I want to know if $\mathbb{R}^n \cup \{\infty\}$ is a Hausdorff topological space. Here: Show that $\mathbb{R}$ is Hausdorff. it is shown that the real line is hausdorff. Here: https://webspace.maths.qmul.ac.uk/b.jackson/topchapter3.pdf it is shown that $\mathbb{R}^n$ when treated as a metric space is Hausdorff. Via web browsing, I can't find much on if if it's true that
the one point compactifications of these spaces are Hausdorff. I want to know this for a separate proof. Do we have to define a $\mathbb{R}^n \cup \{\infty\}$ as a metric space in order for it to be Hausdorff? Can't it be Hausdorff as purely a topological space?
Let's suppose we have the topological space $(\mathbb{R}^n \cup \{\infty\}, \mathcal{T})$ I'm not sure if $\mathcal{T}$ has to be the way I'm about to define it, but for some reason I assume for it to be true. Let's let $\mathcal{T} = \{U\subseteq \mathbb{R}^n: \forall u \in U, \exists r > 0 : B_r(u) \subseteq U \} \cup\{U\subseteq \mathbb{R}^n \cup \{\infty\}: U = (\mathbb{R}^n \backslash C) \cup \{\infty\}\}$ where $C$ represents a compact subset of $\mathbb{R}^n$. In more general terms, I read $\mathcal{T}$ as "the standard set of all open subsets along with all possible open subsets containing the point at infinity." I also used balls of radius r inside the definition of $\mathcal{T}$. I find it weird how we use these balls with distance to define topology, yet topological spaces are independent of distance, so I would like some help clarifying the way I should be thinking about things.
This is my best attempt at the proof, pretty much copying it from the second link:
suppose we have some two elements $x, y \in \mathbb{R}^n$ where $x \neq y$
I don't know how to prove this without distance, so I'll guess i'll let the distance between two elements $a, b \in \mathbb{R}^n = d(a, b) = \sqrt{(b_1 – a_1)^2 +(b_2 – a_2)^2+…+(b_n – a_n)^2}$
let $\delta = d(x, y)$
let $U = B_x(\delta/2)$ and let $V = B_y(\delta/2)$
Let's suppose $U\cap V \neq \emptyset $ is true. If this gives a contradiction, then we know that $U\cap V = \emptyset $ is true.
If $U\cap V \neq \emptyset $ is true, $\exists z \in U\cap V$
$\implies$ $z \in U$ and $z \in V$
$\implies$ $z \in B_x(\delta/2)$ and $z \in B_y(\delta/2)$
$\implies$ $d(x,z) < \delta/2$ and $d(y,z) < \delta/2$
$\implies$ $d(x,z) + d(y, z) < \delta/2 + \delta/2$
$\implies$ $d(x,z) + d(y, z) < \delta$
from the triangle inequality:
$\implies$ $\delta = d(x, y) \leq d(x, z) + d(y, z) < \delta$
Therefore, it's possibly true that $\delta < \delta$ and this is a contradiction.
Therefore, $\mathbb{R}^n$ is a Hausdorff
Now, I assume all that's left to do is show that distinct open sets can be formed around $\{\infty\}$ an arbitrary element of $\mathbb{R}^n$
suppose we have $\{\infty\}$ and some element $x \in \mathbb{R}^n$ do we prove that there are always distinct open sets around these points the same way as we did as before?
Best Answer
In fact your question only makes sense if you define a topology on $\mathbb R^n \cup \{\infty\}$. There are many ways to topologize this set, but the standard interpretation is to regard it as the one-point compactification of $\mathbb R^n$. This construction works for all locally compact Hausdorff spaces and gives a compact Hausdorff space. Let us do it in your example.
As open sets of $\mathbb R^n \cup \{\infty\}$ we take all open subsets of $\mathbb R^n$ and all complements $(\mathbb R^n \cup \{\infty\}) \setminus C$ of compact subsets $C \subset \mathbb R^n$. This is what you have done in your question.
$\mathbb R^n$ is an open subset of $\mathbb R^n \cup \{\infty\}$ whose subspace topology inherited from $\mathbb R^n \cup \{\infty\}$ is its original topology. Thus any two distinct points $x, y \in \mathbb R^n \subset \mathbb R^n \cup \{\infty\}$ have disjoint open neigborhoods because $\mathbb R^n$ is an open Hausdorff subspace. It remains to show that each $x \in \mathbb R^n$ and $\infty$ have disjoint open neigborhoods. Take any open ball $B_r(x)$. This is an open neigborhood of $x$ in $\mathbb R^n \cup \{\infty\}$. The closure $C$ of $B_r(x)$ is a compact subset of $\mathbb R^n$, thus $(\mathbb R^n \cup \{\infty\}) \setminus C$ is an open neighborhood of $\infty$ which is disjpint from $B_r(x)$.