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Let $f:[0,1] \to \{0,1\}$ be continuous, where the spaces have the usual topology inherited by $\Bbb R$. Must $f$ be constant? I think it should because $[0,1]$ is connected and it can't be divided into two open disjoint sets, which should be the preimages of $0$ and $1,$ but I'd like to know if that is correct or not.

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    $\begingroup$ Your argument is correct. $\endgroup$ – Clement Yung Jan 7 at 3:53
  • $\begingroup$ How can a function be constant when the range contains more than one value? How can it be continuous when the domain is compact and the range is discrete? $\endgroup$ – WindSoul Jan 7 at 4:20
  • $\begingroup$ I think you need to mention that $\{0\}$ and $\{1\}$ are open since you're using it. $\endgroup$ – pigeon Jan 7 at 4:27
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Yep, your argument is sound. To put it more succinctly: the set $[0,1]$ is connected, its image under $f$ must be nonempty and connected, and the only connected nonempty subsets of $\{0,1\}$ are singletons. So the image of $f$ is a singleton, ie $f$ is a constant map.

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