# Injective with Finite Discontinuities Mapping from $\mathbb R^n$ to $[0,1]$

Hi,

as a continuation to the fully answered question:

Injective&Intregrable Mapping from $\mathbb R^3$ 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?

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I extended my answer to injective maps $f:\mathbb{R}^n\to\mathbb{R}^{n-1}$. –  GH from MO Jan 2 '13 at 7:30

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, the number of discontinuities of any injective map $f:\mathbb{R}^n\to\mathbb{R}^{n-1}$ is continuum.

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Thanks! Please note that in the mentioned question (in that link), there is an example with only countably many discontinuities. –  Ohad Asor Jan 2 '13 at 7:12
No, in the answer to the mentioned question the discontinuities lie on the union of boundaries of cubes, which is a two-dimensional object of cardinality continuum (and measure zero). As I proved above, there is no injective map with less than continuum many discontinuities. –  GH from MO Jan 2 '13 at 7:20
In fact Gerald Edgar commented on your original question (in that link): "Cannot be done with countably many discontinuities." And you responded: "Right. and thanks for your answer. if this is indeed the problem, so make it zero measure of discontinuities." –  GH from MO Jan 2 '13 at 7:24
i agree i dont have a full understanding, but this will come with time... i'm doing my best –  Ohad Asor Jan 2 '13 at 7:25
So, can we say this... if $A \subset \mathbb R^n$ has Hausdorff dimension ${}\lt n-1$, then $\mathbb R^n \setminus A$ is connected? –  Gerald Edgar Jan 2 '13 at 14:04