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Let $\{0,1\}$ be equipped with the topology suggested by Scientifica in this post, ie: $\tau=\{\emptyset, \{0,1\},\{0\}\}$. What are the continuous functions from $\mathbb{R}$ to $(\{0,1\},\tau)$?

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    $\begingroup$ $(\{0,1\},\tau)$ is called Sierpinski space. $\endgroup$ Mar 27, 2020 at 6:42
  • $\begingroup$ Does the $n$-fold product of such spaces also have a name? $\endgroup$
    – user683848
    Mar 27, 2020 at 9:09
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    $\begingroup$ Not that I know of. $\endgroup$ Mar 28, 2020 at 3:50
  • $\begingroup$ The aswer is very short: this set of functions is bijectively parametrized by the open subsets of $\mathbb{R}$. For such $U\subset \mathbb{R}$, define $\varphi_U$ as the characteristic function of $\mathbb{R}\setminus U$, then the collection $$\{\varphi_U\}$$ is exactly the set you're looking for. $\endgroup$ Apr 8, 2020 at 21:28

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A function is continuous iff the inverse image of all open sets are open. Since $f^{-}(\varnothing)=\varnothing$ and $f^{-1}(\{0,1\}=\mathbb{R}$, a function $f:\mathbb{R}\to\{0,1\}$ will be continuous iff $f^{-1}(0)$ is open in $\mathbb{R}$, therefore $f^{-1}(0)$ has to be a countable union of disjoint open intervals, so the only continuous functions $f:\mathbb{R}\to\{0,1\}$ with that topology are the characteristic functions of complements of open sets of $\mathbb{R}$, i.e. characteristic functions of closed sets in $\mathbb{R}$, i.e. characteristic functions of complements of countable disjoint unions of open intervals.

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