# 1-point compactification and embedding

I'm totally stuck with the following question, I even don't know how to start:

For each natural number $n$ we consider a space $X_n$ that is obtained by removing $n$ distinct points from $\mathbb{R}^2$. We consider the 1-point compactification $X_n^+$ and we denote by $\infty_n\in X_n^+$ the point at infinity (so that $X_n^+=X_n\cup \{\infty_n\}$). Show that

1. $X_n^+$ can be embedded in $\mathbb{R}^3$. I do not have to write down explicit formulas for the embedding, but I have to explain my reasoning using pictures and mention what result(s) I use in order to reach my final conclusion
2. If $X_n$ and $X_m$ are homeomorphic, then $n=m$.

Both questions I find very hard to answer. I really hope someone can help me!

• Start by figuring out what the compactifications for $n=0$ and $n=1$ look like. – user98602 Jan 9 '16 at 17:29
• For 1: Maybe a good way of starting is to think of $X_n$ as $S^2$ minus $n+1$ points? – peter a g Jan 9 '16 at 17:29
• Start by determining the compact subspaces of $X_n$. These are exactly the complements of open neighborhoods of $\infty_n$. Knowing what those are will help you imagine the compactification. – Alex G. Jan 9 '16 at 17:41
• Thanks for your hints, I think the answer helps me even more :) I still find the second question very hard to answer, maybe you could help? Thanks! – jbuser430 Jan 10 '16 at 10:56

HINT: Start by showing that $X_0^+$ is homeomorphic to $S^2$. Then convince yourself that $X_1$ is homeomorphic to the cylinder $S^1\times\Bbb R$. If you added two points at infinity instead of one, one at each end of the cylinder, you would compactify it into $S^2$ by in essence closing off the ends; if you now identify the two new points, you are in effect starting with $S^2$ and identifying the north and south poles. The result is a torus whose inner radius has been shrunk to $0$, so that instead of a hole in the centre it has a single point. Even though we did this in two steps, first adding two points and then identifying them, it turns out that we really have got $X_1^+$.

To extend the basic idea, try to convince yourself that $X_n$ is homeomorphic to the space obtained by removing $n+1$ points from $S^2$; $X_n^+$ is then obtained by filling all $n+1$ holes with a single point.

• Thank you so much for this hint, I think I am going to understand it! Though I still got one question.. Is this enough to say: thus, $X_n^+$ can be embedded in $\mathbb{R}^3$? Because for me, this is not a logical/trivial next step in the proof. Can you explain why this is helpful to prove the statement? – jbuser430 Jan 10 '16 at 10:35
• In my book I found a corollary: If $f:X\rightarrow Y$ is a continuous injection of a compact space into a Hausdorff one, then $f$ is an embedding. Can I use this or is this not useful? – jbuser430 Jan 10 '16 at 12:09
• @JBIBB: Yes, each $X_n^+$ can be embedded in $\Bbb R^3$. Whether you want to mention the result in your second comment probably depends on how much detail you’re supposed to give. I probably wouldn’t, because I’d consider the pictures sufficient, but you might want to use it in connection with an explanation of which points of the compactification go to which points of the object in $\Bbb R^3$ — not in terms of a formula, but in terms of pictures. Since drawing three-dimensional objects is difficult for most of us, I have one suggestion for the pictures. Once you convince yourself that ... – Brian M. Scott Jan 10 '16 at 20:30
• ... $X_n$ is $S^2$ with $n+1$ points removed, imagine spacing those points uniformly around the equator. Now you can draw a planar cross-section through the equator — it will be a circle with $n+1$ points missing — and then draw its one-point compactification, a rose with $n+1$ petals. Then you can explain that the rest of $X_n^+$ consists of two distorted hemispheres, one above and one below this cross-section. – Brian M. Scott Jan 10 '16 at 20:33

(I). To embed $X_n^+$ with $n\geq 1$ in $\Bbb R^3$ first embed $\Bbb R^2$ into $\Bbb R^2$ as the surface of a sphere with one point removed. Deform it by stretching out $n+1$ tentacles that taper at the ends to all meet at a point $p$ outside the sphere.... $p$ is the "point at infinity".

(II).If $n>m\geq 0$ and $X_n$ is homeomorphic to $X_m$ then $X_n^+$ is homeomorphic to $X_m^+.$

But when $n>0$ the point $\infty_n$ has a nbhd base $B$ of open subsets of $X^+_n$ such that for every $b\in B$ the set $b \setminus \{\infty_n\}$ is the union of exactly $n+1$ pair-wise disjoint non-empty open sets. No point in $X_m^+$ has this property, so $X_n^+$ is not homeomorphic to $X_m^+$ for $n>m\geq 0.$

I'm unsure whether or not to make a new topic, but I'm trying to solve the second part of this question:

If $X_n$ and $X_m$ are homeomorphic, show that x = m.

My attempt at a solution: (basically got nowhere with this, just trying to write things down)

This is the same as showing that if $x \ne y$, then $X_n$ and $X_m$ are not Isomorphic. Suppose $x \ne y$, then without loss of generality, $n > m$, $n = m'+a$ for $m' = m$ and some $a$. Because $X_m$, $X_{m'}$ are isomorphic to $S^2$ with m points removed, $X_m$ is isomorphic to $X_{m'}$. Here $X_n = X_{m'} - p_{m+1} - ... - p_{n}$, where $p_{m+1}$ etc. denote the removed points. $X_n$ is isomorphic to $S^2$ with m points removed, and another $a = n - m$ points removed.

Anyway, I have no idea how to solve the question. Perhaps I should prove that one of the spaces $X_n$ and $X_m$ has a topological property that the other one does not have, if $x \ne y$. perhaps I shouldn't use a proof by contrapositive. I hope someone could give me some insight.

• I conjecture that the Euler chacteristics of $X^+_n$ and $X^+_m$ won't be equal if $n\ne m$. – DanielWainfleet Jan 28 '18 at 20:10
• Not something I heard of before, but I just read something about it: I came across the following lemma: The Euler characteristic of a polyhedron does not change when we delete an edge bordering distinct faces, or a vertex bordering 2 distinct edges. So removing a single point shouldn't change it then, right? – Timon Groen Jan 28 '18 at 21:44
• So my off-the-cuff conjecture is false. – DanielWainfleet Jan 29 '18 at 0:44
• I now have an answer which I added as Part II of my earlier answer to the first part of the Q. – DanielWainfleet Jan 29 '18 at 5:43
• I'm unable to comment directly to your answer, but I have trouble figuring out why $b - {∞_{n}}$ is the union of exactly $n+1$ pair-wise disjoint non-empty open sets. Suppose $b = X_n^+$, then $b - {∞_{n}} = X_n$? What would the $n+1$ pair-wise disjoint non-empty open sets in $X_n$ be? – Timon Groen Jan 29 '18 at 9:48