Let $f:ℝ^{m}→ℝ$ is a continuous function. Let us consider the set $$A=\{x∈ℝ^{m} \ |\ f(x)=0\} = f^{-1}[\{0\}]$$

My question is: Under what conditions the set $A$ is connected?, simply connected?


closed as too broad by user99914, Moishe Kohan, user91500, Qiaochu Yuan, user370967 Sep 23 '17 at 15:32

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Convexity might be relevant here. Not sure though. $\endgroup$ – Shalop Sep 22 '17 at 9:28
  • 4
    $\begingroup$ The exact conditions are of course whenever $f$ has the property that $f^{-1}(0)$ is connected/simply connected. I doubt there is some other meaningful property of $f$ that is equivalent to that... $\endgroup$ – skyking Sep 22 '17 at 9:31
  • 1
    $\begingroup$ Convexity of $f$ is not enough. For example $f(x)=x^2-2$ would be counterexample. In higher dimensions, this example does not work though. $\endgroup$ – humanStampedist Sep 22 '17 at 9:32
  • 1
    $\begingroup$ Not sure if we can conclude (any topological properties of $A$) if $m > 1$ and if $f$ isn't bijective, if I'm wrong I'm really interested to see a counter example $\endgroup$ – Perturbative Sep 22 '17 at 9:45