If $x_n\in B_n$ and $\bigcap_nB_n=\{\tilde x_0\}$, does it follow $x_n\to\tilde x_0$? Let $E$ be a topological space, $E^\ast=E\cup\{\infty\}$ denote the Alexandroff one-point extension, $(x_n)_{n\in\mathbb N}\subseteq E^\ast$ and $x\in E^\ast$ with $x_n\xrightarrow{n\to\infty}x$, $B_n\subseteq E^\ast$ be open with $x_n\in B_n$ for $n\in\mathbb N$ and $\bigcap_{n\in\mathbb N}B_n=\{\infty\}$. Moreover, assume $(B_n)_{n\in\mathbb N}$ is nonincreasing.
Are we able to conclude that $x_0=\infty$? If not, does the situation change if $E$ is a (locally compact second-countable) metric space?
Remark: If necessary, assume $B_n=E^\ast\setminus K_n$ for some compact $K_n\subseteq E$ with $K_n\subseteq K_{n+1}^\circ$ and $\bigcup_nK_n=E$.
 A: Let $U$ be any open neighborhood of $\infty$ in $E^*$. Then $E^*\setminus U$ is a compact subset of $E$. Since
$$ \bigcup_{n\in\mathbb{N}} K_n^{\circ} = \bigcup_{n\in\mathbb{N}} K_{n+1}^{\circ} \supseteq \bigcup_{n\in\mathbb{N}} K_{n} = E, $$
it follows that $\{K_n^{\circ}\}_{n\in\mathbb{N}}$ is an open cover of $E^*\setminus U$. So, there exists $N$ such that $E^* \setminus U \subseteq K_{N}^{\circ}$. This then implies $B_N \subseteq U$, and so, $x_n \in U$ whenever $n \geq N$. Therefore $x_n \to \infty$ in $E^*$.

Remark. The condition $K_n \subseteq K_{n+1}^{\circ}$ is crucial in this setting, for otherwise we have the following counter-example:
$$ K_n = [-n, 0] \cup [\tfrac{2}{n}, n], \qquad B_n = \mathbb{R}^*\setminus K_n, \qquad\text{and}\qquad x_n = \tfrac{1}{n}. $$
A: Assume that $E$ is a locally compact second-countable metric space. We know that there is a $(K_n)_{n\in\mathbb N}\subseteq E$ such that $K_n$ is compact and $K_n\subseteq K_{n+1}^\circ$ for all $n\in\mathbb N$ and $\bigcup_{n\in\mathbb N}K_n=E$.
Let $\tau$ denote the topology on $E$ and $\tau^\ast$ denote the topology on $E^\ast$. By definition, $$\tau^\ast=\tau\cup\{E^\ast\setminus B:B\subseteq E\text{ is }\tau\text{-compact}\}.$$
Now let $(x_n)_{n\in\mathbb N}\subseteq E^\ast$ with $$x_n\in E^\ast\setminus K_n\;\;\;\text{for all }n\in\mathbb N.$$ We will show that $$x_n\xrightarrow{n\to\infty}\infty.$$
Let $N\in\tau^\ast$ with $\infty\in N^\ast$. By definition of $\tau^\ast$, $$N=E^\ast\setminus B$$ for some $B\subseteq E$. Since $E=\bigcup_{n\in\mathbb N}K_n$ and $(K_n)_{n\in\mathbb N}$ is nondecreasing, there is a $n_0\in\mathbb N$ with $$B\subseteq K_n$$ and hence $$x_n\in E^\ast\setminus K_n\subseteq E^\ast\setminus B=N.$$
A: If $x\neq\infty$. Then $x\in E$. So $x\in\bigcup_{n=1}^{\infty} K_{n}$.
Thus there exists $m\in\Bbb{N}$ such that $x\in K_{m}\implies x\in K_{n}\,,\forall n\geq m$
Thus $x\notin B_{n}\,,\forall n\geq m$ .
WLOG take $m$ to be the smallest integer $m$ such that $x\in K_{m}$.
Then $\text{int}(K_{m+1})$ is an open set that contains $x$. (As by assumption $K_{m}\subset \text{int}(K_{m+1})$ )
But $x_{m+1}\in B_{m+1}\implies x_{m+1}\notin K_{m+1}\implies x_{m+1}\notin \text{int}(K_{m+1}) .$ .
$x_{m+2}\in B_{m+2}\implies x_{m+2}\notin K_{m+2}\implies x_{m+2}\notin K_{m+1}\implies x_{m+2}\notin \text{int}(K_{m+1}) $ and so on .
Thus $x_{n}\notin \text{int}(K_{m+1})\,,\forall n\geq m+1$ . Thus $x_{n}$ cannot converge to $x$ as $\text{int}(K_{m+1})$ becomes an open set containing $x$ with only finitely many points of the sequence which is a contradiction.
Thus $x=\infty$
