Equivalence classes of $\mathbb R$ Let $X$ be a locally compact, connected, locally connected, Hausdorff space. Considder $U_1\supseteq U_2\supseteq\cdots$ of open and non-empty connected subsets with compact frontiers such that $\cap\overline{U_i}=\emptyset$. We call two sequences are equivalent if and only if every set of one sequence contains some set of the other sequence and vice versa.
I have some troubles with the definition. For instance, consider real number with standard topology. How can we get the equivalence classes of $\mathbb R$?
John suggested $U_n=(n,\infty)$. It is not hard to see that $U_n$ is satisfied by all condition. With same argument We can see that $V_n=(\infty,-n)$ is another sequance which is not equivalent with $\{U_n\}$. Now, is there other equivalence class?
 A: If $X=\mathbb R$, there are only two equivalent sequences: $\{(n, +\infty)\}$ and $\{(-\infty, -n)\}$. The reason is the following: 
Let $\{U_i\}$ be one such sequence. First of all, $U_i$'s are all connected. So $U_i$ has to be some intervals $U_i = (a_i, b_i)$. But as
$$(*) \bigcap_i \overline U_i = \emptyset$$
Then either $a_i = -\infty$ or $b_i = \infty$. (Or $\overline U_i$ would be compact and so are all $\overline U_j$ for $j \geq i$)
If they are both infinities, then $U_i = \mathbb R$. Thus ther exists some $i$ so that $a_i \neq -\infty$ or $b_i \neq \infty$ (or all of them are just $\mathbb R$). 
If $a_i \neq -\infty$, then $b_i = +\infty$. Then $a_j \neq -\infty$ and $b_j = +\infty$ for all $j\geq i$. Using (*) again, we see that $(a_i)$ has to converge to $+\infty$. Thus $\{U_i\}$ is equivalent to $\{(n, \infty)\}$. 
Similarly if $b_i \neq +\infty$, then $\{U_n\}$ is equivalent to $\{(-\infty, -n)\}$. 
Let's also consider $X = \mathbb R^n$. In this case we can show that there is only one equivalent class given by 
$$\bigg\{U_n  = \{x : n<|x|\}\bigg\}_{n=1}^\infty.$$
First, let $V_i$ be an open set with compact boundary. For our consideration we assume that $V_i$ is unbounded (If $V_i$ is bounded, it is impossible that $\cap \overline V_i = \emptyset$)
As the boundary is compact, there is $n_i$ so that $|x| < n_i$ for all $x \in \partial V_i$. 
Claim $U_{n_i} \subset V_i$. 
To show this, let $x \in V_i \cap U_{n_i}$. For any $y\in U_{n_i}$, connect $x$ to $y$ by a path $\gamma$ so that $\gamma$ lies in $U_{n_i}$. In paricular, $\gamma$ does not pass through the $\partial V_i$. Thus $y\in V_i$. This finishes the proof of the claim. 
So for any $V_i$, there is $U_{n_i}$ so that $U_{n_i} \subset V_i$. On the other hand, consider the set $U_n$. Let $K = \{x: |x| \leq n+1\}$. Then $K$ is compact and so are $\overline V_i \cap K$. But 
$$\bigcap_{i=1} \overline V_i = \emptyset \Rightarrow \bigcap_{i=1}^\infty \big( \overline V_i \cap K\big) =\emptyset \Rightarrow  \overline V_i \cap K = \emptyset$$ 
for some $i$. That is $|x| > n$ for all $x\in V_i$ and so $V_i \subset U_n$. Thus $\{V_i\}$ is equivalent to $\{U_n\}$. 
