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I want to show that, if $A$ is an infinite set and $F$ is a field, then the direct sum of copies of $F$ indexed by $A$ has strictly lower dimension than the corresponding direct product.

I know that if I can show that $A^k$, for any positive integer $k$, has the same cardinality as $A$, and if I can also show that $card(A)card(F)<card(F)^{card(A)}$, I'll be one, but how exactly do I do this?

Nishant
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  • It's not true, the cardinalities can be equal $$\operatorname{card} \left(\mathbb{R}^{\mathbb{N}}\right) = \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0} = \operatorname{card} \bigoplus_{\mathbb{N}}\mathbb{R}.$$ – Daniel Fischer May 21 '14 at 23:24
  • Oh, I see. The problem was asking about the dimensions, not cardinalities, of the vector spaces. – Nishant May 21 '14 at 23:30
  • See http://mathoverflow.net/questions/49551/dimension-of-infinite-product-of-vector-spaces – egreg May 21 '14 at 23:53
  • Note that the direct product is in fact the dual space of the sum. Then read the wonderful answers to this question. – Asaf Karagila May 22 '14 at 00:21

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