2
$\begingroup$

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! :)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hayden Apr 10 at 0:45
  • $\begingroup$ Absolutely correct, I typed it out wrong. $\endgroup$ – thewonderfulwizardofoz Apr 10 at 0:49
  • 2
    $\begingroup$ 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 $A_x=(\leftarrow,x)$. $\endgroup$ – Brian M. Scott Apr 10 at 0:51
  • $\begingroup$ 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. $\endgroup$ – chi Apr 10 at 9:36
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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 $J=\lbrace i\in I:A_i\not\subseteq \cup_{j<i }A_j\rbrace$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.