On arguing why the the inverse of the mapping $c(t) = \left(\frac{t}{1 + t^4}, \frac{t}{1 + t^2}\right)$ is not continuous

Question

I was wondering whether the reason why $c(t) = \left(\frac{t}{1 + t^4}, \frac{t}{1 + t^2}\right)$ is not continuous is because i.) under a continuous mapping $f$, any open set in the range of $f$ has an open pre-image under $f$, ii.) $c: \mathbb{R} \to \mathbb{R}^2$ is a curve whose image set does not contain open sets, i.e. $\forall p \in c[\mathbb{R}]: \not \exists \epsilon > 0: \mathcal{B}_\epsilon(p) \subseteq c[\mathbb{R}]$. Thus, while $\mathbb{R}$ is an open set, its pre-image under the inverse of $c$ cannot possibly be open.

Answer 1

One way of seeing that it is discontinuous is to note that $\lim_{n\to\infty}c(n)=(0,0)=c(0)$, but you don't have $\lim_{n\to\infty}n=0$, that is$$\lim_{n\to\infty}c^{-1}\bigl(c(n)\bigr)\ne c^{-1}(0,0).$$