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Let $\{ h_n :X \to Y\}_{n \in \mathbb{Z^+}} $ be a sequence of continuous functions from a topological space $X$ to another topological space $Y$, and for each $n$ let $U_n$ be an open subset of $Y$.

Does $$ \bigcup_{n=1}^{\infty} {h_n}^{-1} (\overline{U_n}) \subset \overline{ \bigcup_{n=1}^{\infty} {h_n}^{-1} (U_n) } $$ holds?

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  • $\begingroup$ $U_n\subset Y$, or not? $\endgroup$ – Irddo Jun 8 '15 at 6:05
  • $\begingroup$ $U_n$ is subset of 'Y', not 'X'. Sorry for typo. I fixed it. $\endgroup$ – Guldam Jun 8 '15 at 6:09
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No, even for one function $h$. Let $X$ be $\Bbb N$ endowed with the discrete topology, $Y$ be $\Bbb R$ endowed with the standard topology, $h(n):X\to Y$ be a enumeration of the rationals and $U=\Bbb R\setminus\{0\}$. Then $$h^{-1}(\overline{U})= h^{-1}(\Bbb R)=\Bbb N\not\subset \Bbb N\setminus\{ h^{-1}(0)\}=h^{-1}(U)= \overline{ h^{-1}(U)}.$$

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A counterexample with one function is as follows. Let $X = \mathbb R$ and $Y = [0,+\infty)$, both with standard topologies. Let $h : X \to Y$ be defined by
$$ h(x) = \max\{ 0, x^2-1\}. $$ Let $U = (0,1)$. Then $\overline{U} = [0,1]$ and $$ h^{-1}([0,1]) = [-\sqrt{2}, \sqrt{2}] $$ and $$ \overline{h^{-1}((0,1))} = [-\sqrt{2},-1] \cup [1, \sqrt{2}] $$.

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