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Corollary 4.8: If $|\Delta|=n<\infty$, then $\newcommand{\Hom}{\operatorname{Hom}}\Hom(\prod_{i\in\Delta}V_i,W)\cong\prod_{i\in\Delta}\Hom(V_i,W)$.

In general, Corollary 48 [sic] is false when $|\Delta|=\infty$. For example, if $W=F$ and $V_i=F$ for all $i\in\Delta$, then the reader can easily see that $|\Hom_F(\prod_{i\in\Delta}F,F)|>|\prod_{i\in\Delta}F|$ when $\Delta$ is infinite. Since $\Hom_F(F,F)\cong F$, we see that $\Hom(\prod_{i\in\Delta}F,F)$ cannot be isomorphic to $\prod_{i\in\Delta}\Hom(F,F)$.

Brown, William C., A second course in linear algebra, Wiley-Interscience Publication. New York etc.: Wiley. x, 264 p., £ 30.50 (1988). ZBL0642.15001.

Ironically, I fail to see that $|\Hom_F(\prod_{i\in\Delta}F,F)|>|\prod_{i\in\Delta}F|$ when $\Delta$ is infinite.

I interpret $\Hom_F(\prod_{i\in\Delta}F,F)$ to mean the set of all $F$-linear transformations mapping from $\prod_{i\in\Delta}F=F\times F\times\cdots$ to $F$, but I do not understand what $|\Hom_F(\prod_{i\in\Delta}F,F)|$ and $|\prod_{i\in\Delta}F|$ actually mean.

If $|\cdot|$ allows me to obtain the dimension of the $\Hom$ and the dimension of the vector space $F\times F\times\cdots$, then shouldn't $|\prod_{i\in\Delta}F|$ be $\infty$ because $F\times F\times\cdots$ is a vector space of infinite dimension? From this, wouldn't it hold that $|\Hom_F(\prod_{i\in\Delta}F,F)|\leqslant|\prod_{i\in\Delta}F|$?

From another viewpoint, is it not possible to define a meaningful linear transformation $\Psi:\Hom(\prod_{i\in\Delta}V_i,W)\to\prod_{i\in\Delta}\Hom(V_i,W)$ by $T\mapsto(T\theta_1,T\theta_2,\cdots)$ where $\Psi$ is both injective and surjective, for infinite $\Delta$? Here, $\theta_q:V_q\to V$ is the $q$-th injection.

What am I doing wrong here?

Community
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Frenzy Li
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    $|\cdot|$ means cardinality. For every set $,\Delta$ it is true that $Hom(\bigoplus_{i \in \Delta}V_i,W)\cong\prod_{i \in \Delta}Hom(V_i,W)$, but, when $\Delta$ is infinite $\bigoplus_{i \in\Delta}V_i \subsetneq \prod_{i \in\Delta}V_i$ and 'intuitively' $Hom(\prod_{i \in\Delta}V_i,W)$ should be bigger than $Hom(\bigoplus_{i \in\Delta}V_i,W)$, if you can prove that $|Hom(\prod_{i \in\Delta}F,F)|>|\prod_{i \in\Delta}F|$ ($F \cong Hom(F,F)$), then those spaces cannot be isomorphic. Corollary 4.8 is true assuming that $\Delta$ is finite since finite direct products and finite direct sums coincide. – Gilberto López Jul 19 '17 at 05:10
  • @GilbertoLópez I have not yet learnt $\oplus$. Is there another path to feel the intuition? I promise to read your comment again once I read through $\oplus$. – Frenzy Li Jul 19 '17 at 05:21
  • in that case, Lord Shark the Unknown gives a better idea. – Gilberto López Jul 19 '17 at 05:28
  • @FrenzyLi : Sorry for this very late comment but since a few years have passed, I really wanted to know what was your ultimate experience with these gaps in explanation in Brown's book. I recently picked up this book and had this same question and was baffled by the complicated explanation below. – math-physicist Jul 07 '22 at 10:53
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    @sadman-ncc Hello and I am happy to share some thoughts. I studied this book for my qualifying exam and I compiled a list (e.g. "How to understand the spaces B and U on p. 64?") of statements that I cannot fathom after 20 minutes of concentrated construction on paper. I later asked a (now committee member) professor for their intuition, and they quite happily explained to me everything. The construction in this book is definitely worth it and can greatly sharpen your understanding. I consider asking for occasional assistance a must as these gaps reflected my lack of background knowledge. – Frenzy Li Jul 08 '22 at 03:31
  • @FrenzyLi : Thanks for your reply. Would you want to continue this conversation in chat? I had some other "logistic" questions regarding Brown's book and its exercises. – math-physicist Jul 08 '22 at 05:07
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    @sadman-ncc Don't know how much I could help but I am glad to move over to Linear & Abstract Algebra – Frenzy Li Jul 08 '22 at 06:03

