So, we are considering the subset

$$ S = \{(x, y) \in \mathbb{R^2} | (x \text{ and } y \in \mathbb{Q}) \text{ or } (x \text{ and } y \notin \mathbb{Q})\} $$

And consider its complement $$ T = \mathbb{R^2} \backslash S $$

The set T is disconnected, actually I am fairly certain it is totally disconnected. I am just having problems showing that rigorously. I was trying to show it using straight lines but I don't think I was getting anywhere. I know that a totally disconnected set's only connected sets are the one point sets. I've been trying to show that given two arbitrary points, that a separation exists between them. It is more difficult since this is in the plane.

Any hints at all would be a great help. Maybe I'm making mountains of molehills.

  • $\begingroup$ as for the source, I have not yet found a similar problem in any book, so as far as I know my professor (Jack Conn) thought this one up for us. $\endgroup$ – Tyler Nov 19 '10 at 6:33
  • 1
    $\begingroup$ The title, like, sounds like Valleyspeak. $\endgroup$ – copper.hat Oct 25 '13 at 2:19

I had written out a full solution, but since this is homework, I've removed it and replaced it with this suggestion.

One way to prove that $T$ is totally disconnected is to show that whenever $p$ and $q$ are distinct points of $T$, then there are open sets $A$ and $B$ of $\mathbb{R}^2$ such that $T\subseteq A\cup B$, $(A\cap B)\cap T=\emptyset$, $p\in A$, and $q\in B$. This will show that there is a disconnection of $T$ in which $p$ and $q$ are in distinct components. In particular, $p$ and $q$ cannot be in the same connected component of $T$. If this holds for all pairs of points $p$ and $q$, then the connected components of $T$ must be single points.

So, pick two distinct points $p$ and $q$ in $T$. Try to find a line that is completely contained in $S$ and which separates $p$ and $q$. One way to achieve a line completely contained in $S$ is to have it go through a rational point with rational slope. Then throw away the line to get your two sets $A$ and $B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.