# Getting rid of unnecessary sets from uncountable union of subsets

So if $$X$$ is a set and $$\lbrace A_i : i\in I\rbrace$$ is an uncountable collection of subsets of $$X$$, I want to simplify $$\cup_{i\in I}A_i$$ by getting rid of all sets that do not change the union i.e. the ones that are contained in the union of all of the other sets.

How can I show that $$\cup_{i\in I}A_i = \cup_{i\in J}A_i$$, where $$J=\lbrace i\in I:A_i\not\subseteq \cup_{j\neq i}A_j\rbrace$$ ?

Does it involve something along the lines of Zorn's Lemma/Axiom of Choice? I'm sure there's something really silly that I'm missing. A hint would be very much appreciated! :)

• Did you want $J = \{ i \in I : A_i \nsubseteq \bigcup_{j\neq i}{A_j}\}$? The given definition of $J$ doesn't seem to match what you've described. – Hayden Apr 10 at 0:45
• Absolutely correct, I typed it out wrong. – thewonderfulwizardofoz Apr 10 at 0:49
• In general you can’t do it: your set $J$ may be empty. For instance, take $X=\Bbb R$, and for each $x\in\Bbb R$ let $A_x=(\leftarrow,x)$. – Brian M. Scott Apr 10 at 0:51
• What if $X=I=\mathbb R$ and $A_i = [-i,i]$? Removing any single $A_k$ does not affect the union. Similarly for finitely many $A_k$s. However, you can't remove them all. – chi Apr 10 at 9:36

Counterexample: Let $$I=\mathbb R$$ and $$A_i=(-\infty,i)$$.

Then $$\bigcup_{i\in I}A_i=\mathbb R$$ and $$J=\{i\in I:A_i\not\subseteq\bigcup_{j\ne i}A_j\}=\emptyset$$ so $$\bigcup_{i\in J}A_i=\emptyset$$. In fact, this family $$\{A_i:i\in I\}$$ has no minimal subfamily which covers $$\mathbb R$$.

Another example: Consider the family $$\{A_i:i\in I\}$$ of all $$2$$-element subsets of $$\mathbb R$$. Again each $$A_i$$ is "unnecessary", but in this case there is a minimal subcover, e.g., $$\{\{x,x+1\}:\lfloor x\rfloor\text{ is even}\}$$.

In fact, a family of sets of bounded finite size always has a minimal subfamily with the same union, though it can't be obtained by simply throwing out all the "unnecessary" sets from the original family; see Taras Banakh's answer to this Math Overflow question. See this other question for some related stuff.

I think that you have a cover $$(A_i)_{i\in I}$$ and you want to find a subcover that is minimal $$(A_i)_{i\in J}$$. As the example of @bof: shows, that is not possible: $$J$$ forms a subcover if and only if $$\sup J= \infty$$ and removing a finite subset of $$J$$ still gives a subset $$J'$$ with $$\sup J' = \infty$$.

The problem with a minimal cover is that the interserction of a decresing family of covers may not be a cover, so Zorn lemma does not work.

The statement

$$\cup_{i\in I}A_i = \cup_{i\in J}A_i$$

with the given definition of $$J$$ is not true. Counterexample: $$I=\{1;2\}, \quad A_1=A_2\neq\emptyset$$ Then $$J=\emptyset$$ and $$\cup_{i\in I}A_i = A_1\neq\emptyset= \cup_{i\in J}A_i$$.

If $$I$$ is infinite, again take $$A_i=A\neq\emptyset$$ for all $$I$$. Then $$J=\emptyset$$ and the statement fails.

It would work if the indexes $$i$$ are in a well ordered set and you write $$J=\lbrace i\in I:A_i\not\subseteq \cup_{j

• $I$ has to be uncountablely infinite. – HiMatt Apr 10 at 0:41
• Ok, let it be; but the $A_i$ can be equal, so the counterexample works – DiegoG7 Apr 10 at 0:42
• Ok, to clarify: With no duplicates i.e. $A_i\neq A_j$ for all $i\neq j$ – thewonderfulwizardofoz Apr 10 at 0:47
• Anyway, I think that your definition of the set $J$ is not appropriate – DiegoG7 Apr 10 at 0:47
• Corrected definition, sorry for the confusion. Should be 'not a subset of union of all other sets' – thewonderfulwizardofoz Apr 10 at 0:50