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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.

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    $\begingroup$ Why do you think $[0.5, 1)$ is open? $\endgroup$
    – Robert Israel
    Feb 6, 2019 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, 2019 at 5:30
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    $\begingroup$ 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, 2019 at 5:37

1 Answer 1

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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.

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  • $\begingroup$ 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, 2019 at 15:22
  • $\begingroup$ Yes it is, and the proof is essentially the same. $\endgroup$
    – Robert Israel
    Feb 6, 2019 at 17:48

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