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.
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 $\infty$.