The "point at infinity" in Alexandroff compactification I am confused that from where the extra element $\omega$, "the point at infinity" come from in Alexandroff compactification.
For example, in the stereographic projection of $\Bbb R$ onto $S^1$, we take this $\omega$ as the set $\{\infty, -\infty\}$. But I think of this as an educated guess for the $\omega$ which works out to be correct.
But in the proof of this compactification, we assume from the start that $\omega$ exists. But I find this confusing that how can we know beforehand that such a point will exist for all the locally compact Hausdorff spaces? I feel that we need to prove it but I couldn't find this explicitly mentioned in any text that I searched.
My professor gave me the argument that we can take $\omega$ as some point from the space itself if needed and then compactify the complement of $\omega$ but I find many difficulties with this approach such as:
the original space $X$ may be finite in which case $X$ cannot be homeomorphic to $X\backslash\omega$
$X\backslash \omega$ may not be locally compact Hausdorff subspace of $X$.
I am sorry if I am being vague but feel free to suggest edits to make the question better.
 A: We are given a set $X$ with some topology. We're working in a universe of sets, so there are always sets $A$ that are not an element of $X$ (there is no universal set, by Russell's paradox, or $\mathscr{P}(X)$ is a strictly larger set than $X$, etc. There are several argument depending on your axioms, e.g. if Regularity holds, then $A=X$ itself will work, as $X \notin X$ in that case). However we get it, we have some point (or set, as everything is a set) $\infty \notin X$. 
We then proceed to define a topology on the larger set $X \cup \{\infty\}$ etc. 
It's somewhat similar to working inside $\Bbb R$ and declaring the existence of some abstract number $i$ that satisfies $i^2=-1$ (and so cannot be a member of $\Bbb R$) and constructing the complex numbers from that. In that case we can also find another "model" for it, by using the product $\Bbb R^2$ and seeing $i$ as $(0,1)$ etc. In topology we can also often find a model for the Alexandroff compactification of $X$ inside some other space we already know (like $\Bbb S^2 \subseteq \Bbb R^3$ for $\Bbb R^2$ and $\Bbb S^1$ for $\Bbb R$, or a "figure 8" in the plane for $(0,1) \cup (1,2)$ etc. We can then show that this concrete space is homeomorphic to the abstract construction $X \cup \{\infty\}$ we start with. But we don't need that model to define the Alexandroff compactification, it's a helpful intuition, sometimes.
