Question on compactification I was studying for quals and had trouble with this question. Any help would be great, thanks.
A two-point compactifcation of a Hausdorff space $X$ is a compact Hausdorff space
$Y$ such that $X$ is a dense subspace of $Y$, and $Y \setminus X$ consists of exactly two points.
Prove that no two-point compactification of the Euclidean plane $\mathbb{R}^2$
exists.
 A: This is a cool question. Hausdorff rules out a Riemann sphere with "two north poles," analogous to the real line with two origins. I think you want to use the fact that the complement of a closed disk in the plane is homeomorphic to a punctured disk.
A: (After some consideration, I notice that the below just fleshes out ideas in Ted Shifrin's answer above/below.)
Suppose that $Y = \mathbb{R}^2 \cup \{ \infty_1 , \infty_2 \}$ is a two-point compactification of $\mathbb{R}^2$.  Pick disjoint open neighbourhoods $U_1$ and $U_2$ for $\infty_1$, $\infty_2$, respectively.  By compactness of $Y$ it follows that $Y \setminus ( U_1 \cup U_2 )$ is compact, and hence must be a closed and bounded subset of $\mathbb{R}^2$, say $$Y \setminus ( U_1 \cup U_2 ) =\mathbb{R}^2 \setminus ( U_1 \cup U_2 ) \subseteq \{ ( x,y) \in \mathbb{R}^2 : \| (x,y) \| \leq M \}$$ for some $M > 0$.  Note that $\{ ( x,y) \in \mathbb{R}^2 : \| (x,y) \| > M \}$ is a connected subset of $\mathbb{R}^2$, however $U_1 \setminus \{ \infty_1 \}$ and $U_2 \setminus \{ \infty_2 \}$ are disjoint nonempty open subsets of $\mathbb{R}^2$ which cover it!
