# Question on one point compactification

I was given the following question in my general topology class assignment which is multi parts - most of which I managed alright by myself some of which I need help on. We are given a non compact topological space $(X,\tau)$. Then we expand $X$ by a point not in $X$, call this point $\infty$ and define $X^*=X\cup\{\infty\}$ with the following topology $$\tau^* =\tau\cup\{V\subset X^*|\infty \in V \text{ and }X\setminus V\text{ is closed and compact}\}.$$

I am given this line before the questions itself: "Denote the embedding map by $i:X\to X^*$". I do not understand this but assume they mean the inclusion map, i.e., sending every member of $X$ to itself in $X^*$. Do you think that's what is meant by embedding map? I know any map that is a homeomorphism is an embedding by definition. I think this is a clue meaning the inclusion map.

Here is the question itself in all parts I am asked to prove:

1. Show that $(X^*,\tau^* )$ is a topological space. (DONE)
2. Show that $i:X\to im(X)$ is a homeomorphism with $im(X)$ endowed with the subspace topology. (DONE, assuming $i$ is indeed the inclusion map.)
3. Show that $im(X)$ is open and dense in $X^*$ and its complement is a singleton . (STUCK HERE)
4. Show that $(X^*,\tau^*)$ is compact. (DONE)
5. Assume that $(X^*,\tau^*)$ is Hausdorff. Prove the following: for any Hausdorff topological space $(X',\tau')$ and a mapping $i':X\to X'$ such that properties 2 3 4 above hold, there exists a homeomorphism $f:X*\to X'$ such that $f\circ i=i'$. (REALLY STUCK)

The moral of this as they say is that $i$ and $X^*$ are uniquely determined up to homeomorphism.

1. Show that the one-point compactification of $\mathbb{R}^n$ is homeomorphic to $\mathbb{S}^n$ by looking at the stereographic projection $s:\mathbb{S}^n\setminus (1,0,...,0)\to\mathbb{R}^n$, $$s(x_0,x_1,...,x_n) = (x_1/(1-x_0),...,x_n/(1-x_0)).$$

We are asked to prove that its inverse is an embedding of $\mathbb{R}^n$ in $\mathbb{S}^n$ and use part 5 after verifying its conditions. (REALLY REALLY STUCK)

I realize this might be a very long question but I am hoping someone could help me on this very important question on parts I couldn't do and to tell me if $i$ is indeed the inclusion map.

Thank you all very kindly

• Yes, "the" embedding here is the inclusion map – Hagen von Eitzen May 26 '15 at 20:01
• Thanks @HagenvonEitzen for that – kroner May 26 '15 at 20:01

1. Since $X\in\tau\subseteq\tau^*$, $X$ is open in $X^*$. For density, the only point in $X^*\setminus X$ is $\infty$. Let $V$ be an open neighbourhood (in $X^*$) of $\infty$. Then $X\setminus V$ is closed and compact in $X$, but since we know that $X$ is non compact, then $X\setminus V\neq X$, so that $V\cap X\neq\varnothing$. This proves that $X$ is dense in $X^*$. The complement of $i(X)=X$ is the singleton $\left\{\infty\right\}$/
1. Note that $X^*$ is made up of two different parts: $X$ and the point $\infty$. We have to identify these parts in $X'$. It should be clear that $i'(X)$ acts as $X$, and that the remaining point in $X'$ acts as $\infty$. More precisely, let $\left\{\infty'\right\}=X'\setminus i'(X)$ (by property 3, the complement of $i'(X)$ is a singleton). Define $f:X^*\to X'$ by setting $f(x)=i'(x)$ for $x\in X$ and $f(\infty)=\infty'$.
The main problem here is to check that $f$ is continuous at $\infty$. Let $V'$ be an open neighbourhood of $\infty'$ in $X'$. Let's show that $i'(X)\setminus V'$ is closed and compact in $i'(X)$. Closedness is immediate. For compactness, let $\mathcal{U}$ be an open cover of $i'(X)\setminus V$ (in the relative topology of $i'(X)$. Then $\mathcal{U}\cup\left\{V'\right\}$ is an open cover of $X'$ (in $\tau'$), so there exists a finite subcover $\mathcal{U}'\subseteq \mathcal{U}$ such that $\mathcal{U}'\cup\left\{V'\right\}$ covers $X'$, so $\mathcal{U}'$ covers $i'(X)\setminus V$.
Since $i'$ is a homeomorphism, $X\setminus(i')^{-1}(V')$ is closed and compact. Let $V=((i')^{-1}(V'))\cup\left\{\infty\right\}$. Then $V\in \tau^*$ is an open neighbourhood of $\infty$ in $X^*$ and $f(V)\subseteq V'$.
1. The inverse of the stereographic projection is the embedding of $\mathbb{R}^n$ in $\mathbb{S}^n$. The inverse of the stereographic projection (which is already continuous) is $$i'(x_1,\ldots,x_n)=\left(\frac{-1+x_1^2+\cdots+x_n^2}{1+x_1^2+\cdots+x_n^2},\frac{2x_1}{1+x_1^2+\cdots+x_n^2},\ldots,\frac{2x_n}{1+x_1^2+\cdots+x_n^2}\right)$$ so it is also continuous. Thus, $i'$ is a homeomorphism of $\mathbb{R}^n$ onto its image, $\mathbb{S}^n\setminus(1,0,\ldots,0)$, which is open and dense in $\mathbb{S}^n$, and whose complement is the singleton $(1,0,\ldots,0)$. Moreover, $\mathbb{S}^n$ is compact. By 5., there exists an homeomorphism between $\mathbb{S}^n$ and $(\mathbb{R}^n)^*$.