2
$\begingroup$

In the Euclidean plane $R^2$,consider the subset $$ A=\{(x,y)\in \Bbb R^2|\text{Either $x$ or $y$, but not both, is a rational number}\} $$ Is $A$ connected? Is $\Bbb R^2$\A connected?

I have tried many methods, but I still have no idea how to solve it. Can someone tell me about how to prove it? I try to prove given any continuous function two valued function $f :A\to\{0,1\}$, the $f$ function will be one-valued function, but still don't know how to finish it. Maybe my thoughts are wrong. Hope someone can help me! Thank you

$\endgroup$
  • $\begingroup$ Path-connectedness may be of help $\endgroup$ – Ilya Nov 28 '14 at 7:30
0
$\begingroup$

The line $y=x$ is disjoint to $A$, that is you can consider the disjoint open sets $U=\{\,(x,y)\mid y>x\,\}$ and $V=\{\,(x,y)\mid y<x\,\}$ that cover $A$ (or use $f(x,y)=\begin{cases}1&\text{if $y>x$}\\0&\text{if $y<x$}\end{cases}$ as your continuous function).

In fact you can find rational $a\ne 0$, $b$, such that $y=ax+b$ separates any two given points of $A$, i.e., $A$ is totally disconnected.

The set $$ B:=\{\,(x,y)\mid x+y\in\mathbb Q\lor x-y\in\mathbb Q\,\}$$ is disjoint from $A$ and is path connected (for any two points in $B$ you can combine a northeast/southwest and a northwest/souteast line to a path). Since $B$ contains $\mathbb Q^2$, it is also dense in $\mathbb R^2\setminus A$. We conclude that $\mathbb R^2\setminus A$ is connected.

$\endgroup$
0
$\begingroup$

Consider:

$$f \colon \mathbb{R}^2 \longrightarrow \mathbb R ~~\text{as}~~ f(x,y) = x+y$$

Notice that $f(\mathbb{R}^2-A)=\mathbb{Q} \subset \mathbb{R}$ so by continuity $\mathbb{R}^2-A$ can't be connected, etc.

$\endgroup$

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.