If $B_{n}=\bigcup _{i=1}^{n}A_{i}$, prove that $\bar{B}_{n}=\bigcup _{i=1}^{n}\bar{A}_{i}$.

Question

If $B_{n}=\bigcup _{i=1}^{n}A_{i}$, prove that $\bar{B}_{n}=\bigcup _{i=1}^{n}\bar{A}_{i}$, where $\bar{E}$ is defined as the closure set of $E$, which is the union of $E$ and its limit points $E'$.

I am trying to prove the reverse containment and I saw a proof online.

I don't understand what is underlined in red. How come $p\in \bar{B}_{n}$? Isn't $p$ a limit point of $A_{i_{0}}$ and so it's not necessarily contained in $A_{i_{0}}$?

Answer 1

If $p \in A_{i_o}$, we have $p \in \bar{B}_n$.

Suppose not, then $p \in A_{i_0}'$, then for every neighborhood, we can find a point , $x \in B_n$, hence $p$ is a limit point of $B_n$. Hence $p \in B_n'$.

Hence $p\in \bar{B}_n$