Let $X$ be a subspace of $\Bbb R^2$ consisting of points whose coordinates are both irrational. Prove that $X$ is path-connected

Question

I have to prove that if $X$ is a subspace of $\Bbb R^2$ consisting of points whose coordinates are both irrational then $X$ will be path-connected.

I think that it isn't connected by arcs, but I don't know how prove it.

Answer 1

As $X$ is disconnected by

$$ \{ (x,y) \in X : 0 < y \} $$

and
$$ \{ (x,y) \in X : 0 > y \} $$

$X$ cannot be path connected.

Answer 2

Recall that $\sqrt2$, $\sqrt 3$, $e$, $\pi$ are irrational and that $e<3<\pi$.

Assume $\gamma\colon[0,1]\to X$, $t\mapsto \langle \gamma_1(t),\gamma_2(t)\rangle$ is a path from $\gamma(0)=\langle\sqrt 2,e\rangle\in X$ to $\gamma(1)=\langle \sqrt 3,\pi\rangle$. Then $\gamma_2$ is continuous and by the Inermediate Value Theorem, there exists a $t^*\in[0,1]$ with $\gamma_2(t^*)=3$. It follows that $\gamma(t^*)\notin X$, contradiction.