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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 $A_i\subset X_i$?2011-02-18

2 Answers 2

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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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    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.2011-02-19
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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 $A_j$.