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


*

*$X_n^+$ 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

*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! 
 A: 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.
A: (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 $n>m\geq 0.$
A: 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.
