# Elementary set theory problem and what's my misconcpetion?

This is my proof for No.6, and tell me what's wrong.

1. Let $f\in \left(\prod_{i\in I}A_i)\cup(\prod_{j\in J}B_j\right)$

$\Rightarrow$ $(f\in \prod_{i\in I}A_i)\lor(f\in \prod_{j\in J}B_j)$

$\Rightarrow$ $(f:I\to$ $\bigcup_{i\in I}A_i\;\land\ f(i)\in A_i\;\forall\;i\in I)$ $\lor$ $(f:J\to$ $\bigcup_{j\in J}B_j\;\land\; f(j)\in B_j\;\forall\;j\in J)$

$\Rightarrow$ $(f: I\times\bigcup_{i\in I}A_i$ $\lor$ $f: J\times\bigcup_{j\in J}B_j)$$\land (f: I\times\bigcup_{i\in I}A_i$$\lor$$f(j)\in B_j\;\forall\;j\in J)$$\land$ $(f(i)\in A_i\;\forall\;i\in I$$\lor$$f:J\to$ $\bigcup_{j\in J}B_j)$$\land$$(f(i)\in A_i\;\forall\;i\in I$$\lor$$f(j)\in B_j\;\forall\;j\in J)$

Here is where I have been jammed in.

I thought that the first bracket at the last line like this

$(f: \left(I\cup J\right)$$\times$$\bigcup_{i\in \left(I\cup J\right)}\left(A_i\cup B_j\right))$

But I couldn't deal with the remained brackets.

then how can I deal with those brackets to make them like

$\prod_{(i,j)\in I\times J}\left(A_i\cup B_j\right)$

1. Let $\{A_i\}_{i\in I}$ and $\{B_j\}_{j\in J}$ be families of classes. Prove that $$\left(\prod_{i\in I}A_i\right)\cup\left(\prod_{j\in J}B_j\right)=\prod_{(i,j)\in I\times J}\left(A_i\cup B_j\right)\;.$$
• Are you sure that $\prod\limits_{i\in I}A_i$ is an indexed cartesian product? With that interpretation the two sides of the equals sign in your goal are not even the same kind of thing -- on the left is a set of functions with domain $I$ or $J$; on the right is a set of functions with domain $I\times J$. On the other hand, if we consider it to be a nonstandard way of writing $\bigcap\limits_{i\in I}A_i$, then the conclusion would actually be true. – Henning Makholm Jan 13 '17 at 19:24
• x @Grimza: Don't discount the possibility that there may be a typo in your source. Sometimes textbooks and other material are typeset from handwritten manuscripts by people who don't fully understand the math in them. – Henning Makholm Jan 13 '17 at 19:38
• @HenningMakholm but the problem came out immediately after defining the product of sets... so I don't think this is typo. – glimpser Jan 13 '17 at 19:43

It would be easier, not to mention easier to read, if you used more words and didn’t try to do everything symbolically. The real problem, however, is that the result is false. Let $I=J=\{0,1\}$, $A_0=\{0\}$, $A_1=\{1\}$, $B_0=\{2\}$, and $B_1=\{3\}$. Then

$$\prod_{i\in I}A_i=\{0\}\times\{1\}=\{\langle 0,1\rangle\}$$

and

$$\prod_{j\in J}B_j=\{2\}\times\{3\}=\{\langle 2,3\rangle\}\;,$$

so

$$\left(\prod_{i\in I}A_i\right)\cup\left(\prod_{j\in J}B_j\right)=\{\langle 0,1\rangle,\langle 2,3\rangle\}\;,\tag{1}$$

but

$$\prod_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)=\{0,2\}\times\{0,3\}\times\{1,2\}\times\{1,3\}\;.\tag{2}$$

The set in $(1)$ has two members, each of which is an ordered pair; the set in $(2)$ has $16$ members, each of which is an ordered $4$-tuple. Clearly the sets are not equal.

Is it possible that your source is using the very old-fashioned notation in which $\prod_{i\in I}A_i$ means what we nowadays write $\bigcap_{i\in I}A_i$, the intersection of the sets $A_i$? Because under that interpretation the result is true.

• I wish I could upvote more than once for that first sentence alone. – Henning Makholm Jan 13 '17 at 19:40
• @Brian M.Scott I've been thinking by checking some specific examples like what you did. But at the textbook, ∏i∈IAi is a product of the sets Ai and defined to be the class of functions domain with I to codomain the generalized union of Ai and f(i) should belongs to Ai for all i in I. and the intersection of sets had been defined at the other chapter already. – glimpser Jan 13 '17 at 19:41
• @Henning, you could easily afford a bounty... – Mark S. Jan 13 '17 at 19:44
• @Grimza: In that case the problem is simply wrong. – Brian M. Scott Jan 13 '17 at 19:45
• @Mark: Better he should save it for someone who could really use a boost! – Brian M. Scott Jan 13 '17 at 19:47