How to describe the one point compactification of a space 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?
 A: 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.
A: Intuitively, (and let me stress intuitively) you want to add a single new point that so that all sequences that don't converge to anything have somewhere to go, at least on a subsequence. That might help you figure out what space to try. Of course, you then have to prove your guess is right! You might want to note that the one point compactification is unique. So if you can find any compact space that does the trick, you're done.
1) Is already compact and Hausdorf, so finding its one-point compactification is straightforward. (What would be the first thing you would try?)
2) Now, for this space I can build sequences that march off to the "ends" of the interval but have nowhere to converge to. If I add a single point and let the ends of the interval go towards that single point, what space have I made?
3) For this one, sequences can go off "to infinity". So I add a point $\infty$ and I want all sequences that used to go "to infinity" to converge to this point. What should the topology on this new space be? Is it actually compact? 
A: The first thing you have to do to identiify the one-point compactification is to uidentify the compact subsets of the space in question.


*

*$[ 0 , 1 ]$ is a compact (Hausdorff) space, and so we know that the compact sets are simply the closed subsets of $[0,1]$.  Therefore if $[ 0 , 1 ] \cup \{ * \}$ is the one-point compactification of $[0,1]$, then the open neighbourhoods of $*$ look like $\{ * \} \cup U$ where $U$ is any open subset of $[0,1]$.  In particular, as $\varnothing$ is a open subset of $[0,1]$ it follows that $\{ * \} \cup \varnothing = \{ * \}$ is an open nieghbourhood og $*$ in $[0,1] \cup \{ * \}$; i.e., $*$ is an isolated point in $[ 0,1 ] \cup \{ * \}$.

*$(0,1)$ is actually homeomorphic to the real line $\mathbb{R}$, and so it doesn't hurt to instead consider the one-point compactification of $\mathbb{R}$.  By the Heine-Borel Theorem the compact subsets of $\mathbb{R}$ are the bounded closed sets.  In particular, every compact subset of $\mathbb{R}$ is a subset of a compact set $[a,b]$ for $a < b$.  It follows that the sets of the form $( b , + \infty ) \cup \{ * \} \cup ( - \infty , b )$ (where $a < b$) form a basis for the open neighbourhoods of $*$ in the one-point compactification.

*$\mathbb{N}$ is a discrete space, and so the compact subsets are the finite subsets.  In particular, every compact set is a subset of a set of the form $[n] = \{ 0 , \ldots , n \}$ where $n \in \mathbb{N}$.  It follows that a basis of the open neighbourhoods of $*$ in the one-point compactification consists of all sets of the form $\{ n+1 , n+2 , \ldots \} \cup \{ * \}$ where $n \in \mathbb{N}$.
