Suppose we consider a function $f(x)$ defined from an open interval say $(a,b)$ to some set $T$. When would the set $T$ be an open interval?

  • 1
    Functions that verify the condiction "the image of an open set is itelf an open set" are called open functions. You may verify that homeomorphisms are open functions, but there are several functions which are open without being homeomorphisms. – Marra Feb 24 '13 at 13:59
  • 2
    When would that open set be an open interval? – Manjil P. Saikia Feb 24 '13 at 14:00
  • 1
    If you're using $\mathbb{R}$ with the standard topology (I guess you are, since it's a real analisys question ;) ), then open sets are the unions of open intervals and open intervals are open sets. – Marra Feb 24 '13 at 14:03
  • 2
    Certainly when $f$ is continuous it maps intervals to intervals (so continuous and open implies maps open intervals to open intervals). – Ben Millwood Feb 24 '13 at 14:05
  • 3
    I very much doubt that one can find a nontrivial necessary and sufficient condition for the problem as stated. – Harald Hanche-Olsen Feb 24 '13 at 14:15

To give some sense to how little can be said without any further restrictions on the function, consider this: There exists a map $f\colon(0,1)\to(0,1)$ which maps every nonempty open subinterval of $(0,1)$ onto $(0,1)$.

I don't know if such a function can be given explicitly, but here is an existence proof ultimately relying on the axiom of choice:

Write $x\sim y$ if $x-y$ is rational. Then $\sim$ is an equivalence relation on $(0,1)$. Write $E$ for the set of equivalence classes. Then the cardinality of $E$ equals that of $(0,1)$, because each equivalence class is countable. In particular there is an onto map $F\colon E\to(0,1)$. Define $$f(x)=F([x])$$ where $[x]$ is the equivalence class of $x$.

Now given any $y\in(0,1)$ and $0<a<b<1$ there is some $x\in(0,1)$ with $F([x])=y$. We can certainly find some $x'\in(a,b)$ with $x'\sim x$, so that $f(x')=F([x'])=F([x])=y$. Hence $f$ maps $(a,b)$ onto $(0,1)$ as claimed.

  • Thanks a lot for the explanation. Wasn't looking for an explicit example though. – Manjil P. Saikia Feb 25 '13 at 6:15

The set $T$ would be an open interval if the function is also injective, that is, supposing that T is a subset of R and $f$ is continuous.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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