Do we have $\overline{\bigcup\{A_\beta: \beta<\alpha\}}=\bigcup\{ \overline{A_\beta}: \beta<\alpha\}$

Question

Let $\alpha$ be any ordinal, and for the family $\{A_\beta: \beta< \alpha\}$, we have $A_{\beta_1}\subset A_{\beta_2}$ when $\beta_1<\beta_2<\alpha$, then do we have $$\overline{\bigcup\{A_\beta: \beta<\alpha\}}=\bigcup\{ \overline{A_\beta}: \beta<\alpha\}$$

Thanks ahead:)

Answer 1

Let $A_\beta=[0,\beta)$, then $\bar A_\beta=[0,\beta]$.

If $\alpha$ is a limit ordinal then $$\bigcup_{\beta<\alpha}\bar A_\beta=[0,\alpha)\neq[0,\alpha]=\overline{\bigcup_{\beta<\alpha} A_\beta}.$$