One point compactification of uncountable set with discrete topology cannot be imbedded into $\mathbb{R}^{2}$? Let $X$ be an uncountable set with the discrete topology. Prove that the one point compactification $Y$ of $X$ cannot be imbedded into the plane $\mathbb{R}^{2}$.
I'm assuming that there exists an imbedding $f: Y \rightarrow f(Y)$ of $Y$ into $\mathbb{R}^{2}$ and trying to get a contradiction. If $f$ exists, it is a continuous, bijective open/closed map. My thought was that somehow I could obtain a contradiction using that $f$ has to be an open map. We know that the open sets of $Y$ are all of the open sets of $X$ (so every subset of $X$ as it has the discrete topology) and all sets of the form $Y-C$ where $C$ is a compact subspace of $X$; that is, all sets of the form $Y-C$ where $C$ is a finite subset of $X$ since a subset of $X$ is compact iff it is finite. I am stuck. 
 A: You can look at cardinalities of the sets, in the first topological space you have the cardinality of the power set of the Real numbers, in the second you can use the fact that the real numbers have a countable base therefore there can be at most the cardinality of the power set of N of open sets.
A: $X$ (and therefore also $Y$) has an uncountable family of disjoint open sets.  No subspace of $\mathbb R^2$ does, because $\mathbb R^2$ is second countable.
A: Suppose you can embed $Y$ in $R^2$ by $f$, then $f(Y)$ is a compact, it is thus closed and bounded and contained in a ball $B(0,r)$. Consider $f(X)-f(\infty)$ where $Y=X\bigcup\{\infty\}$, it is a discrete subset of $B(0,r)$ with only one limit point $f(\infty)$. Let $n>0$ and integer, the intersection of  the complementary of $B(f(\infty),1/n)$ and $Y$ is thus finite. This implies that $Y$ is countable. Contradiction.  
A: The space $Y$ can not be embedded in any metric space, because it is not first countable; the one-point set $Y\setminus X$ is not a $G_\delta.$
