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Let $S = \displaystyle \prod_{n \ge 1} \{ 0, 1\}$ be the set of binary sequences, so $S$ with the product topology is homeomorphic to the Cantor set. Endow $\mathbb{Z}_{\ge 1} \cup \{ \infty \}$ with the subspace topology from the extended real line, so basic open subsets are either singletons of integers, or sets of the form $U_n = \{ a > n : n \in \mathbb{Z} \} \cup \{ \infty \}$.

Define an "index function" $d:S \to \mathbb{Z}_{\ge 1} \cup \{ \infty \}$ by taking $d(s)$ to be the position of the first 1, and $d(000...) = \infty$. For example, $d(00100111...) = 3$. Is $d$ continuous? The preimage of a singleton integer set $d^{-1}(n)$ is a basic open subset in the product topology, but $$ d^{-1}(U_n) = \{000...\} \cup \bigcup_{m\ge n} d^{-1}(n) $$ Since $S$ is homeomorphic to the Cantor set, $S$ has no isolated points so $\{000...\}$ is not open. Is $d^{-1}(U_n)$ open?

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    $\begingroup$ "is there a way to write X as a union of open sets, or is X not open?" is just a length way to mean "is X open?" $\endgroup$ Commented Apr 3, 2024 at 17:07
  • $\begingroup$ In any case, you should be able to fully describe what $d^{-1}(U_n)$ is, and then figure out whether it is open. $\endgroup$ Commented Apr 3, 2024 at 17:09
  • $\begingroup$ I have edited the question to be less redundant. $\endgroup$ Commented Apr 3, 2024 at 18:22

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Yes, $d$ is continuous.

Another description of $d^{-1}(U_n)$ is that it is all sequences that start with $n$ zeros. $$ d^{-1}(U_n) = \{ 00 \cdots 00a_{n+1} a_{n+2} \cdots \} $$ This is also a basic open subset of $S$. Since the preimage of any basic open subset is open, $d$ is continuous.

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