Let $f\in C^0(\mathbb R^n,\mathbb R)$, with $n>1$ integer, and $f(\mathbb R^n)=\mathbb R$.

Is it true that the cardinality of $f^{-1}(\{c\})$ must be the same as the cardinality of $\mathbb R$ (the continuum) for any $c\in\mathbb R$?

Source : les dattes à Dattier

  • $\begingroup$ Of course not. What if $f\equiv 1$ and $c=0?$ And what is Q6 supposed to mean? $\endgroup$ – zhw. Jan 23 '17 at 18:06
  • $\begingroup$ If that inverse set has 2 or more points in it, then yes, same card. $\endgroup$ – coffeemath Jan 23 '17 at 18:07
  • $\begingroup$ sorry, I have change the hypothesis. $\endgroup$ – Dattier Jan 23 '17 at 18:07
  • $\begingroup$ I guess you mean $f(\mathbb R^n)=\mathbb R$, not $f(\mathbb R)=\mathbb R$ ? $\endgroup$ – MPW Jan 23 '17 at 18:08
  • $\begingroup$ To give the source in this form is a bit strange. I don't mind enough to do something about it, but it's strange and you likely decrease the reception you get. $\endgroup$ – quid Feb 28 '18 at 0:05

Let's take two points $z_1$, $z_2\in\mathbb R^n$ such that $f(z_1)<c$, $f(z_2)>c$ (they exist because $f$ is surjective). Denote $L\subset\mathbb R$ a hyperplane for which $z_1$ and $z_2$ are laying on the different sides. Then for any point $x\in L$ consider the set consisting of two straight line segments $Z_x=[z_1,x]\cup[x,z_2]$. This set is connected, so there exists a point $z_x\in Z_x$ for which $f(z_x)=c$. Since for all the $Z_x$ their pairwise intersection is $\{z_1,z_2\}$, we can conclude that there's a continuum set of points where $f(x)=c$.

  • $\begingroup$ Nice, +1. Note that the $Z_x$ are not pairwise disjoint, but the $(z_1,x]\cup [x,z_2)$ are. $\endgroup$ – zhw. Jan 23 '17 at 18:29
  • $\begingroup$ You're right. I've edited the answer. $\endgroup$ – Sergei Golovan Jan 23 '17 at 18:29

Let $E = f^{-1}(\{c\}).$ Because $f$ is continuous, $E$ is closed. Also, $f^{-1}((c,\infty)),f^{-1}((-\infty,c))$ are disjoint, nonempty, and open. Thus $\mathbb R^n \setminus E$ is not connected. But as is well known, if $E$ is countable, then $\mathbb R^n \setminus E$ is connected (actually path connected); here we use the fact that $n>1.$ It follows that $E$ is uncountable. So then $E$ is closed and uncountable, and I think this implies the cardinality of $E$ is that of $\mathbb R.$ But I'm not sure how to get this last bit yet.


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.