The intermediate value property (IVP): A function has the intermediate value property on an interval $[a,b]$ if for all $x I think that $f$ has IVP if and only if $f^{-1}$ maps connected sets to connected sets. Please, can you tell me if what I think is correct?
Answer
Take $f(x)=x^2$. Then $f$ is continuous, and thus it has the IVP property but $$f^{-1}\big([1,4]\big)=[-2,-1]\cup[1,2].$$
No comments:
Post a Comment