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Let $X_\delta$ and $Y_\delta$ be index families of sets with the index set $\Delta$.

Show the following:
$$\begin{align} \prod_{\delta \in \Delta} X_\delta \cup \prod_{\delta \in \Delta} Y_\delta \subseteq \prod_{\delta \in \Delta} X_\delta \cup Y_\delta & \tag{i}\\ \prod_{\delta \in \Delta} X_\delta \cap \prod_{\delta \in \Delta} Y_\delta = \prod_{\delta \in \Delta} X_\delta \cap Y_\delta &\tag{ii} \end{align}$$

When working on this problem I arrived at a small issue. For finite intersections and unions the proofs are straightforward. However, when using arbitrary indices, how would my strategy change. I'm a bit gun shy about starting proofs for these because I'm starting to think that the arbitrary index changes everything.

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$\LaTeX$ is great, and $\Pi\neq\prod$. And display mode makes the change even more apparent! –  Asaf Karagila Feb 18 '13 at 6:39
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2 Answers 2

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This is really just about chasing after the definitions.

Recall that $f\in\prod_{\delta\in\Delta} X_\delta$ if $f$ is a function whose domain is $\Delta$ and for every $\delta$, $f(\delta)\in X_\delta$.

So suppose that $f\in\prod X_\delta\cup\prod Y_\delta$ then $f$ is a function whose domain is $\Delta$ and $f(\delta)$ is in $X_\delta$ for all $\delta$; or $f(\delta)\in Y_\delta$ for all $\delta$. In particular this means that $f(\delta)\in X_\delta\cup Y_\delta$ for all $\delta$, and so $f\in\prod(X_\delta\cup Y_\delta)$.

To see that equality may fail to holds, note that the union of the product does not contain "mixed" choice functions, that is functions that part of their images are elements from $X_\delta$ and part are from $Y_\delta$ and so if $X_\delta\cap Y_\delta=\varnothing$ equality might not hold.

The part about the intersection is very similar. One has to trace back what does it mean to be in the intersection, and what does it meant to be in the product of the intersections, but it's more or less as above.

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It's nor really an issue of whether or not $X_\delta\cap Y_\delta$ is empty. Even if it is so, the equality can still hold, and if it is not, it might not hold. The equality will not hold iff there are some $\delta_1,\delta_2$ such that $X_{\delta_1}\nsubseteq Y_{\delta_1}$ and $Y_{\delta_2}\nsubseteq X_{\delta_2}$. –  tomasz Feb 18 '13 at 15:32
    
Yes, you are right. I was not trying to be exact on when equality fails to hold, just to give an example. The simplest being where $X_\delta\cap Y_\delta$ are disjoint. –  Asaf Karagila Feb 18 '13 at 15:34
    
I got that part, but my point was, even if they are disjoint, the equality might still hold if one of them is empty, so the equality holding or not does not really depend on disjointness, only noncontainment. This is a trivial case, but still... –  tomasz Feb 18 '13 at 20:23
    
Actually if some of the $X_\delta$ are empty and some are not then equality doesn't hold either. –  Asaf Karagila Feb 18 '13 at 20:48
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The strategy should be the same, and I can't think of a proof that would change depending on whether or not $\Delta$ is finite.

Just take an arbitrary element of the LHS and show that is also an element of the RHS (and the opposite, for the second one).

Also, the unions and intersections are always finite, it's the products that can be potentially infinite in this exercise.

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