TITLE: Find $f:C\to\mathbb{R}^2$ continuous and bijective but not open, $C\subset\mathbb{R}^2$ is closed
QUESTION [12 upvotes]: Are there a closed subspace  $C$ of $\mathbb{R}^2$ and a continuous, bijective function  $f:C\to\mathbb{R}^2$ that is not open?
I cannot think about a continuous and bijective function $f:C\to\mathbb{R}^2$.
My best candidate was the set $C=[0,+\infty)\times[0,+\infty)$ and the function 
$$f(s,t)=\Bigg(s\cos\bigg(2\pi\frac{t}{t+1}\bigg),s\sin\bigg(2\pi\frac{t}{t+1}\bigg)\Bigg),$$
but this is not injective in $\{0\}\times[0,+\infty)$.
The only hint that I can think about is that $C$ cannot be compact, because $f$ is bijective. Therefore, $C$ cannot be bounded.
An idea that I've been trying is to work in the sphere $S^2$. There, we can think of $C$ as a subset of $S^2$ with N=(0,0,1) as a limit point such that $C\cup\{N\}$ is compact.
Then the function $f$ takes $C$ onto $S^2-\{N\}$. But in this way I only see open functions.
Any suggestions?

REPLY [2 votes]: The answer of @Jesus RS is brilliant! It shows the more important point of the problem: $C$ does not have to be connected.
If we have a closed subspace $C$ of $\mathbb{R}^2$ and a continuos and bijective function $g:C\to\mathbb{R}^2\setminus\{p\}$, where $p\in\mathbb{R}^2$, then we just have to consider a point in the complement of $C$, say $q$, and the function
$$f:C\cup\{q\}\to\mathbb{R}^2$$
\begin{equation}
f(z)=
\begin{cases}
&g(z)&,&x\neq q\\
&p&,&x=q
\end{cases}  
\end{equation}
will be continuous and bijective, but is not open because $\{q\}$ is open in $C\cup\{q\}$ but $f(\{q\})=\{p\}$ is not open in $\mathbb{R}^2$.
Now it is very easy to show an explicit function satisfying these conditions.
The function $f:\big([0,+\infty)\times\mathbb{R}\big)\cup\{(-1,0)\}\to\mathbb{R}^2$ defined by
\begin{equation}
f(x,y)=
\begin{cases}
&\bigg(e^y\cos\big(\frac{2\pi x}{x+1}\big),e^y\sin\big(\frac{2\pi x}{x+1}\big)\bigg)&,&(x,y)\neq (-1,0)\\
&(0,0)&,&(x,y)=(-1,0)
\end{cases}  
\end{equation}
is continuous, bijective and not open, where the domain is a closed subspace of $\mathbb{R}^2$.