Is $\mathbb{Q}^2 \cup \mathbb{I}^2 $ disconnected?

Question

My intuition says that it is but I'm not entirely sure. I thought about using the projection map since it is continuous and surjective and also because I know that the rationals and irrationals are disconnected in the reals. However, there isn't a fixed set to be projected in this case so I figure that not to be route to take in this case.

Something is telling me that disconnectedness follows straight from the definition although I'm not sure how.

Any help is appreciated.

Nice question. After playing around a bit, my intuition is that it is connected, but I don't see a proof right now. — Cheerful Parsnip, Apr 13 '16 at 13:19

score 5 · Accepted Answer · answered Apr 13 '16 at 13:19

5

$\mathbb{Q}^2 \cup \mathbb{I}^2$ is connected.

Every line of finite, nonzero rational slope passing through $(0,0)$ is contained in $\mathbb{Q}^2 \cup \mathbb{I}^2$, because if the quotient of two real numbers is a nonzero rational number then either both are rational or both are irrational. The union of these lines is path connected. Thus $\mathbb{Q}^2 \cup \mathbb{I}^2$ contains a dense connected subset, so it is connected.

answered Apr 13 '16 at 13:19

Lee Mosher

120,280

That makes perfect sense. Thanks! – user328442 Apr 13 '16 at 13:25

Is $\mathbb{Q}^2 \cup \mathbb{I}^2 $ disconnected?

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