0
$\begingroup$

I am trying to define a continuous function $f:(0,1) \mapsto \{0,1\}$. Intuitively, it seems that the possible continuous functions are the trivial functions of $f(x) = 0$ or $f(x) =1$.

Based on the definition of continuity (topology), I first set topologies on $(0,1)$ and $\{0,1\}$. Let $(0,1)$ be equipped the standard Euclidean metric space and $\{0,1\}$ be equipped with the discrete topology.

[Editted: [0.5,1) open -> closed]

Let define $f_{0.5}(x) = 1_{x < 0.5}(x)$. Then since $f_{0.5}^{-1}(0) = [0.5,1)$ is closed and $f_{0.5}^{-1}(1) = (0,0.5)$ is open, $f_{0.5}(x)$ is not continuous.

Can we prove that the only continuous function is a trivial one?

I am looking for some clear explanations or intuition behind this. Any comments/answers/suggestions will be very appreciated.

$\endgroup$
  • 1
    $\begingroup$ Why do you think $[0.5, 1)$ is open? $\endgroup$ – Robert Israel Feb 6 at 5:28
  • $\begingroup$ @RobertIsrael Oh you're right. [0.5,1) is closed as the complement (0,0.5) is open. $\endgroup$ – induction601 Feb 6 at 5:30
  • 1
    $\begingroup$ Yes! Every continuous function $f$ from $(0,1)$ to $\{0,1\}$ is either $f \equiv0$ or $f\equiv1$, since $(0,1)$ is connected! $\endgroup$ – Chinnapparaj R Feb 6 at 5:37
2
$\begingroup$

This follows from the connectedness of $(0,1)$. The sets $f^{-1}(0)$ and $f^{-1}(1)$ are open (by continuity) and clearly disjoint. If they were both non-empty, $(0,1)=f^{-1}(0)\cup f^{-1}(1)$ would be disconnected.

$\endgroup$
  • $\begingroup$ Thanks for your answer. By the way, is the same true for any discrete set? That means, any continuous function $f:(0,1) \to A$ where $A$ is a set containing only a finite number of elements, then $f$ should be a constant. $\endgroup$ – induction601 Feb 6 at 15:22
  • $\begingroup$ Yes it is, and the proof is essentially the same. $\endgroup$ – Robert Israel Feb 6 at 17:48

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.