Can a neighborhood of $\mathbb{R}^n$ be countable? If so, how to prove it? Ok, so I'm having a little trouble with a bigger proof. A smaller piece of it is if a neighborhood of $\mathbb{R}^n$, $N(x,r,\mathbb{R}^n)$, is countable and, if so, how to prove it.
I'm trying to show that a subset of $\mathbb{R}^n$ is countable knowing that the intersection of the subset and a neighborhood of $\mathbb{R}^n$ is countable.
Any help would at all would be very appreciated.
 A: Every non-empty open set in $\Bbb R^n$ has cardinality $2^\omega=\mathfrak c$, the cardinality of the real line. This follows immediately from the fact that if $\varnothing\ne U\subseteq\Bbb R^n$ is open, then there are open intervals $(a_k,b_k)\subseteq\Bbb R$ for $k=1,\dots,n$ such that $$\prod_{k=1}^n(a_k,b_k)\subseteq U\;.$$ To see this, fix $c_k\in(a_k,b_k)$ for $k=1,\dots,n-1$, and observe that 
$$\{c_1\}\times\{c_2\}\times\ldots\times\{c_{n-1}\}\times(a_n,b_n)\subseteq \prod_{k=1}^n(a_k,b_k)\subseteq U\;,$$
and there is an obvious bijection between $\{c_1\}\times\{c_2\}\times\ldots\times\{c_{n-1}\}\times(a_n,b_n)$ and $(a_n,b_n)$, which is well-known to have the same cardinality as $\Bbb R$. 
In short, every non-empty open set in $\Bbb R^n$ contains an open line segment.
A: I assume $n\ge 1$. Any neighborhood of $x$ will contain an open ball centered at $x$. It is not hard to show this open ball is uncountable. Assume without loss of generality that the point is $0$. We can then consider the uncountable subset $(w,0,\dots)$ with $0<w<r$, where $r$ is the radius of the ball. This is clearly uncountable, so the ball is too. 
