Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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

1 Answer 1

up vote 8 down vote accepted

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.

share|improve this answer
    
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
1  
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.