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Inspired by this question I asked myself the question which I am going to describe:

Let $\mathbb {I}$ be the set of all irrational numbers. Let $\mathbb {I}^2$ be the Cartesian product of the set $\mathbb {I}$ with itself, in other words, the set of all ordered pairs of numbers such that both numbers are irrational.

Is the set $\mathbb {R}^2 \setminus \mathbb {I}^2$ path connected?

I believe that this is not so hard and that it is a known fact but I do not have at the moment some useful idea that could solve this problem.

Any ideas?

Community
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A. P.
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2 Answers2

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Yes. This space is path connected. You can think of this as the set of all point $(x,y)$ where at least one of $x,y$ are rational. Pick $(x_1,y_1)$ and $(x_2,y_2)$. We have three cases:

  • If the first coordinate of both are rational, the path travels vertically to a point $(x_1,a)$ where $a$ is rational, travel along the line $y=a$ to the point $(x_2,a)$ and then travel vertically again to $(x_2,y_2)$.
  • If the last two coordinates are both rational, do essentially the same thing flipped 90 degrees
  • If $x_1$ is rational and $y_2$ is rational, travel along the line $x=x_1$ to the point $(x_1,y_2)$, then to $(x_1,y_2)$.
Plutoro
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Yes, it is path connected.

You can connect any $(a,b)$ with $a\in\Bbb Q$ to $(a,0)$ by a straight segment, and also $(a,0)$ to $(0,0)$.
Similarly, if you start out from $(a,b)$ with $b\in\Bbb Q$.

Berci
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