The plane minus a countable set homeomorphic to the plane minus an uncountable set?

Is it possible that $\Bbb R^2-C$ can be homeomorphic to $\Bbb R^2-U$ where $C$ is countably infinite and $U$ is uncountable?

Intuitively I believe the answer is no, but I'm having difficulty showing this in full generality.

I know that $\Bbb R^2-C$ is always path-connected when $C$ is countable. And I know that $\Bbb R^2-U$ consists of two path-connected components when $U$ is the graph of a continuous function from $\Bbb R$ to $\Bbb R$ or when $U$ is a simple closed curve. But these are very limited cases. I know that if $U$ is uncountable, then uncountably many points of $U$ are limit points of $U$, but that hasn't been helping much.

Any help in answering this general problem is appreciated.

• Well, $\mathbb{R}^2\setminus \{(0,0)\}$ is homeomorphic to $\mathbb{R}^2\setminus B[(0,0),1],$ isn't it? – mfl Oct 19 '14 at 17:33
• @mfl: Assuming that you mean the closed ball, yes. – Brian M. Scott Oct 19 '14 at 17:34
• @BrianM.Scott Yes, I wanted to write the closed ball. Thank you for noticing it. I have edited the comment. – mfl Oct 19 '14 at 17:36
• @mfl thanks for that example. I've altered the question to something less trivial. – Robert Wolfe Oct 19 '14 at 17:37
• Working again with the same idea. Consider $C=\mathbb{Z}\times \mathbb{Z}$ and $U=\cup_{(m,n)\in C} B[(m,n),1/3].$ We have that the desired spaces are homeomorphic. – mfl Oct 19 '14 at 17:49