Hi,


as a continuation to the fully answered question:


[Injective and Integrable Mapping from $\mathbb R^3$ to $\mathbb R$](https://mathoverflow.net/questions/117609/injectiveintregrable-mapping-from-mathbb-r3-to-mathbb-r)


Can one think of an injective $f:\mathbb R^n\rightarrow[0,1]$ that has only a finite number of discontinuities? Or maybe one can come with some topological claim showing it is impossible?



Injective with Finite Discontinuities Mapping from $\mathbb R^n$ to $[0,1]$
Assume $f:\mathbb{R}^n\rightarrow[0,1]$ is an injective map. Take any circle $C\subset \mathbb{R}^n$. If $f$ is continuous on $C$, then the restriction of $f$ to $C$ is a homeomorphism $C\to f(C)$ (because $C$ is compact), which is a contradiction, because $C$ minus any point is connected, while $f(C)$ minus any point in $f(C)\cap(\inf f(C),\sup f(C))$ is disconnected. This shows that $f$ has a discontinuity on any circle in $\mathbb{R}^n$. In particular, when $n>1$, the cardinality of the discontinuities is continuum.


**Added:** By a similar argument combined with the [Borsuk-Ulam theorem](http://en.wikipedia.org/wiki/Borsuk%25E2%2580%2593Ulam_theorem), the number of discontinuities of any injective map $f:\mathbb{R}^n\to\mathbb{R}^{n-1}$ is continuum.


