Munkres exercise 17.6 asks in part (b) to prove that for subspaces $A,B$ of a topological space $X$, $\overline{A \cup B} = \overline{A} \cup \overline{B}$. I didn't have any trouble with this proof. In part (c), he generalizes the result to arbitrarily many sets, but states that equality does not hold. Now, we have only
$$\overline{\bigcup\limits_{\alpha \in I} A_{\alpha}} \supset \bigcup\limits_{\alpha \in I} \overline{A}_{\alpha}.$$
I didn't have trouble coming up with a counterexample to show that $\subset$ doesn't hold, but I'm concerned only with where my proof for two sets breaks down. The proof relies on there being only finitely many sets, but my proof doesn't make mention of the fact that there are only finitely many (e.g., I'm not taking a max or a min, which can only, for sure, be done for a finite set), which leads me to believe that it is insufficiently rigorous. Below is my proof of part (b) showing $\overline{A \cup B} = \overline{A} \cup \overline{B}$.
By definition, we have $\overline{A\cup B} \supset A \cup B$, so $\overline{A\cup B} \supset A$ and $\overline{A\cup B} \supset B$. That is, $\overline{A \cup B}$ is a closed set containing $A$, so $\overline{A} \subset \overline{A\cup B}$. Analagously, $\overline{A\cup B}$ is a closed set containing $B$, so $\overline{B} \subset \overline{A\cup B}$. Therefore, $\overline{A}\cup \overline{B} \subset \overline{A\cup B}$. Conversely, $\overline{A}$ and $\overline{B}$ are closed sets which contain $A$ and $B$, respectively. Therefore, $\overline{A} \cup \overline{B}$, the union of closed sets, is a closed set containing $A \cup B$, so $\overline{A\cup B} \subset \overline{A} \cup \overline{B}$. Therefore, $\overline{A \cup B} = \overline{A}\cup \overline{B}$, as required.
If I were to try to generalize the backward inclusion, I would proceed as follows. We have $\overline{\bigcup\limits_{\alpha \in I} A_{\alpha}} \supset \bigcup\limits_{\alpha \in I} A_{\alpha}$. For every $\alpha$, $\overline{\bigcup\limits_{\alpha \in I} A_{\alpha}}$ is a closed sets containing $A_{\alpha}$, so $A_{\alpha} \subset \overline{\bigcup\limits_{\alpha \in I} A_{\alpha}}$. therefore, $\bigcup\limits_{\alpha \in I} \overline{A}_{\alpha} \subset \overline{\bigcup\limits_{\alpha \in I} A_{\alpha}}$.
This proof is certainly incorrect, but I'm not sure where. I'm considering an "arbitrary number of set inclusions" in the line beginning "for each $\alpha$." My guess is that this line of reasoning is disallowed by the laws of set theory. Though I can provide a counterexample to this backward inclusion, I would really like to know why this proof is wrong and which step is illegal.
Best Answer
This proof seems correct and the conclusion is correct. What is incorrect is the opposite inclusion. The proof from two sets breaks because the union of infinitely many closed sets don't need to be closed.