I need to prove whether or not $X=\Bbb R^2 -\Bbb Q^2$ is connected. The definition of connectedness I am using is a space X is connected if it is not the union of two disjoint nonempty open sets. This is a very new concept for me so I am not exactly sure how to set it up. I believe we will prove it by contradiction, but I do not know where to start.
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5Do you know the concept of path connected-ness? A space $X$ is said to be path-connected if there is a continuous path (continuous map from $[0,1]$ to $X$) between any two points of $X$. A path-connected space is connected (prove this!), and this fact may help here. – Dorebell Nov 04 '15 at 04:30
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Related: Does every plane curve contain a rational point? – Eclipse Sun Nov 04 '15 at 05:00
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Take two points $x,y \in X$. Since $\Bbb{Q}^2$ is countable and the number of lines going through $x$ is uncountable (in bijection with $[0,\pi)$), there are uncountably many lines going through $x$ and contained in $X$. The same applies to $y$. Therefore, you can find lines going through $x$ and $y$ respectivly that are not parralel, and thus intersect each other. This proves that $X$ is connected.
Nitrogen
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Fix two points $(x,y), (u,v) \in A:= \mathbb{R}^2\setminus \mathbb{Q}^2$.
Suppose $x, v\notin \mathbb{Q}$ then the lines joining $(x,y)\to (x,v)$ and $(x,y) \to (u,v)$ are both in $A$.
Something similar happens if $y,u \notin \mathbb{Q}$.
If $x,u\notin \mathbb{Q}$, then assume WLOG that $v<y, u<x$ and choose $z \in (v,y)\setminus \mathbb{Q}$ and $w\in (u,x)\setminus \mathbb{Q}$. So the lines $(x,y) \to (z,w), (z,w) \to (u,w)$ and $(u,w) \to (u,v)$ are all in $A$.
Similarly if $y,v\notin \mathbb{Q}$. Hence $A$ is path-connected, and so connected.
Prahlad Vaidyanathan
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2Not quite. It could be that, say, $y \in \mathbb Q$ while $x \notin \mathbb Q$. Then the line joining $(x,y)$ to $(u,y)$ contains points of $\mathbb Q^2$. – Robert Israel Nov 04 '15 at 04:38
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