How can I prove that $\mathbb Q \times \mathbb Q$ is connected?

Question

How can I prove that $\mathbb Q \times \mathbb Q$ is connected? I was trying to prove that it isn't because $\mathbb Q$ is not connected but I can't find two open $A,B$ to prove it.

Answer 1

You know that $\mathbb Q$ is not connected. That means that there is a set $A\subset\mathbb Q$ such that $A$ is open and closed in $\mathbb Q$ and that furthermore $A\neq\emptyset,\mathbb Q$. So, consider the set $A\times A$.

Answer 2

The map $f(x,y)=x$ is continuous on $\mathbb R^2.$ If $\mathbb Q\times \mathbb Q $ were connected, then $f(\mathbb Q\times \mathbb Q)$ would be connected. But $f(\mathbb Q\times \mathbb Q)=\mathbb Q,$ contradiction.

Answer 3

Your initial intuition was correct - $\mathbb{Q}^2$ is indeed not connected.

Think about it intuitively. $\mathbb{Q}^2$ is basically a bunch of points sprinkled in the plane $\mathbb{R}^2$. The easiest way to show that it's disconnected would be to "cut it in half with a line" - that is, find a line $l$ such that $(1)$ there are rational points (= elements of $\mathbb{Q}^2$) on both sides of $l$ and $(2)$ there are no rational points actually on $l$ itself.

• Do you see why this would in fact disconnect $\mathbb{Q}^2$?

Now $(1)$ is basically trivial - $\mathbb{Q}^2$ is scattered throughout $\mathbb{R}^2$ so that there are rational points on both sides of any line. So we're just down to $(2)$: can you think of a line in $\mathbb{R}^2$ not containing any rational points? HINT: there's even a horizontal (or vertical, if you prefer) example - think about $\pi$ ...)