Let $n\geq 2$. Are injective functions dense in $C([0,1]^n,\mathbb R^n) $ with the uniform norm?


No. Identify $\mathbb{R}^2$ with $\mathbb{C}$ and consider $f(z) = z^2$. If $g$ is close enough to $f$ then $\alpha \mapsto g(e^{i \alpha})$ stays in an annulus and winds around the origin twice, which cannot be done injectively.

  • $\begingroup$ Even simpler: use $f: [0,1] \to \mathbb{R}$ defined by $f(x) = x(x-1)$. $f(1/2) = 1/4$, $f(0) = f(1) = 0$. If $g$ is continuous and within $0.01$ of $f$, then $-0.01 \leq g(0) \leq 0.01$, $-0.01 \leq g(1) \leq 0.01$, and $0.24 \leq g(1/2) \leq 0.26$. By intermediate value theorem, $g$ must attain the value $0.2$ twice, once in each subinterval. This seems like it might have been an undergraduate real analysis homework problem :(. $\endgroup$ – Steven Gubkin Apr 22 at 12:44
  • 3
    $\begingroup$ To be fair, the OP specified $n\ge 2$ $\endgroup$ – Alessandro Della Corte Apr 22 at 12:44
  • $\begingroup$ @A.DellaCorte Ah, overlooked this. Martin probably has the simplest example then. $\endgroup$ – Steven Gubkin Apr 22 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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