One-point compactification for $(1,\infty)$? Let $f:(0,1) \rightarrow \mathbb{R}$ be given by the function $f(x) = 1/x$. I would like to construct a one-point compactification of the range of this function such that I can convert $(1,\infty) \mapsto S^1$. I know that I can sort of union a point at infinity to $(1,\infty)$ and  wrap the whole graph of the function up into a circle and indicate the infinite point as the point we added to the sphere. However, how should I do this properly? Should I build up a homeomorphism like the stereographic projection that compactify $\mathbb{R}^2$?
Thanks!
 A: Add a point $\infty$ with basic open neighbourhoods $\{\infty\}\cup(1,a)\cup (b,\infty)$, $1<a<b$.
A: $(1,\infty)$ is homeomorphic to $\Bbb R$, and one-point compactifications of homeomorphic spaces are homeomorphic, $X$ and $Y$ are homeomorphic. Show that also their one-point compactifications are homeomorphic. .
But, one-point compactification of $\Bbb R$ is $\Bbb S^1$, what is the one point compactification of R?
A: A more detailled approach not using much external results (and thus a little bit cumbersome) goes as follows:
Your intuition is right, you need to add a point $p$ at infinity. But then you only have the set $X := (1,\infty)  \cup \{p\}$ , but what you want is not a set but a compact topological space.
Hence you need to define a topology on $X$. The topology  $\mathcal T$ we define is the one which formalizes your intuition of "wrap the whole graph of the function up into a circle and indicate the infinite point as the point we added to the sphere".
Hence we take the standard topology on $(1,\infty)$, i.e. the subset topology of $\mathbb R$ and expand it by sets containing $p$ which we want to be open. At this step you have to think a little, but intuitvely it should be clear which sets we chose.
Now we obtained a topological space $(X, \mathcal T)$ and it remains to show that it homeomorphic to $S^1$ (and compact). Thus we do the straight forward thing: We define a map $g:S^1 \to X$ and show that it is a homeomorphism. For obtianing $g$ we can extend $f$ to $[0,1]$ by $f(0)=f(1)=p$, which descends to the quotient $S^1 = [0,1] / \{0\sim 1\}$.

So did we really get a compaticifaction of $(1,\infty)$?
A standard definition of a compactification of a space $Y$ is an embedding $i:Y\to Z$ in a compact space $Z$ such that the image $i(Y)$ is dense in $Z$, i.e its closure is $\overline{ i(Y)} = Z$.
In our case, $Y = (1,\infty)$, $Z = (1,\infty) \cup \{p\} = X$ with the respective topologies. $i$ is the standard inclusion, i.e. the identity on $(1,\infty)$.
We need to show three things:

*

*The map $i$ is an embedding

*Its image $i(Y)$ is dense in $Z$.

*The space $Z$ is compact.

The first is clear, $i$ is injective, continuous and a homeomorphism on its image (its the identity).
For the second part, observe that $i(Y)$ is not closed (as $\{p\}$ is not open by definition of $\mathcal T$), so its closure has to be $X$.
So the only not obvious part is the third one, showing that $X$ is compact, which I did in a previous part of the answer.
Also note that in order to satisfy properties 1 and 2, it suffices to unifiy $Y$ with a point $p$ with the set $\{p\}$ not open. So the hard part when constructing a compactification is 3. (Which is why often only 3. is proved explicitely).
