Uncountable chain of closed sets or real numbers stabilizes I need to prove the following statement:
Let $\left\{ F_{\alpha} \right\}_{\alpha < \omega_1}$ ($\omega_1$ being the first uncountable ordinal) is a descending chain of closed sets of real numbers, i.e. if $\alpha < \beta \leq \omega_1$, then $F_{\alpha} \supseteq F_{\beta}$. Then there is an ordinal $\delta < \omega_1$, such that for every $\delta < \alpha$, $F_{\delta} = F_{\alpha}$.
Actually, I don't think this is true, because if we take the closed intevals $\left[ a, 2 \right]$, for $a \in [0, 1]$, then this is uncountable descending chain of closed sets in $\mathbb{R}$ and it doesn't stabilize.
I know, that $\omega_1$ and continium, which is the cardinality of $[0,1]$ are different things and the ordering of $\omega_1$ doesn't coincide with the ordering of $[0,1]$, so I am pretty sure that I'm not on the right track in solving the problem. Please, give me some hints!
 A: The trick is to change to open sets and a countable basis of the topology.
Let $G_\alpha=\overline{F_\alpha}$; the chain $\{G_\alpha\}$ is nondecreasing and we have to prove that it stabilizes from somewhere.
Let $B=\{(p,q): p,q\in\mathbb{Q}, ~p<q\}$ be the set of rational intervals; this is a countable basis in the sense that every open set is the union of such intervals.
For every $\alpha<\omega_1$, let $D_\alpha=\{I\in B:~ I\subset G_\alpha\}$ be the set of base intervals that are contained in $G_\alpha$. Since $G_\alpha$ is an open set, we have $G_\alpha=\bigcup_{I\in D_\alpha}I$. The chain $\{G_\alpha\}$ is non-decreasing, so $\{D_\alpha\}$ is a non-decreasing chain of subsets of $B$.
For every $I\in\bigcup_{\alpha<\omega_1}D_\alpha$, let $\beta_I<\omega_1$ be the first index such that $I\in D_{\beta_I}$ and let $\gamma=\sup_I \beta_I$. For every $I\in\bigcup_{\alpha<\omega_1}D_\alpha$ we have $I\in D_{\beta_I}\subset D_\gamma$, so $D_\gamma=\bigcup_{\alpha\le\omega_1}D_\alpha$; the chain $\{D_\alpha\}$ is stable for $\alpha\ge\gamma$.
$\gamma$ is the supremum of countably many countable orders, so $\gamma<\omega_1$.
A: Here is a proof that a well ordered descending
nest of bounded intervals stablizes.
Let a$_k$ = inf F$_k$, b$_k$ = sup F$_k$.
If F$_k$ doesn't stabilize, either a$_k$ or b$_k$ doesn't.  
Assume a$_k$ doesn't stabilize.
Define recursively
g(k) = min{ j : for all w < k, a$_{g(w)}$ < a$_j$ }.
Show f:$\omega_1$ -> R, k -> a$_{g(k)}$ is an order embedding.  
As that contradicts the theorem $\omega_1$
does not embed into R, a$_k$ stabilizes.  
Likewise show the order dual statement b$_k$ stabilizes.
As both a$_k$ and b$_k$ stablize, so does F$_k$.
A: For the sake of simplicity, we replace $\omega_1$ with a totally ordered set $I$ with uncountable cofinality (thus, the proof does not work when $I$ is any subset of $\mathbb{R}$). 
Let $F_I=\bigcap_{i \in I}{F_i}$. Let, for each $n \geq 1$, $U_n$ be the set of $x \in \mathbb{R}$ at distance less than $1/n$ from $F_I$. Let $T_n=[-n,n] \backslash U_n$, so that $T_n$ is a compact subset of $\mathbb{R}$. 
Now fix $n \geq 1$. The $F_i \cap T_n$, for $i \in I$, are a decreasing sequence of compact subsets of $\mathbb{R}$ with empty intersection. So there is $j_n \in I$ such that for all $i \geq j_n$, $F_i \cap T_n=0$. 
As $I$ has uncountable cofinality, there exists $k \in I$ greater than every $j_n$. 
Let $s \geq k$ be an element of $i$. Then $F_s$ doesn’t meet any of the $[-n,n] \backslash U_n$. Let $x \in F_s \backslash F_I$. Then there is $n \geq 1$ such that $d(x,F_I) > 1/n$. Let $m$ be an integer greater than $n+|x|+1$: then $x \in T_m$, so $F_s \cap T_m$ is nonempty, a contradiction. 
Thus for all $s \geq k$, $F_s=F_I$. 
