Looking for a special kind of injective function

Does there exist an injective function $f:\mathbb R \to \mathbb R$ such that for every $c \in \mathbb R$ , there is a real sequence $(x_n)$ such that $\lim\big(f(x_n)\big)=c$ but $f$ is neither continuous nor surjective ? If I remove the injectiveness condition then I can find such a function $f(x)=x$, for $x \ne 0$ $f(0)\ne0$ ; this is neither continuos nor surjective but this does not work with injectivity assumed as it is not injective

Function $f:\mathbb R\rightarrow\mathbb R$ prescribed by:

$x\mapsto x$ if $x\notin\mathbb{N}$ and $x\mapsto x+1$ otherwise.

To avoid confusion let us say explicit that $0\notin \mathbb N$. Note that $f(0)=0$ and $f(1)=2$

• $f$ is injective (straightforward).
• $f$ is not continuous (if $x_n\notin \mathbb N$ with $x_n\rightarrow 1$ then $f(x_n)=x_n\rightarrow 1\neq2=f(1)$).
• $f$ is not surjective ($1$ is not in the image of $f$).
• $f(\mathbb R)\subset\mathbb R$ is dense (straightforward).
• $f(0)=1=f(1)$ so $f$ is not injective and I don't see why $f$ is not surjective .... – user123733 Dec 20 '14 at 13:32
• $f(1)=2$ (not $f(1)=1$) also $f(0)=0$ (I explicitly said that here $0\notin \mathbb N$). On surjectivity: for wich $x$ do you have $f(x)=1$? – drhab Dec 20 '14 at 13:37
• okay but what about not attaining $1$ ? – user123733 Dec 20 '14 at 13:40
• $1$ is sent to $2$. Wich element is sent to $1$? – drhab Dec 20 '14 at 13:40
• ok sorry ; please just make any minor edit ; I will upvote and accept – user123733 Dec 20 '14 at 13:48