Discontinuous inverse function

I have shown that the map $ c\text{:}\ \mathbb{R}\rightarrow \mathbb{R}^2 $ given by $$ c(t)=\left(\frac{t}{1+t^4}, \frac{t}{1+t^2}\right) $$ is continuous and bijective. Intuitively, it's clear that its inverse $ c^{-1}$ is discontinuous at the origin since $\lim_{t \rightarrow \infty} c(t)=\overline{0} $ and $c^{-1}(\overline{0}) =0<\infty$. I was still hoping to have some more rigorous argument (maybe $\delta, \varepsilon$) to verify the claim.

So I know that $\lim_{t \rightarrow \infty}c(t)=\overline{0}$, i.e. $$\forall \varepsilon>0\text{: }\exists N>0\text{: }\forall t>N\text{: } \Vert c(t)\Vert_2 <\varepsilon. $$

Let $a>0$ and define $\tilde{c}\text{: } \left[a, \infty \right) \rightarrow \mathbb{R}^2$ by $\tilde{c}(t)=c(t)$. Now I wish to show that $$ \lim_{\tilde{c}([a, \infty)) \ni p \rightarrow \overline{0}} \tilde{c}^{-1}(p) =\infty,$$ i.e. $$ \forall N>0\text{: } \exists \delta >0\text{: } \forall p\in B(\overline{0}, \delta)\cap \tilde{c}([a, \infty))\text{: } \tilde{c}^{-1}(p)>N.$$ And that is exactly the part which I can't figure out how to do. First I was thinking about showing that $ t\mapsto \Vert \tilde{c}(t)\Vert_2 $ is decreasing when $t \ge t_0 >0$ for some $t_0$ but the task became too complicated. Any ideas would be appreciated. I guess that it's not possible/easy to find an explicit formula for the inverse function.

Solution 1:

It's not 100% clear to me where this is going. That is, at what point does "therefore $c^{-1}$ is not continuous at $0$" fit in?

Here's my perspective on the problem:

Let $X$ be the image of $c$. Then, in order to discern whether or not $c^{-1} : X \to \mathbb{R}$ is continuous or not, we need to know what the topology of $X$ is. This, of course, is induced from $\mathbb{R}^2$ and the interesting point is $0 \in X$.

Now, every open neighbourhood around $0 \in X$ contains a ball $B(0, \varepsilon) \cap X$. And now, since $\lim\limits_{t \to \pm \infty} c(t) = 0$, it must be the case that for $|t| \ge N$ that $c(t) \in B(0, \varepsilon)$.

Using the general definition of "continuous," $c^{-1}$ is continuous if the image under $c$ of every open set $U \subseteq \mathbb{R}$ is an open subset of $X$. Thus, let us take $U = (-1, 1)$. The image of $U$ contains $0$, so if $c^{-1}$ is continuous, $c(U)$ must be an open set and it must contain a neighbourhood $B(0, \varepsilon) \cap X$ for small enough $\varepsilon$.

From here, let us fix a sequence $x_1, x_2, \dots$ such that $x_k \to \infty$ and $c(x_k) \in B(0, \frac1k)$. It follows that $x_k \notin U$ for sufficiently large $k$ and since $c$ is one-to-one, $c(x_k) \notin c(U)$. Thus $B(0, \frac1k) \cap X \not\subseteq c(U)$.