Example of no inclusion in set of closures

Question

Let $(X,d)$ be a metric space and $A_j \subseteq X$ for $j = 1,2,...$ . Let $B = \bigcup_{j=1}^\infty A_j$.

Find an example for which $\bigcup_{j=1}^\infty \bar{A_j} \neq \bar{B}$.

Also, in general what $A_j$ needs to be so that $\bigcup_{j=1}^\infty \bar{A_j}$ is a proper set of $ \bar{B}$.

Answer 1

What about $X=\mathbb R$ and $A_j=\{\frac1j\}$?

This gives $B=\{\frac1j; j=1,2,\dots\}$ and $\overline B\ne B$.

Each $A_j$ is closed, so $A_j=\overline A_j$ and $B=\bigcup_j A_j= \bigcup_j\overline{A_j}$, but $\overline B=\{0\}\cup B$.

Answer 2

$$A_j= \frac{1}{j} \mathbb Z$$

Answer 3

Take any separable metric space $X$ with a proper countable dense subset $B=\{a_1,a_2,\ldots\}$ of it. Let $A_j=\{a_j\}$. Then $\overline{A_j}=A_j$ for all $j$, $\bigcup_j\overline{A_j}=\bigcup_jA_j=B$, and $\overline B=X$.

For example, you could take $X=\mathbb{R}$, $B=\mathbb{Q}$.