TITLE: Connectedness property of $R^2$
QUESTION [12 upvotes]: My class teacher proposed this problem which seem very interesting.

If we remove countably many open disc from $R^2$. Is the remaining 
   space still be path connected. 

I have done the problem for the case when disc are not disjoint and find out that space need not be path connected.
But I am wondering what happen if disc are 
 disjoint.my guess is space should be path connected any help will be appreciated.

REPLY [6 votes]: This might be a counter example where I replace open balls by open rectangles.
let consider the set {$1/n$}... if $n$ is odd  thne consider the open  rectangles $(1/n,1/n+1) \times (2k, 2k+2)$ s.t $ k \in \mathbb{Z}$ ...and if $n$ is even then consider open rectangles  $ (1/n,1/n+1) \times (2k+1, 2k+3)$ s.t $ k \in \mathbb{Z}$... now my claim is that if I take out all such rectangles from $\mathbb{R^2}$ , then the remaining space cannot be be path connectedto prove this we can use some similar kind of argement which we usually do while proving that topological sine curve is not path-connected...