1. ## Compactness question

I'm trying to figure out a problem here and I don't quite get it.

Let $n \leftrightarrow q_{n}$ be a bijection between $N$ and $Q\cap[0, 1]$. For $\epsilon\in(0, 1)$, define $G_{n}$ to be the open set $B_{\frac{\epsilon}{2^n}}$-ball around the point $q_{n}$. Define $F=G^{c}\cap[0, 1]$. Show that F is compact.

Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set $F$ is empty, which I know is compact. However, the next part of the question asks me to show that $0 and that $sup(F)<1$.

Basically, it looks to me like $G_{n}$ completely engulfs $[0, 1]$ and then some, so $G^{c}\cap[0, 1]$ wouldn't contain any values. Thus the set F is empty.

2. Originally Posted by miatchguy
I'm trying to figure out a problem here and I don't quite get it.

Let $n \leftrightarrow G_{n}$ be a bijection between $N$ and $Q\cap[0, 1]$. For $\epsilon\in(0, 1)$, define $G_{n}$ to be the open set $B_{\frac{\epsilon}{2^n}}$-ball around the point $q_{n}$. Define $F=G^{c}\cap[0, 1]$. Show that F is compact.

Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set $F$ is empty, which I know is compact. However, the next part of the question asks me to show that $0 and that $sup(F)<1$.

Basically, it looks to me like $G_{n}$ completely engulfs $[0, 1]$ and then some, so $G^{c}\cap[0, 1]$ wouldn't contain any values. Thus the set F is empty.
No, that's not necessarily true. You can, for example, sum all the diameters of these balls (geometric series), and that sum just might turn out to be strictly smaller than the length of the $[0;1]$.

3. Originally Posted by miatchguy
Let $n \leftrightarrow G_{n}$ be a bijection between $N$ and $Q\cap[0, 1]$. For $\epsilon\in(0, 1)$, define $G_{n}$ to be the open set $B_{\frac{\epsilon}{2^n}}$-ball around the point $q_{n}$. Define $F=G^{c}\cap[0, 1]$. Show that F is compact.
What do you know about a closed subset of a compact set?

4. Originally Posted by miatchguy
I'm trying to figure out a problem here and I don't quite get it.

Let $n \leftrightarrow G_{n}$ be a bijection between $N$ and $Q\cap[0, 1]$. For $\epsilon\in(0, 1)$, define $G_{n}$ to be the open set $B_{\frac{\epsilon}{2^n}}$-ball around the point $q_{n}$. Define $F=G^{c}\cap[0, 1]$. Show that F is compact.

Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set $F$ is empty, which I know is compact. However, the next part of the question asks me to show that $0 and that $sup(F)<1$.

Basically, it looks to me like $G_{n}$ completely engulfs $[0, 1]$ and then some, so $G^{c}\cap[0, 1]$ wouldn't contain any values. Thus the set F is empty.
A) Apparently Plato and Failure get it, but I don't understand what $G$ is? You've never said.

B) What topology is this? Is this the usual real analysis kind of spaces?

5. Originally Posted by Drexel28
A) Apparently Plato and Failure get it, but I don't understand what $G$ is? You've never said.

B) What topology is this? Is this the usual real analysis kind of spaces?
So sorry. Made a horrible typo. At the beginning I meant $n \leftrightarrow q_{n}$ rather than $n \leftrightarrow G_{n}$
$G_{n}$ is the collection of balls around the points in the sequence $q_{n}$

6. Originally Posted by miatchguy
So sorry. Made a horrible typo. At the beginning I meant $n \leftrightarrow q_{n}$ rather than $n \leftrightarrow G_{n}$
$G_{n}$ is the collection of balls around the points in the sequence $q_{n}$
Ok, so I get what $G_n$ is. What's $G$???

7. Originally Posted by Drexel28
Ok, so I get what $G_n$ is. What's $G$???
$G$ is the set of all balls around $q_{n}$ defined by the sequence $G_{n}$. Of course $G^{c}$ is the complement of that set.

8. Because I know this problem well, I just assumed things that were not explicitly stated.
Actually $G = \bigcup\limits_n {B\left( {q_n ;\frac{\varepsilon }{{2^n }}} \right)}$ so this is an open set covering the rationals in $[0,1]$.
But $\sum\limits_{n = 1}^\infty {\frac{\varepsilon }{{2^n }}} = \frac{\varepsilon }{{2 - \varepsilon }} < 1$, so $G^c\cap [0,1]$ is not empty.
Moreover $G^c$ is closed and $[0,1]$ is compact.
So?

9. Originally Posted by Plato
Because I know this problem well..
Good, because I had no idea what the hell was going on. So $\varepsilon$ was fixed?

10. Originally Posted by Plato
Because I know this problem well, I just assumed things that were not explicitly stated.
Actually $G = \bigcup\limits_n {B\left( {q_n ;\frac{\varepsilon }{{2^n }}} \right)}$ so this is an open set covering the rationals in $[0,1]$.
But $\sum\limits_{n = 1}^\infty {\frac{\varepsilon }{{2^n }}} = \frac{\varepsilon }{{2 - \varepsilon }} < 1$, so $G^c\cap [0,1]$ is not empty.
Moreover $G^c$ is closed and $[0,1]$ is compact.
So?
Thanks. I got that part now. I still don't quite understand why 0 and 1 are not elements of $G^{c}$, but I'll try and figure it out.

11. Originally Posted by miatchguy
why 0 and 1 are not elements of $G^{c}$, but I'll try and figure it out.
Well both 0 & 1 are rational.

12. Originally Posted by Plato
Well both 0 & 1 are rational.
Well yeah, but the can't be included in $G^{c}$ because $inf(G^{c})>0$ and $sup(G^{c})<1$ and since $G$ is open over $[0, 1]$ it can't include 0 or 1 either. So that's what I'm having trouble figuring out.

13. Originally Posted by miatchguy
Well yeah, but the can't be included in $G^{c}$ because $inf(G^{c})>0$ and $sup(G^{c})<1$ and since $G$ is open over $[0, 1]$ it can't include 0 or 1 either. So that's what I'm having trouble figuring out.
1) $G$ is not a cover of $[0,1]$. Just the rationals in $[0,1]$

2) $0=q_j~\&~1=q_k$ for some $j~\&~k$.
So $0\in B(0;\epsilon/2^j)$ and $1\in B(1;\epsilon/2^k)$.
Thus $\inf(F)\ge\epsilon/2^j$ and $\sup(F)\le1-\epsilon/2^k$.

14. Originally Posted by Plato
1) $G$ is not a cover of $[0,1]$. Just the rationals in $[0,1]$

2) $0=q_j~\&~1=q_k$ for some $j~\&~k$.
So $0\in B(0;\epsilon/2^j)$ and $1\in B(1;\epsilon/2^k)$.
Thus $\inf(F)\ge\epsilon/2^j$ and $\sup(F)\le1-\epsilon/2^k$.
Cool, this is exactly what I was looking for.