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

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?

• $U_n\subset Y$, or not? – Irddo Jun 8 '15 at 6:05
• $U_n$ is subset of 'Y', not 'X'. Sorry for typo. I fixed it. – Guldam Jun 8 '15 at 6:09

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 $$h^{-1}(\overline{U})= h^{-1}(\Bbb R)=\Bbb N\not\subset \Bbb N\setminus\{ h^{-1}(0)\}=h^{-1}(U)= \overline{ h^{-1}(U)}.$$
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 $$\overline{h^{-1}((0,1))} = [-\sqrt{2},-1] \cup [1, \sqrt{2}]$$.