Let $A_1, A_2, A_3, \dots$ be subsets of a metric space.
If $B=\bigcup_{i=1}^\infty A_i$, prove that $\overline{B}\supset \bigcup_{i=1}^\infty \overline{A_i}.$
Show, by an example, that this inclusion can be proper.
Proof: We'll prove that $\left(\overline{\bigcup_{i=1}^\infty A_i}\right)^c\subset \left(\bigcup_{i=1}^\infty \overline{A_i}\right)^c.$ Let $x\in\left(\overline{\bigcup_{i=1}^\infty A_i}\right)^c$ then $x\notin\overline{\bigcup_{i=1}^\infty A_i}=\left(\bigcup_{i=1}^\infty A_i\right)\cup\left(\bigcup_{i=1}^\infty A_i\right)'$. Hence $x\not\in \bigcup_{i=1}^\infty A_i$ and $x\not\in \left(\bigcup_{i=1}^\infty A_i\right)'$ then $x\notin A_i$ for any $i\geqslant 1$ and $x$ is no limit point of $\bigcup_{i=1}^\infty A_i$. Then exists deleted neighborhood of point $x$ such that $U'(x)\cap\left(\bigcup_{i=1}^\infty A_i\right)=\varnothing$ then $U'(x)\cap A_i=\varnothing$ for all $i\geqslant 1$. Hence $x\notin A_i$ and $x\notin A_i'$ for all $i$. Then $x\notin \overline{A_i}$ for all $i\geqslant 1$ and $x\notin \bigcup_{i=1}^\infty \overline{A_i}$. Then $x\in \left(\bigcup_{i=1}^\infty \overline{A_i}\right)^c$. Therefore, $\left(\overline{\bigcup_{i=1}^\infty A_i}\right)^c\subset \left(\bigcup_{i=1}^\infty \overline{A_i}\right)^c.$ QED
My example is such: Let $A_i=(-\infty; -1/i)\cup(1/i; +\infty)$ then $\overline{A_i}=(-\infty; -1/i]\cup[1/i; +\infty)$. Then $\bigcup_{i=1}^\infty \overline{A_i}=\mathbb{R^1}\setminus\{0\}$ but $\overline{\bigcup_{i=1}^\infty A_i}=\overline{\mathbb{R^1}\setminus \{0\}}=\mathbb{R^1}$. Therefore $\mathbb{R^1}\setminus\{0\}$ is a proper subset of $\mathbb{R^1}$.
Are my proof and example true?
