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).
  • $\begingroup$ $f(0)=1=f(1)$ so $f$ is not injective and I don't see why $f$ is not surjective .... $\endgroup$ – user123733 Dec 20 '14 at 13:32
  • $\begingroup$ $f(1)=2$ (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 $f(x)=1$? $\endgroup$ – drhab Dec 20 '14 at 13:37
  • $\begingroup$ okay but what about not attaining $1$ ? $\endgroup$ – user123733 Dec 20 '14 at 13:40
  • $\begingroup$ $1$ is sent to $2$. Wich element is sent to $1$? $\endgroup$ – drhab Dec 20 '14 at 13:40
  • $\begingroup$ ok sorry ; please just make any minor edit ; I will upvote and accept $\endgroup$ – user123733 Dec 20 '14 at 13:48

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