How can I show that $\mathbb{R}^n \cup \{\infty\}$ is a Hausdorff topological space?

Question

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?

Answer 1

In metric spaces, a set say $U$ is called open if $∀x∈U,∃ε>0$ such that the disc $D(x,\epsilon)\subseteq U$. Also in a metric space, if we choose any two arbitary points, say $x,y$, ($x≠ y$) we have that $d(x,y)>0$. Its easy to see that every metric space is a Hausdorff (or $T_2$) topological space. You only have to choose two points, $x,y:x≠y$ with $d(x,y)=\epsilon$ where $\epsilon>0$ and then the discs $D(x,\epsilon/2),B(y,\epsilon/2)$ are open and $D(x,ε/2)∩ B(y,ε/2)=\emptyset$. We know that open discs are also open sets themselves in a metric topological space.

Answer 2

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 $B_r(x)$.