Union of closure of two set and closure of union of two sets: what's wrong in general case

Question

I know that there are many question with the same title here on MSE, but I still confuse with my approach to solve this, so I still put it here for proof verification and I do want to know is there anything wrong with part b and why it fails in part c, in the case of arbitrary family of subsets.

b. $\overline{A\cup B}=\bar{A}\cup \bar{B}$

Let $x\in \overline{A\cup B}$ then for any open neighborhood $U_x$; $U_x\cap \left(A\cup B\right)\neq \emptyset$.

Then $U_x\cap A\neq \emptyset$ or $U_x\cap B\neq \emptyset$ hence $x\in \bar{A}$ or $x\in \bar{B}\Rightarrow x\in \bar{A}\cup \bar{B}$. So $\overline{A\cup B}\subseteq \bar{A}\cup \bar{B}$.

Similarly, if $x\in \bar{A}\cup \bar{B}$ then $x\in \bar A$ or $x\in \bar B$. Thus for any open neigborhood $U_x$ of x: $U_x\cap A \neq \emptyset$ or $U_x\cap B\neq \emptyset$

$\Rightarrow U_x\cap (A\cup B)\neq \emptyset$ or $x\in \overline{A\cup B}$.

Therefore, $\overline{A}\cup \overline{B}\subseteq \overline{A\cup B}$. So $\overline{A\cup B} = \overline{A}\cup \overline{B}$.

c. $\overline{\cup_\alpha A_\alpha}=\cup_\alpha\overline{A_\alpha}$ For this part, the same method applies here.

If x belongs to the LHS, then any neighborhood $U_x$ of $x$ intersects $\cup_\alpha A_\alpha$ equivalently $U_x$ intersects one of $A_\alpha$ for some $\alpha$ therefore $x\in \overline{A_\alpha}$ and hence $x\in \cup_\alpha\overline{A_\alpha}\Rightarrow \overline{\cup_\alpha A_\alpha}\subseteq \cup_\alpha\overline{A_\alpha}$.

If $x$ belongs to the RHS then $x\in \overline{A_\alpha}$ for some $\alpha$. So for any neighborhood $U_x$ of $x$: $U_x\cap A_\alpha\neq \emptyset\Rightarrow U_x\cap (\cup A_\alpha) \neq \emptyset$ so $x\in \overline{\cup_\alpha A_\alpha}\Rightarrow \cup_\alpha \overline{A_\alpha}\subseteq \overline{\cup_\alpha A_\alpha}$

Therefore $\overline{\cup_\alpha A_\alpha}=\cup_\alpha\overline{A_\alpha}$.

Please explain to me. Thank you very much.

Answer 1

The b part is fine.

I recommend you to write c with quantifiers to see it clearly. The wrong content is $\subset$. I name $\mathcal{T}$ the topology of your space.

Let be $x\in \overline{\cup_\alpha A_\alpha} \Rightarrow \forall U \in \mathcal{T}, x \in U : U \cap (\cup_\alpha A_\alpha) \neq \emptyset \Rightarrow \forall U \in \mathcal{T}, x\in U \, \, \exists \alpha \, : U \cap A_\alpha \neq \emptyset $. Due to the order of the quantifiers, that is not equivalent to $\exists \alpha \, \forall U \in \mathcal{T}, x\in U : U \cap A_\alpha \neq \emptyset $, what it does mean be in the clausure of some $A_\alpha$, so you cannot deduce that.

In fact, you can think in the sets $(-1+1/n,1-1/n)$ with $n\in \mathbb{N}, n\geq 2$. We have that $$\bigcup_{n\geq 2} (-1+1/n,1-1/n) = (-1,1)$$ so $$\overline{\bigcup_{n\geq 2} (-1+1/n,1-1/n)} = \overline{(-1,1)}=[-1,1]$$ but $$\bigcup_{n\geq 2} \overline{(-1+1/n,1-1/n)} = \bigcup_{n\geq 2} [-1+1/n,1-1/n] = (-1,1)$$ and you have a counterexample.

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Answer 2

• Part b is already flawed (I mean: the result is true but your proof is wrong).

  • Your proof in b is flawed at the following point: "for any open neighborhood $U$ of $x$ [...] $U∩A≠∅$ or $U∩B≠∅$ hence $x∈\bar A$ or $x∈\bar B$" is not correct because, denoting by $O_x$ the set of open neighborhoods of $x$, from $$\forall U\in O_x(U∩A≠∅\lor U∩B≠∅)$$ alone, you cannot logically derive $$(\forall U\in O_x,U∩A≠∅)\lor (\forall U\in O_x,U∩B≠∅).$$
  • To summarize, you are pretending that $\forall U,(P(U)\lor Q(U))$ implies $(\forall U,P(U))\lor(\forall U,Q(U)),$ which is a (frequent) mistake. For instance: $$[\forall t,(t=\varnothing\lor t\ne\varnothing)]\not\Rightarrow[[(\forall t,t=\varnothing)\lor(\forall t,t\ne\varnothing)].$$
  • The proof for b can be repaired, as you can see in many posts on this site.

• The flaw in your similar part c is perhaps more blatant because there, the result itself is false.

  • The mistake is at the following point: "any neighborhood $U$ of $x$ [...] intersects one of $A_\alpha$ for some $\alpha$ therefore $x\in \overline{A_\alpha}$". The problem is that $α$ depends on $U.$
  • Here you are pretending that $\forall U\exists\alpha,P(U,\alpha)$ implies $\exists\alpha\forall U, P(U,\alpha).$ This is the same kind of mistake as in b. The existential quantifier $\exists$ is like an "infinite $\lor$", and this analogous claim is similarly wrong. For instance: $$[\forall s\exists t(t=s)]\not\Rightarrow[\exists t\forall s(t=s)].$$ (Think also to the difference between continuity and uniform continuity, in the $\epsilon-\delta$ definition.)
  • The proof of c cannot be repaired, as you already know (since this is the reason of your post), and there are counterexamples in many posts on this site.