Gretchen Schwartz

Answered

2022-07-01

I am reading one book and the author presents what he calls a distributivity formula:

$\bigcup _{j\in J}\bigcap _{i\in {I}_{j}}{F}_{i}^{j}=\bigcap _{\{{i}_{j}\}\in K}\bigcup _{j\in J}{F}_{{i}_{j}}^{j}$

where $K={\mathrm{\Pi}}_{j\in J}{I}_{j}$ (the set of all sequences {${i}_{j},j\in J$}).

My question is regarding the notation. The LHS is quite clear to me. However, I'm now sure what are actually the members of K. Say $J=\{1,2,\dots \}$ and ${I}_{j}=\{1,2,\dots \}$. What is actually ${i}_{j}$ here?

$\bigcup _{j\in J}\bigcap _{i\in {I}_{j}}{F}_{i}^{j}=\bigcap _{\{{i}_{j}\}\in K}\bigcup _{j\in J}{F}_{{i}_{j}}^{j}$

where $K={\mathrm{\Pi}}_{j\in J}{I}_{j}$ (the set of all sequences {${i}_{j},j\in J$}).

My question is regarding the notation. The LHS is quite clear to me. However, I'm now sure what are actually the members of K. Say $J=\{1,2,\dots \}$ and ${I}_{j}=\{1,2,\dots \}$. What is actually ${i}_{j}$ here?

Answer & Explanation

Marisol Morton

Expert

2022-07-02Added 13 answers

On the RHS, we are finding the intersection of $\bigcup _{j\in J}{F}_{{i}_{j}}^{j}$. Here, the ${i}_{j}$ is the values for which we are intersecting; it says $\{{i}_{j}\}\in K$. If $J=\{1,2,3,\cdots \}$, then $K={I}_{1}\times {I}_{2}\times {I}_{3}\times \cdots $

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