Condition on $(x_n)$ equivalent to $\lim x_n \in U$

Question

Let $(x_n)_{n=1}^\infty \subseteq \mathbb{R}$ be a Cauchy sequence of real numbers. Let $U \subseteq R$ be an open set - for simplicity, we can suppose $U$ is an open interval $(a,b)$. Is there a condition only involving the sequence $(x_n)$ that is equivalent to the limit $x = \lim_{n \to \infty} x_n$ being inside $U$?

The obvious candidate would be: "there is $N \in \mathbb{N}$ such that for all $n \geq N$, we have $x_n \in U$". While this is necessary, it is not sufficient - e.g. if $U = (0,1)$, the sequence $x_n = 1/n$ satisfies this condition, but the limit $x = 0 \notin U$.

Answer 1

Yes. One could be "there exists $\;\epsilon>0\;$ s.t. for any

$$n\in\Bbb N\;,\;\;x_n\in(a+\epsilon,\,b-\epsilon)"$$

In fact, the above condition can be weakened to

$$\;\exists\,\epsilon>0\;\,\exists\,N\in\Bbb N\;\;s.t.\;\;\forall\,n>N\,,\,\,x_n\in(a+\epsilon,\,b-\epsilon)\;$$

Answer 2

A necessary and sufficient condition is $\lim \inf d(x_n, U^{c})>0$. This works for any open set $U$. Here $d(x,A)$ denotes $\inf \{|x-y|: y \in A\}$.

Note that $\lim \inf d(x_n, U^{c})=d(x,U^{c})$ where $x =\lim x_n$. Since $U^{c}$ is closed, $d(x,U^{c})>0$ iff $x \notin U^{c}$ iff $x \in U$.