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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!

Edi
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3 Answers3

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Add a point $\infty$ with basic open neighbourhoods $\{\infty\}\cup(1,a)\cup (b,\infty)$, $1<a<b$.

Hagen von Eitzen
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$(1,\infty)$ 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?

Sumanta
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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:

  1. The map $i$ is an embedding
  2. Its image $i(Y)$ is dense in $Z$.
  3. 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).

klirk
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  • Say if we have defined $X$ as our new compact set, do we simply show that $S^1$ is homeomorphic to $X$ and as $S^1$ is compact, $X$ is also compact? Also, how should I properly derive a set IS the compactification of my set? – Edi Sep 22 '20 at 10:48
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    This would be one way to show that $X$ is compact. In fact, for this you only need that $g$ is continuous. Of course, you could also work directly with the topology on $X$, assume that there is a cover of $X$ which has no finite subcover and then derive a contradiction. Regarding properly showing that this is a compactification, I edited my answer – klirk Sep 22 '20 at 11:26
  • Appreciated, really detailed answer :) – Edi Sep 22 '20 at 11:28
  • One more question, Hagen mentioned that the basic open neighborhoods are ${∞}∪(1,)∪(,∞), 1<<$, is this because any open sets containing ${\infty}$ must be of the form $K^c \cup {\infty}$ for some compact $K$ (which is closed interval in our case)? How about the open sets that do not contain ${\infty}$? – Edi Sep 22 '20 at 12:37
  • And for the homeomorphism mapping to $S^1$, are we including 0 and 1 in our domain so that we have the point "$1/0 = \infty$" which is the point we want as our compactification? – Edi Sep 22 '20 at 12:53
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    The open sets which do not contain $\infty$ (or $p$ as i called it) are just the open sets in the subset topology on $(1, \infty)$, i.e. the intersection of open sets in $\mathbb R$ with $(1,\infty)$. What you say about the form of the open sets is true, but in this simple example, when we don't really need it because we already have an intuition. As you said in your question: We want to glue both ends of the interval $(1,\infty)$ to $p$. Chosing the open sets containing $p$ to be of this form ensures that $p$ is arbitrarily close (in the sense of continuity) at $1$ and $\infty$. – klirk Sep 22 '20 at 14:52
  • So if we define a map $g:X \rightarrow S^1$ by $x \mapsto (cos(\frac{2\pi}{p}x),sin(\frac{2\pi}{p}x))$ will we get the compactification we want...? Or we are not supposed to treat $p$ like a number (well, it is $\infty$)? So sorry for keep asking stupid questions but this really confuses me when we are not using $(0,1)$ as our target for the compactification... – Edi Sep 22 '20 at 17:57
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    @Edi: Often, $p$ is treated as if there was a number $\infty$, but in my opinion this disguises whats really going on. In the end it makes sense, but you should be aware why this is the case. – klirk Sep 22 '20 at 18:58
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    I already (but not detailly) wrote down the correct map $g$ in the answer (by treating $S^1$ as a quotient of the interval). Of course, we can also think of $S^1$ as a subset of $\mathbb R^2$ and work with angles as you do, but there are some pitfalls: $X=(1,\infty) \cup {p}$ so in order to define a map $g: X\to S^1$, we need to define $g$ on $(1,\infty)$ and on $p$, in a continuous way. So, $g|{(1,\infty)}$ has to be cont, a good choice would be $x \mapsto (\cos(\frac{ 2 \pi} x), \sin(\frac{ 2 \pi} x)) $. Additionally, you need to define $g|{{p}}$, (by $p \mapsto (1,0)$). – klirk Sep 22 '20 at 19:02
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    In order to check continuity, you could then look at arbitrary sequences in $X$ converging to $p$ (for this you need the introduced topology on $x$, i.e. the neighbourhoods of $p$) and check if their images converge to $g(p)$. – klirk Sep 22 '20 at 19:07