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Let $T:E\to F$ be a continuous map between topological spaces. For an ordinal $\beta$ suppose that $\{B_\alpha:\alpha<\beta\}$ is a collection of open subsets of $E$ such that:

  1. $\alpha_1<\alpha_2<\beta\Rightarrow B_{\alpha_2}\subset B_{\alpha_1}$
  2. $\bigcap_{\alpha<\beta}B_\alpha\neq\emptyset$
  3. $\overline{T(B_\alpha)}$ is open for each $\alpha<\beta$. (Here $\overline{A}$ denotes the closure of $A$.)
  4. $\bigcap_{\alpha<\beta}\overline{T(B_\alpha)}\neq\emptyset$
  5. Both sets $\overline{T(\cap_{\alpha<\beta}B_\alpha)}$ and $\bigcap_{\alpha<\beta}\overline{T(B_\alpha)}$ are open.

Question: Does the following equality holds? $$\overline{T(\cap_{\alpha<\beta}B_\alpha)}=\bigcap_{\alpha<\beta}\overline{T(B_\alpha)}$$

I have the inclusion in one direction: since $T(\cap_{\alpha<\beta}B_\alpha)\subseteq\bigcap_{\alpha<\beta}T(B_\alpha)\subseteq\bigcap_{\alpha<\beta}\overline{T(B_\alpha)}$ holds, and the last set is closed, we have that $\overline{T(\cap_{\alpha<\beta}B_\alpha)}\subseteq\bigcap_{\alpha<\beta}\overline{T(B_\alpha)}$. But the other inclusion is hard to get.

Proofs, counterexamples and/or references are welcome.

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1 Answer 1

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Note that 2. implies 4. Also note that if the spaces $E$, $F$ are discrete, then you have 3. and 5. for free, and your question reduces to whether $T(⋂_{α < β} B_α) = ⋂_{α < β} T(B_α)$. So try to solve this reduced situation (if you find a counterexample here, you will be done, if you prove the equality here, than you will have a partial answer and you can try to generalize your proof). The point is that looking at a simplified special situation is a good idea in general when you have no idea how to prove a general theorem.

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