TITLE: Is $f: \mathbb Q_{>0} \to \mathbb Q, f(x)=x^2$ an open map?
QUESTION [5 upvotes]: Suppose $f: \mathbb{Q_{>0}} \to \mathbb{Q} $ maps every $x$ in its domain to $x^2$. And by an open map I mean a function which maps open sets to open sets. Given the natural topology on $\mathbb{Q}$, prove or disprove: $f$ is an open map.
Edit: I think $f$ cannot be open because in the range of $f$ we don't have the numbers whose square roots are irrational. But I don't know how to say this formally.
2nd edit: I think since a base of the topology on $\mathbb{Q}$ consist of open intervals, it is suffices to show that every open interval in $\mathbb{Q}$ has a point whose square root is not rational.

REPLY [3 votes]: A similar, but alternate view using sequences (we are blessed with sequences since we have usual topology on $\mathbb Q$, with usual metric):
Let $f:\mathbb Q _{>0} \to \mathbb Q$ be as above, $f(x) = x^2$. We show that it is not an open map by showing its image $f(\mathbb Q_{>0})$ is not an open set in $\mathbb Q$ (Note $\mathbb Q _{>0}$ is open in $\mathbb Q _{>0}$!). Denote the image $W = f(\mathbb Q_{>0}) = \{x^2 : x\in \mathbb Q_{>0}\}$.
Showing $W$ is not open in $\mathbb Q$ is equivalent to showing its complement $\mathbb Q - W$ is not closed in $\mathbb Q$. Since we have a natural metric, to show $\mathbb Q - W$ is not closed in $\mathbb Q$ is to exhibit a sequence whose terms are in $\mathbb Q - W$ and it converges in $\mathbb Q$ but the limit is not in $\mathbb Q - W$ (i.e. the limit is in $W$). [This is because closed sets contain all its limit points.]
Indeed, consider the sequence $(a_n)$ given by $a_n = n / (n + 1)$, $n\in \mathbb N$. Note that each $a_n$ is a rational number but is not a square of a rational number [you might want to prove this. I have edited to give one below]. Hence $a_n \in \mathbb Q - W$. But the limit of $(a_n)$ exists in $\mathbb Q$, namely $a_n \to 1$. Finally, note that $1\in W$. Thus $(a_n)$ is a convergent sequence in $\mathbb Q - W$ with limit in $W$, that is, $\mathbb Q - W$ is not closed. 
Hence $W$ is not open, and that $f$ is not an open map. ////
[Proof of $n/(n+1)$ not a square of rational number, for $n\in\mathbb N$:
 Suppose to the contrary that $n/(n+1) = a^2 / b^2$, for some $a,b,\in \mathbb N$ with $gcd(a,b) =1$, $b>a$. Then we have $b^2 n = a^2n +a^2$, which implies $n = a^2/(b^2 - a^2) = a^2/(b+a)(b-a)$, an integer. This shows that $(b+a)|a^2$, say $a^2 = x(b+a) = xb + xa$, for some integer $x\ge 1$. But this means $a^2 - xa = a(a-x) = xb$, or $a|xb$. But since $gcd(a,b)=1$, we must have $a|x$, say $x = ya$, some integer $y\ge 1$. Then by substitution, we see that $a^2 = ya(b+a) = yab +ya^2$, namely $a^2(1-y) = yab\le 0$, a contradiction.] (Does anyone else have an alternate proof to this "simple" statement?)