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In my notebook, something is very briefly, not in detail whatsoever, path connectedness mentioned, and two assumptions are made about $x_1, x_2 \in \mathbb Q.$ If anyone can prove this I would greatly appreciate it... I tried myself, but got nowhere.

Asaf Karagila
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  • Consider vertical and horizontal line segments which stay in the set to prove path connectedness. – Cameron L. Williams Jul 23 '15 at 22:32
  • What is the reasoning behind this, it says this is the answer in my notebook aswell, I do not understand it –  Jul 24 '15 at 10:52
  • WHy cant you connect tthe two points directly ? –  Jul 24 '15 at 10:54
  • Because you might pass through points where neither coordinate is rational. Consider $(\pi,0)$ and $(0,\pi)$. If you drew a straight line between them, then you'd hit the point $(\pi/2,\pi/2)$. – Cameron L. Williams Jul 24 '15 at 12:11

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Let $(x_1,y_1),(x_2,y_2)\in B$. Define $f:[0,1]\rightarrow B$ as follows: if $x_1\in\Bbb Q$, set $f$ on $[0,1/4]$ to form a line segment from $(x_1,y_1)$ to $(x_1,0)$, and on $[1/4,1/2]$ to form a line segment from $(x_1,0)$ to $(0,0)$. Otherwise $y_1\in\Bbb Q$, and we can set $f$ on $[0,1/2]$ similar to above, except going from $(x_1,y_1)$ to $(0,y_1)$ to $(0,0)$.

On $[1/2,1]$ define $f$ similarly as on $[0,1/2]$ but in reverse and to $(x_2,y_2)$.

I'll leave it to you to check that $f(x)\in B$ for all $x\in[0,1]$ and that $f$ is continuous. Them it follows that $(x_1,y_1),(x_2,y_2)$ are connected by a path, so $B$ is path-connected, and thus connected.

Wojowu
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