Metric spaces and ultrafilters Assume $X_n$ are proper metric spaces and $X$ is a compact metric space.
Fix an non-principal ultrafilter $\omega$ on $\mathbb{N}$.
Assume we have a metric space $\mathbb{X}$ such that:
(1) $\mathbb{X}=X\coprod X_1\coprod X_2\coprod...$
(2) For each $j=1,2,3,...$ the canonical injection $i_j:X_j\rightarrow \mathbb{X}$ is distance preserving.
(3) The canonical injection $i:X\rightarrow \mathbb{X}$ is distance preserving.
(4) Each $X_1,X_2,...$, is open in $\mathbb{X}$.
(5) For each $x\in \mathbb{X}$, $dist(x,X_n)\rightarrow dist(x,X)$.
Let $x_1,x_2,....$ where $x_n\in X_n$. Why does it follow that the $\omega-limit$ of that sequence exists and is in $X$?
Comments:

*

*Clearly $(x_n)_n$ is in $\mathbb{X}$.


*It is not true that for large enough $n$, $x_n\in X$ (by disjointness).


*In compact metric space, the $\omega$-limit for any sequence in $X$ exists and is unique.


*Proper metric spaces are complete.
 A: This is not true.  For a counterexample, consider $X_n=\mathbb{R}\times\{1/n\}$ and $X=[0,1]\times\{0\}$.  Give $\mathbb{X}$ the following metric: the distance between two points is the shortest path between them where you are allowed to move horizontally in $\mathbb{R}^2$ and vertically only within $[0,1]\times\mathbb{R}$.  So restricted to $[0,1]\times\mathbb{R}$ this is the $\ell^1$ metric, but for instance $d((2,1),(2,1/2))=5/2$ since you have to first move from $(2,1)$ to $(1,1)$ and then down to $(1,1/2)$ and then back out to $(2,1/2)$.  This satisfies all your assumptions, but the sequence $x_n=(2,1/n)$ has no accumulation point in $\mathbb{X}$.
To make this true you need to strengthen (5) to say that $X_n$ converges to $X$ with respect to the Hausdorff metric of $\mathbb{X}$.  With that assumption, for each $n$ we can pick a point $y_n\in X$ such that $d(x_n,y_n)\leq d_H(X_n,X)\to 0$ as $n\to\infty$.  By compactness of $X$, $(y_n)$ converges to some $x\in X$ with respect to $\omega$, and then it follows that $(x_n)$ converges to $x$ with respect to $\omega$ as well.
