Prove or disprove : Complement of $\mathbb N \times \mathbb N$ is open and connected? Consider$\mathbb R^2$ with the usual topology , The complement of $\mathbb N \times \mathbb N$ is 


*

*open but not connented .

*connected but not open.

*both open and connected

*neither open nor connected.
My attempt is :
Define $A = \cup_{n\in \mathbb N \cup \{0\}}(n,n+1)\times(n,n+1+\epsilon) $
and   $B= \cup_{n\in \mathbb N \cup \{0\}}(n ,n+1+\epsilon)\times(n,n+1) $
So complement of $\mathbb N \times \mathbb N$ is $A\cup B$ which is open 
please tell me about the connectedness.
Thank you
 A: I will steal my first line from Darrin:
Since $\mathbb{N} \times \mathbb{N}$ is closed (why?), then its complement is open. 
We need to show $\mathbb{N} \times \mathbb{N}$ includes all its limit points. Given a point in $(\mathbb{R} \times \mathbb{R}) \setminus (\mathbb{N} \times \mathbb{N})$ you can find an $\epsilon$ so that point is more than $\epsilon$ from any point in $\mathbb{N} \times \mathbb{N}$, so it is not a limit point.  Thus $\mathbb{N} \times \mathbb{N}$ includes all its limit points and is closed.  Connected seems obvious.
A: Since $\mathbb{N} \times \mathbb{N}$ is closed (why?), then its complement is open. 
Choose two points in the complement. Can you connect them with a polygonal path? How? (Hint: consider vertical and horizontal segments, and "avoid" elements in $\mathbb{N} \times \mathbb{N}$). Now recall that a set in the plane is connected if and only if it is polygonally connected. 
A: Show that it is path connected. Take two points $x$ and $y$ in $\mathbb{R}^2\setminus A$, where, $A=\mathbb{N}\times \mathbb{N}$ is a countable subset of $\mathbb{R}^2$. If the line segment joining $x$ and $y$, that is $L_1 = \{z\in \mathbb{R}^2|z=tx+(1-t)ty, t\in [0,1]\}$ in $\mathbb{R}^2$, has null intersection with $A$, we are done. If not, since $L_1$ is an uncountable subset of $\mathbb{R}^2$, therefore $L_1\setminus A$ must be non empty ($A$ is countable). Choose $u \in L_1\setminus A$. Let, $v\in \mathbb{R}^2\setminus A$, be another point not in $L_1$, such that the line $L_2$ joining $u$ and $v$ that is $L_2 = \{z\in \mathbb{R}^2|z=ux+(1-t)tv, t\in [0,1]\}$ does not coincide with $L_1$. If the path joining $x,v,y$ with line segments (call it $[x,v,y]$) has empty intersection with $A$, we are done, that is we get a path in $\mathbb{R}^2\setminus A$ joining $x$ and $y$. But on the contrary if each path $[x,v,y]$ intersects $A$, at some point, then we get an uncountable collection of points in $A$, which is not possible. So, the space $\mathbb{R}^2\setminus A$ is path connected.
A: Define the continuous functions $f(x,y) = (\sin (\pi x))^2 + (\sin ( \pi y))^2$, $g_1(x,y) = x $, $g_2(x,y) = y$ and then note that
$\mathbb{N}^2 = f^{-1} \{ 0 \} \cap g_1^{-1} [1,\infty) \cap g_2^{-1} [1,\infty) $. Since the sets $\{ 0 \}$ and $[0,\infty)$ are closed it follows that $\mathbb{N}^2$ is closed.
Now pick two points $(x_1,y_1), (x_2,y_2) \in(\mathbb{N}^2)^c$ and consider the uncountable collection of paths $\phi_\epsilon:[0,1] \to \mathbb{R}^2$ defined by 
$\phi_\epsilon(t) = (x_1+t(x_2-x_1), y_1+t(y_2-y_1)+\epsilon t(1-t))$. Note that if $\epsilon_1 \neq \epsilon_2$, then $\phi_{\epsilon_1}((0,1)) \cap \phi_{\epsilon_2}((0,1)) = \emptyset$, hence there are many $\epsilon$ such that $\phi_\epsilon([0,1]) \cap \mathbb{N}^2 = \emptyset$.
This shows that $(\mathbb{N}^2)^c$ is path connected, hence connected.
A: We can show that $\mathbb R^2\setminus\mathbb N^2$ is connected by showing that any two points of $\mathbb R^2\setminus\mathbb N^2$ lie on a circle which contains no point of $\mathbb N^2$. Let $P$ and $Q$ be two distinct points of $\mathbb R^2\setminus\mathbb N^2$.
Note that there are uncountably many circles that pass through both $P$ and $Q$. (If $d$ is the distance between $P$ and $Q$, then there is one such circle of diameter $d$ and two such circles of diameter $D$ for each real number $D\gt d$.) Since each point of $\mathbb N^2$ lies on at most one of those circles (because three noncollinear points determine a circle), and since $\mathbb N^2$ is countable, at most countably many of those circles pass through a point of $\mathbb N^2$. Therefore, there are uncountably many circles which pass through $P$ and $Q$ and lie entirely in $\mathbb R^2\setminus\mathbb N^2$. (Continuum many, in fact, but we only needed one.)
