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In the Roman's book (Advanced Linear Algebra) he defines the direct product of a family of vector spaves over $\mathbb{F}$ as follows:

Definition: Let $\mathcal{F}=\{V_{i}| i\in K\}$ be any family of vector spaces over $\mathbb{F}$. The direct product of $\mathcal{F}$ is the vector space $$\prod_{i\in K}V_{i}=\{f:K\rightarrow\cup_{{i\in K}} V_{i}|f(i)\in V_{i}\}$$ thought of as a subspace of the vector space of all functions from $K$ to $\cup_{{i\in K}}V_{i}$. ( Here $K$ is a set of indexes).

I don't understand how $V=\{f:K\rightarrow \cup_{{i\in K}}V_{i}\}$ is a vector space over the field $\mathbb{F}$. Is the set $\cup_{{i\in K}}V_{i}$ a vector space over $\mathbb{F}$? I can't see how.

Thanks for your help.

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I've made the corrections. –  spohreis Jan 25 '12 at 21:35

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up vote 4 down vote accepted

You are correct, the direct product should have been defined as in Arturo's answer*, and there is not generally a vector space structure on the set of functions from $K$ to $\bigcup\limits_{i\in K}V_i$. I searched but did not find an online list of errata for the text. For reference, the definition is on page 41 of the text.

*Arturo's answer is now deleted, but briefly, the set is correctly named (it is the Cartesian product), but the operations are defined pointwise (which wouldn't generally make sense for arbitrary functions from $K$ to $\cup_{i\in K}V_i$).

There is a special case where it would work. If $\{V_i:i\in K\}$ is a chain under inclusion as subspaces (or at least directed), then $\cup_{i\in K}V_i$ is a vector space.

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Hmph. I tried to look at the text in the 3rd edition on-line, but the page with this definition is the first one that is not available for preview on Springer's site. I do remember that I found even the second edition contained a lot of errors... –  Arturo Magidin Jan 25 '12 at 21:45

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