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The notation $f:E\times F \to G$ means that $f$ is a function that needs two arguments, one from $E$, one from $F$, and the image is in $G$. This is how the function $\text{Union}$ is defined: the two arguments $A$, $B$ are in $\mathcal{P}(X)$ and the image $\text{Union}(A, B) = A\cup B$ is in $\mathcal{P}(X)$.


The product of a power set to a power set is, indeed, "a set of tuples where the first of the ordered pair is a subset from first power set and the second is a subset from the second power set", to use your own words. What you reported, though, is the definition of the Union operation. You can think of it as a function that associates an element of ...


As you have said, $X\times Y = \{(x,y) \mid x\in X, y\in Y\}$. Thus, a function $f:X\times Y\rightarrow Z$ would get two arguments: one from $X$ and the other from $Y$, and output a value from $Z$. Formally, this is written as $f((x,y))=z$, and to reduce the number of brackets, its usually written as just $f(x,y)=z$. In your case, $f$ is a function that ...


Ok, so the short answer to this question seems to be a "yes, but it grows quickly in size". You can view the bloom filters for each tuple element as vectors of bits. In that case, their tensor product will be a "bloom filter" for the cartesian product, where projection can be done by taking partial traces. The downside of this is that the ...

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