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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.

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  • $\begingroup$ 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
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    $\begingroup$ The title, like, sounds like Valleyspeak. $\endgroup$ – copper.hat Oct 25 '13 at 2:19
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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 $B$.

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