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I presume $F$ is a field. Let's write $V$ instead of $\prod_{i\in \Delta}F$. Then $V$ is an infinite-dimensional vector space over $F$. Suppose the dimension of $V$ is $\frak a$. Then $V^*=\text{Hom}(V,F)$ is isomorphic to $F^{\frak a}$. The cardinality of $V$ is $\max(|F|,\frak a)$ and that of $V^*$ is $|F|^{\frak a}$. Certainly $|F|^{\frak a} \ge 2^{\frak a}>\frak a$ (Cantor's theorem). But is it necessarily the case that $|F^{\frak a}|>|F|$?

I claim that $|{\frak a}|\ge |F|$. There is a well-known trick to prove this: $V$ contains a subspace isomorphic to the set of sequences $(a_0,a_1,a_2,\ldots)$ ($a\in F$). In this space the elements $(1,a,a^2,\ldots)$ for $a\in F$ are linearly independent, so ${\frak a}\ge|F|$. Therefore $|F^{\frak a}|\ge2^{\frak a}>|F|$?

In any case, there are also simple examples. Take $F=\Bbb Q$ and $\Delta=\Bbb N$. Then $|V|=2^{\aleph_0}$ and $|V^*|=2^{2^{\aleph_0}}$.

Angina Seng
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  • Wow, this is a lot more info than I had expected... First, if we let $\dim V=\frak a$, does this mean that $|\Delta|=\frak a$? – Frenzy Li Jul 19 '17 at 05:18
  • @FrenzyLi In general $\frak a$ will not be the same as $|\Delta|$. That's basically the point. – Angina Seng Jul 19 '17 at 05:27
  • So, even when $V=\prod_{i\in \Delta}F$, $|V|=|\Delta|$ does not generally hold? – Frenzy Li Jul 19 '17 at 05:28
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    @FrenzyLi That is never true (Cantor). – Angina Seng Jul 19 '17 at 05:31
  • So, the takeaway lesson is that, $|\hom_F(\prod_{i\in\Delta}F,F)|=|F|^{|F|^{\aleph_{0}}}$, $|\prod_{i\in\Delta}F|=|F|^{\aleph_{0}}$, and as long as $|F|>1$ (well, a field always contains at least 2 elements), the inequality holds? – Frenzy Li Jul 19 '17 at 06:05
  • It is not given that $|\Delta|=\aleph_0$. Anyway, I suspect the author is being lazy; the statement he gives is true, but is non-trivial, not something one can "easily see". – Angina Seng Jul 19 '17 at 06:07
  • If $\Delta$ is finite, does $|\Hom_F(\prod_{i\in\Delta}F,F)|=|\prod_{i\in\Delta}F|$ hold? If it doesn't, wouldn't the same argument work for finite $\Delta$? I ask this because if we pick $F=\mathbb Q$ and $\Delta={1,2,3}$, then $|V|={\aleph_0}^{3}={\aleph_0}$ and $|V^*|={\aleph_0}^{\aleph_0^3}=2^{\aleph_0}$? – Frenzy Li Jul 19 '17 at 06:41
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    When $\Delta$ is finite $\prod_{i \in \Delta}F$ is a finite dimensional vector space over $F$, its dimension is $|\Delta|$, and a finite dimensional vector space is always isomorphic to its dual space, in this case $Hom(\prod_{i \in \Delta} F,F)$, thus both sets have the same cardinality. – Gilberto López Jul 21 '17 at 00:12