Showe that $f \colon \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if the inverse image of every open set is an open set.
This is what I have so far
If: Take an open set $U$ in the target space, since $f$ is continuous at some point $x \in U$, there must exist $\delta$ such that $\| f(x) - f(x^*) \| < \varepsilon$, then $\| x - x^* \| < \delta$. If we let $\varepsilon < R$ (the radius of the open set $U$), then $\delta$ is the radius of some ball in $f^{-1}(U)$ such that all points in the ball are also in the preimage of $U$.
And only if: same set up but this time assume $U$ is an open set in $\mathbb{R}^m$ and its preimage $f^{-1}(U)$ is an open set in $\mathbb{R}^n$. Since this is the case, there exists $r_1$ such that $B(r_1,x)$ is a subset of $f^{-1}(U)$ and $r_2$ such that $B(r_2, f(x))$ is a subset of $U$. Therefore if we let $\delta = r_1$, we can show that if the $\| f(x) - f(x^*) \| < \epsilon < r_2$ then $\| x - x^* \| < \delta = r_1$, and so that the limit exists and thus the function is continuous.
I was wondering if this argument is convincing or if I am misunderstanding a key concept or misusing any terms.
Thanks
