In my Topology course we defined the one point compactification of a Hausdorff space $\left(X,\tau\right) $ to be a compact Hausdorff space $\left(Y,\tau^{'}\right)
$ such that $X\subseteq Y$, $\tau\subseteq\tau^{'}$ and $\left|Y\backslash X\right|=1
$.
More specifically one such construction is given by taking $Y=X\cup\left\{ \infty\right\}$ and defining: $$\tau^{'}=\tau\cup\left\{ U\subseteq Y\;|\;\infty\in U\:\wedge\; Y\backslash U\;\mbox{is compact}\right\}$$
That is all nice and well but it gives me no clue whatsoever as to how to find a "nice representation" to the one point compactification of a given space. In particular I need to describe an embedding in $\mathbb{R}^{n}$ of the one point compactifications of each of the following:
- $\left[0,1\right]$
- $\left(0,1\right)$
- $\mathbb{N}$
Help would be greatly appreciated.
EDIT: I just noticed there was another section to the question about the compactification of $X=\left(0,1\right)\times\left\{ 0,1\right\}$. Since this is all in the subspace topology of $\mathbb{R}^{2}$ which is metric I find it more convenient to use the "heuristic" of trying to answer the questions "what are the sequences in $X$ that don't converge in $X$ and how to add one point to the space so all those sequences would converge to it". In the case of $\left(0,1\right)$ the answer was folding the line segement into a circle. In this case since I essentially have two parallel "line segments" in $\mathbb{R}^{2}$ it would seem the nicest way to achieve what I want would be to fold them both into an $8$ shape but I can't really see what sort of mapping would do that for me…
EDIT 2: I tried tackling the problem in another way, instead of trying to fold $\left(0,1\right)\times\left\{ 0\right\}$ upwards and $\left(0,1\right)\times\left\{ 1\right\}$ downwards to form two ellipses with one point in common I decided to copy $\left(0,1\right)\times\left\{ 0\right\}$ and $\left(0,1\right)\times\left\{ 1\right\}$ into two circles with one point $\left(0,0\right)$ in common, I did this using the following mapping: $$f\left(x,y\right)=\begin{cases}
\left(\sin\left(2\pi x\right),\cos\left(2\pi x\right)-1\right) & \left(x,y\right)\in\left(0,1\right)\times\left\{ 0\right\} \\
\left(\sin\left(2\pi x\right),1-\cos\left(2\pi x\right)\right) & \left(x,y\right)\in\left(0,1\right)\times\left\{ 1\right\}
\end{cases}$$
If I'm not mistaken this should be a homeomorphism between $X$ and the union of two circles of radius 1, one centered at $\left(0,-1\right)$ and one at $\left(0,1\right)$ minus the point $\left(0,0\right)$. Then the one point compactification of said union would be obtained by adding the point $\left(0,0\right)$ (the union of the circles is closed as the union of two closed sets and is also bounded and thus compact by Heine-Borel theorem). This compactification would be in turn homeomorphic to the compactification of $X$.
Could someone confirm if this train of thought indeed arrives at its intended destination?
Best Answer
HINTS large and small:
This one is easy: $[0,1]$ is already compact, so $\{\infty\}$ is an open set, and $\infty$ is an isolated point. Just take the natural copy of $[0,1]$ in $\Bbb R^n$ and add an isolated point. (This space is not normally called a compactification of $[0,1]$, because $[0,1]$ is not a dense subset of it: the usual definition of compactification requires that the original space be a dense subset of the compactification. Thus, compact Hausdorff spaces don’t have properly larger compactifications.)
If $K$ is a compact subset of $(0,1)$, $K$ has both a smallest element $a$ and a largest element $b$, so $K\subseteq[a,b]$. Thus, every open nbhd of $\infty$ contains a set of the form $(0,a)\cup(b,1)$. That means that any sequence in $(0,1)$ that converges to $0$ in $\Bbb R$ must converge to $\infty$ in $Y=(0,1)\cup\{\infty\}$, and so must any sequence in $(0,1)$ that converges to $1$ in $\Bbb R$. What if you bent $[0,1]$ around into a circle and glued $0$ to $1$, renaming the ‘double’ point $\infty$?
Think of a convergent sequence together with its limit point.