Proving Properties of Open Sets in $\mathbb{R}^d$

[nateHI]

Homework Statement

Let $O$ be a proper open subset of $\mathbb{R}^d$ (i.e.$O$ is open, nonempty, and is not equal to $\mathbb{R}^d$). For each $n\in \mathbb{N}$ let
$O_n=\big\{x\in O : d(x,O^c)>1/n\big\}$
Prove that:
(a) $O_n$ is open and $O_n\subset O$ for all $n\in \mathbb{N}$,
(b) $O_1\subset O_2 \subset \dots$, and $\cup_n O_n=O$
(c) If $O_n\neq 0$ then $d(O_n, O^c)\ge \frac{1}{n}$
(d) If $O_n\neq 0$ then $d(O_n, O^c_{n+1})\ge \frac{1}{n(n+1)}$

Homework Equations

The Attempt at a Solution

(a) solved
(b) solved
(c) I'm not sure what the instructor is looking for here since there is no $n$ and no $x\in O_n$ such that $d(x,O^c)=\frac{1}{n}$ since that would contradict the construction of $O_n$. It seems like the problem statement should be
If $O_n\neq 0$ then $d(O_n, O^c)> \frac{1}{n}$.
(d) $d(O_n,O^c_{n+1})=d(O_n,O^c)-d(O_{n+1},O^c)\ge \frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$

 

[Stephen Tashi]

(c) I'm not sure what the instructor is looking for here since there is no $n$ and no $x\in O_n$ such that $d(x,O^c)=\frac{1}{n}$ since that would contradict the construction of $O_n$.

The following just a suggestion, not necessarily a useful "hint":

How do your course materials define the distance between two sets? Perhaps to find $d(O_n, O^c)$ you might have to do something like take a limit of distances $d(x_i,O^c)$ with each $d(x_i,O^c) \gt \frac{1}{n}$ but with the sequence of distances converging to $\frac{1}{n}$.

 

[nateHI]

The following just a suggestion, not necessarily a useful "hint":

How do your course materials define the distance between two sets? Perhaps to find $d(O_n, O^c)$ you might have to do something like take a limit of distances $d(x_i,O^c)$ with each $d(x_i,O^c) \gt \frac{1}{n}$ but with the sequence of distances converging to $\frac{1}{n}$.

The distance between the two sets is $inf d(x_i, y_i)$ where $x_i\in O_n$, $y_i\in O^c$. But the problem I see is that by part (a) $O_n$ is open (but bounded) hence the lower bound of $d(O_n, O^c)$ is not attainable even though $O^c$ is closed. Also, the limits don't coincide since
$lim_{n\to\infty} d(O_n,O^c)\to 1/n$ but $lim_{n\to\infty}1/n\to 0$

 

[Stephen Tashi]

hence the lower bound of $d(O_n, O^c)$ is not attainable even though $O^c$ is closed.

One need not attain an infimum for it to be an infimum. Let $O$ be the open interval $(0,1)$. and let $N = 4$. I think $O_n$ is the open interval $( \frac{1}{4}, \frac{3}{4})$. What is distance between $O^C$ and $( \frac{1}{4}, \frac{3}{4})$ ?.

 

[nateHI]

One need not attain an infimum for it to be an infimum. Let $O$ be the open interval $(0,1)$. and let $N = 4$. I think $O_n$ is the open interval $( \frac{1}{4}, \frac{3}{4})$. What is distance between $O^C$ and $( \frac{1}{4}, \frac{3}{4})$ ?.

OK thanks, I guess saying $d(O_n, O^c)\ge 1/4$ is just another way of writing what the lower bound is. I was probably overthinking ( possibly under-thinking) things.

Does part (d) seem correct?

 

[Stephen Tashi]

You should explain why $d(O_n,O^c_{n+1})=d(O_n,O^c)-d(O_{n+1},O^c)$