb Z^+}V_{\epsilon_n}(r_n))\leq$ $\sum_{n\in \Bbb Z^+}m(V_{\epsilon_n}(r_n))=2,$ but there is considerable overlap so we have $m(O)<2$ rather than $m(O)=2.$ For example we can find some $j\ne 1$ such that $V_{\epsilon_j}(r_j)\subset V_{\epsilon_1}(r_1),$ so $m(O)\leq$ $ m(\cup_{1\ne n\ne j}V_{\epsilon_n}(r_n))+m(V_{\epsilon_1}(r_1))<2.$

If you take any $\delta>0$ and replace each $\epsilon_n $ with $\delta \epsilon_n$ then the measure of $O$ will be less than $2\delta,$ which can be as close to $0$ as you like. So if $S=\{r_n: n\in \Bbb Z^+\}$ is $any$ countable subset of $\Bbb R$ then you can cover $S$ with a family of open intervals whose lengths add to as small a positive value as you like.

I have no other criticisms of your work. Measure is a generalization of the concepts of length, area, volume..... Relevant topics: Lebesgue measure, outer measure, inner measure, Borel sets, $\sigma$-algebras.