Show that the plane with countably many points removed is path-connected under the Euclidean topology. Show that the plane with countably many points removed is path-connected under the Euclidean topology.
This is quite an interesting issue, hopefully everyone will help you
 A: Consider $\Bbb R^2\setminus A$ with $A$ a countable set and take two points $(x_1,y_1),(x_2,y_2)\in\Bbb R^2\setminus A$.
For $(x_1,y_1)$ consider the set of all possible straight lines passing through that point: $\{r: \text{$r$ is a straight line with equation }y-y_1=m(x-x_1)\text{ for some $m\in\Bbb R$}\}$ (actually we are not considering the vertical line $x=x_1$, but it won't matter).
You can see that there is a distinct element in that set for each $m\in\Bbb R$, so it is uncountable. Since every straight line in the set is disjoint from all the others except at the point $(x_1,y_1)$ and there is an uncountable number of them, there has to be some straight line in the set such that it doesn't pass through any point in $A$. Let's say it has slope $m$.
Now consider the set $\{s: \text{$s$ is a straight line with equation }y-y_2=k(x-x_2)\text{ for some $k\in\Bbb R\setminus\{m\}$}\}$. Those are the straight lines passing through $(x_2,y_2)$ excluding the vertical one, and the one with slope $m$. As before there has to be one that doesn't pass through any point in $A$, say with slope $k$.
Since $k\neq m$ then the two straight lines will cut in the plane, so you can construct a path between them using those two lines.
A: Let $X$ be $\mathbb{R}^2$ with countably many points deleted. Let $x,y \in X$. Let us consider all the lines which contain $x$ in $\mathbb{R}^2$: this is an uncountable set, so there exists a line $r$ whose points are not removed; now let us consider the same argument for $y$, and let us consider a line $s$ which contains $y$ and intersects $r$ in a point $z$. Finally, you can connect $x$ and $y$ going along $r$ from $x$ to $z$, and then along $s$ from $z$ to $y$.
