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I know how to handle the 2d case: http://www.proofwiki.org/wiki/Set_Difference_of_Cartesian_Products

But I am having trouble simplifying the following:

Let $X=\prod_{1}^\infty X_i, A_i \subset X_i$

How can I simplify/rewrite $X - (A_1 \times A_2 \times \cdots A_n \times X_{n+1} \times X_{n+2} \cdots)$ with unions/intersections?

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Do you mean $A_i\in X_i$ or $A_i\subset X_i$? –  Christian Blatter Feb 18 '11 at 14:49

2 Answers 2

Try writing

$$\prod_{k=n+1}^{\infty} X_k = X'$$

then you want the difference of

$$(X_1 \times X_2 \times \cdots \times X_n \times X') - (A_1 \times A_2 \times\cdots \times A_n \times X')$$

You can use the rule that you linked inductively to this difference. Then note that in some parts of the expression you will get $X' - X' = \emptyset$.

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Set product is associative. Difference isn't. –  Asaf Karagila Feb 18 '11 at 7:28
    
@Asaf That is just what I thought. I checked wikipedia to make sure, there it says "strictly speaking set product is not associative." I think an example of this would be: how would you add $(x_1,(y_2,z_2)) + ((x_2,y_2),z_2)$ if $(x_1,(y_2,z_2)) \in X \times (Y \times Z), ((x_2,y_2),z_2) \in (X \times Y) \times Z$? I see that you know set-theory well; I would be pleased if you could explain this to me. Thanks! –  milcak Feb 19 '11 at 5:04
2  
This is a good question, actually. Strictly speaking, yes this is not associative since $\langle x,\langle y,z\rangle\rangle\neq\langle\langle x,y\rangle z\rangle$, but there is an isomorphism between the two (which I believe is natural in the categorical sense), so in the usual case we tend to ignore this fact and just treat cartesian products as $n$-tuples and not pairs of pairs of pairs. When treating them as such the product is associative indeed. –  Asaf Karagila Feb 19 '11 at 7:02

You cannot describe the indicated set $S$ without using some negation. So let $A_j':=X_j \setminus A_j$ and $\hat A_j:=\pi_j^{-1}(A_j')=\lbrace x=(x_i)_{i\geq1}\in X | x_j\in A_j'\rbrace$. Then the set $S$ can be written as $S=\bigcup_{1\leq j\leq n} \hat A_j$, because an $x\in X$ is a member of $S$ iff at least one of the first $n$ "coordinates" $x_j$ of $x$ does not belong to the corresponding $A_j$.

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