# Continuous functions from $\mathbb{R}$ to $\{0,1\}$ [closed]

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 $$(\{0,1\},\tau)$$?

• $(\{0,1\},\tau)$ is called Sierpinski space. Mar 27, 2020 at 6:42
• Does the $n$-fold product of such spaces also have a name?
– user683848
Mar 27, 2020 at 9:09
• Not that I know of. Mar 28, 2020 at 3:50
• 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. Apr 8, 2020 at 21:28

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.