I am trying quiz questions of previous years and I couldn't think on how to solve this problem so, I am asking it here.
Question is :
Rationals are dense in $\mathbb{R}$ and closure of $\mathbb{Q} $ × $\mathbb{Q} $ = $\mathbb{R^2} $ and so is closure of complement of $\mathbb{Q} $ × $\mathbb{Q} $ . So, I think (C) , (D) are not true.
But I am not able to decide bw (1) and (2). If I can find 2 disjoint open sets for S such that their union is S then S would be disconnected but I think finding such open sets would be impossible as any open ball around Rational numbers will also contain irrationals.
So, I think A is correct. Am I right?
Please help.
You already explained why (3) is false ($\overline{\mathbb{Q}^2}=\mathbb{R}^2$).
To see why (4) is false, it is equivalent to argue that $S$ isn't open in $\mathbb{R}^2$. This is easy, as $S$ cannot be written as a union of open balls (for example, an open ball around $(0,0)$ will necessarily also contain elements with irrational components).
Now, (1) is also false: $S=A\cup B$, where $A:=\left\{(x,y)\in\mathbb{R}^2: x<\sqrt{2}\right\}\cap \mathbb{Q}^2$ and $B:=\left\{(x,y)\in\mathbb{R}^2: x>\sqrt{2}\right\}\cap \mathbb{Q}^2$. $A$ and $B$ are disjoint and open in $S$.
Lastly, to see why (2) is correct, we can argue that $S^C$ is connected by arcs. I will leave this to you. For $(a,b), (c,d)\in S^C$, try to find a continuous map $\gamma:[0,1]\to S^C$ that satisfies $\gamma(0)=(a,b)$ and $\gamma(1)=(c,d)$. A drawing may help. Try to use straight lines (parallel to the $x$- or $y$-axis).
For example, assume $(a,b)$ and $(c,d)$ are such that $a$ and $c$ are irrational. We can take $\gamma$ such that its image connects the points $(a,b), (a,\sqrt{2}),(c,\sqrt{2}), (c,d)$ via straight lines.
The other cases work similarly.