Kuratowski convergence for unbounded/non-compact sets

Question

Definition of Kuratowski convergence as copied from the Wikipedia page:

Let $(X,d)$ be a metric space, where $X$ is a set and $d$ is the function of distance between points in $X$.

For any $x \in X$ and any non-empty compact subset $A \subseteq X$, define the distance between the point and the subset: $$d(x,A) = \inf\{d(x,a) | a \in A\}.$$

For any sequence of such subsets $A_n \subseteq X$, $n \in \mathbb{N}$, the Kuratowski limit inferior of $A_n$ as $n \rightarrow \infty$ is $$\text{Li}_{n \rightarrow \infty} A_n = \{x \in X | \lim_{n \rightarrow \infty}\sup d(x,A_n) = 0\};$$ the Kuratowski limit superior of $A_n$ as $n \rightarrow \infty$ is $$\text{Ls}_{n \rightarrow \infty} A_n = \{x \in X | \lim_{n \rightarrow \infty}\inf d(x,A_n) = 0\};$$

If the Kuratowski limits inferior and superior agree, then their common value is called Kuratowski limit and $A_n$ converges in the Kuratowski sense.

Q: Why are the sets $A_n$ required to be compact? Specifically, why are the $A_n$ required to be bounded? There is an example on that same Wikipedia page, in which the $A_n$ are closed but not bounded. Could the notion of Kuratowski convergence be extended to include open and unbounded sets?

Background: I am working on a proof for the convergence of a cone that can be expressed by a sequence of sets. Since cones are unbounded and thereby not compact, the conditions for applying Kuratowski convergence are not satisfied.

Let $\mathbf{u} \in \mathbb{R}^m$, then $$D^0_n(\mathbf{u}) := \left\{ \mathbf{v} \in \mathbb{R}^m | \frac{1}{n} \max_{i \in [m]} (v_i - u_i) + \min_{i \in [m]} (v_i - u_i) \geq 0 \right\}$$ and $$D^\infty_n(\mathbf{u}) := \left\{ \mathbf{v} \in \mathbb{R}^m | \max_{i \in [m]} (v_i - u_i) + \frac{1}{n} \min_{i \in [m]} (v_i - u_i) \geq 0 \right\}$$

I want to show that $D^0_{n \rightarrow \infty}(\mathbf{u}) = \{\mathbf{u}\} + \mathbb{R}_+^m$ and $D^\infty_{n \rightarrow \infty}(\mathbf{u}) = \{\mathbf{u}\} + \mathbb{R}_m \setminus \mathbb{R}^m_-$ using Kuratowski convergence, where $\mathbb{R}^m_+$ is the positive and $\mathbb{R}^m_-$ is negative orthant.

Answer 1

This is in answer to the request in the comments, rather than answering the question asked in the post.

The basic idea is to replace the standard Euclidean metric $d$ on $\Bbb R^m$ with an equivalent bounded metric $d'$. For $x \in \Bbb R^m$, let $B(x,r) = \{y \mid d(y,x) < r\}$ denote the open ball of radius $r$ about $x$ under the metric $d$, and let $B'(x,r)$ be the ball under the metric $d'$. What we need from $d'$ is that for all every $x$ and every $r > 0$, there exists an $r' > 0$ with $B'(x, r') \subseteq B(x,r)$. And also for every $s' > 0$, there is an $s > 0$ with $B(x,s) \subseteq B'(x,s')$. Then the two metrics $d$ and $d'$ are equivalent: they produce the same topology on $\Bbb R^m$. Further, for sequences of sets $\{A_n\}$, they give the same Kuratowski limit. I will leave it to you to confirm that.

There are several ways of producing such a bounded metric $d'$. Stereographic projection provides one version: parametrize $\Bbb R^{m+1}$ as $(x_0, x_1, \dots, x_n)$ and embed $\Bbb R^m$ as the plane $x_0 = 0$. On the sphere $S^m = \{\hat x \in \Bbb R^{m+1}\mid \|\hat x\| = 1\}$, label the point $\infty := (1, 0, 0, \dots 0)$. For each $x \in \Bbb R^m$, draw the line from $\infty$ to $x$. The line will intersect $S^m$ in some other point $\hat x$. The map from $x$ to $\hat x$ is a homeomorphism between $\Bbb R^m$ and $S^m \setminus \{\infty\}$. One can then define $d'(x, y) = d(\hat x, \hat y) = \|\hat x - \hat y\|$. A little geometry and algebra can produce an exact formula for $d'$, and allow you to confirm that it does indeed meet the conditions in the previous paragraph.

But I am not going to bother, because you've indicated that you are already familiar with the concept of compactifications. A much simpler approach is to add an arbitrary point $\infty$ to $\Bbb R^m$, and define $$d'(x,y) = \begin{cases}d(x,y) & x\ne \infty, y \ne \infty, d(x,y) < 1\\\frac 1{d(x,0)} & x \ne \infty, y = \infty, d(x, 0) < 1\\d'(y,x) & x = \infty, y\ne \infty\\0&x = \infty, y = \infty\\1& \text{otherwise}\end{cases}$$

You can check that $d'$ is positive, definite and still satisfies the triangle inequality. I.e., it is a metric. And confirming the condtions for equivalence with $d$ on $\Bbb R^m$ is almost trivial. You can also confirm that it provides the normal Alexandroff topology for the one-point compactification. (I.e., the complements of compact sets in $\Bbb R^m$ form a basis of neighborhoods for $\infty$.)

If you then add $\infty$ to all of your cones, they become compact sets, so you can apply the full theory of Kuratowski limits. Since $\infty$ is in each of the cones, it will be in the limit. But for any other point $x$ in the limit, it has a neighborhood that misses $\infty$, and therefore $x$ has to be in the limit of the sets without $\infty$.