Let $(X, \mathcal{T})$ be a non compact topological space, $\infty \notin X$ and $(X^* := X \cup \{\infty\}, \mathcal{T}^* := \{U \subseteq X^*\mid U \cap X \in \mathcal{T} \land (\infty \in U \implies X \setminus U \mathrm{\ compact)}\})$ the Alexandroff compactification of $X$.
Let $(Y, \mathcal{S})$ be a Hausdorff space, and let $f: X \to Y$ be a continuous function. Prove that the following statements are equivalent:
(1) $\operatorname{stack}\{f(X \setminus K) \mid K \subseteq X \mathrm{\ compact}\}$ converges
(2) There exists a unique continuous function $f^*: X^* \to Y$ with $f^*\vert_X = f$
My attempt:
$(1) \implies (2)$ Suppose $\operatorname{stack}\{f(X \setminus K) \mid K \subseteq X \mathrm{\ compact}\} \to y \in Y$
Define $f^*$ as the extention of $f$ such that $\infty \mapsto y$. This seems the only thing I can do now. I now have to show that this function is continuous, and I have no clue how I should do this. I also have to show this is unique, and I think I can use an argument that any image of $\infty$ other than $y$ will make the continuity of $f^*$ fail in that particular point.
$(2) \implies (1)$ Probably this follows from the fact that for any filter $\mathcal{F}$ on $X^*$, $\mathcal{F} \to x \implies \operatorname{stack}f({\mathcal{F}})\to f^*(x)$, by continuity of $f^*$, but I cannot see what filter to use. If I would make an educated guess, I would say that $\{f(X \setminus K) \mid K \subseteq X \mathrm{\ compact}\}$ is a filter that converges to $\infty$.
Can anyone fill in the gaps?
By construction $X$ is a dense open subset of $X^\ast$.
$(1) \Rightarrow (2)$ $f^\ast$ is defined as in your attempt. Since $X$ is open in $X^\ast$, $f^\ast$ is continuous in all points of $X$. To show that it is also continuous in $\infty$, let $V$ be a neighborhood of $y$ in $X^\ast$. Since $\text{stack}\lbrace f(X \backslash K) \mid K \subset X \text{ compact} \rbrace \to y$, we know that $V \in \text{stack}\lbrace f(X \backslash K) \mid K \subset X \text{ compact} \rbrace$. Hence there exists a compact $K$ such that $f(X \backslash K) \subset V$. But $X^\ast \backslash K$ is an open neighborhood of $\infty$ since $X \backslash (X^\ast \backslash K) = K$ is compact, and clearly $f^\ast(X^\ast \backslash K) \subset V$. Uniqueness of $f^\ast$ is due to the well-known fact that the set of points where two continuous maps $u,v : Z \to Y$ agree is closed in $Z$ provided $Y$ is Hausdorff (recall that $X$ is dense in $X^\ast$).
$(2) \Rightarrow (1)$ $\mathcal{F} = \lbrace X^\ast \backslash K \mid K \subset X \text{ compact} \rbrace$ is a filter-base in $X^\ast$ which converges to $\infty$. By continuity of $f^\ast$, $f^\ast(\mathcal{F})$ converges to $y$. But $f^\ast(X^\ast \backslash K) = \lbrace y \rbrace \cup f(X \backslash K)$ so that $\text{stack}\lbrace f(X \backslash K) \mid K \subset X \text{ compact} \rbrace \to y$.
You see that you came very close to this.