Here is a question from the GRE Mock test.
Let $S$ be the subset of $\mathbb{R}^{2}$ consisting of all points $(x, y)$ in the unit square $[0,1] \times[0,1]$ for which $x$ or $y$, or both, are irrational. With respect to the standard topology on $\mathbb{R}^{2}, S$ is
(A) closed
(B) open
(C) connected
(D) totally disconnected
(E) compact
This seems to be a rather famous question. It has been asked Let $S \subset \mathbb{R^2}$ consisting of all points $(x,y)$ in the unit square $[0,1] × [0,1]$ for which $x$ or $y$, or both, are irrational., and Question on a subset $S$ of $[0,1]\times[0,1]$ where for each $(x,y)\in S$ at least one of $x$ and $y$ is irrational. I'm having difficulty more with interpreting this question.
By "with respect to the standard topology on $\mathbb{R}^{2}, S$ is", does this mean that I should determine connectedness while taking open sets to be the open sets of the the standard topology on $\mathbb{R}^{2}$? Or should I consider connectedness as in the original definition, where I take the subspace topology induced by $S$?
Furthermore, if I took open sets to be the open sets of the the standard topology on $\mathbb{R}^{2}$, would $S$ even be connected? Intuitively, I think not, but I have no idea how to prove it.
Best Answer
Clearly, $S$ is not open, since $\mathbb{Q} \times \mathbb{Q}$ is dense in $\mathbb{R}^2$.
$S$ is dense in $[0, 1]^2$. Since $S \neq [0, 1]^2$, clearly $S$ is not closed.
$S$ is connected - in fact, path-connected. Let $w = \frac{\sqrt{2}}{2}$. Then there is a path from any point in $S$ to $(w, w)$. First, translate along the coordinate that is rational until that coordinate is $w$, if there is such a coordinate. Then, translate along the other coordinate.
Since $S$ is path-connected, it is clearly not totally disconnected.
Finally, since $S$ is not closed, it is not compact. This is because in $\mathbb{R}^n$, a subspace is compact if and only if it is closed and bounded.