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Let $K$ be a field, $S$ an arbitrary set, and $K^S$ denote the vector space of functions from $S$ to $K$. What is the dimension of this space?

By dimension, I mean the cardinality of a Hamel basis. I think the answer can only depend on $|S|$ and $|K|$. I am also assuming the axiom of choice, thus the question makes sense.

WillG
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  • This might be interesting because $K^S$ is the dual of $K^{(S)}$ (the functions for $S$ to $K$ with finite support): https://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-i – Jochen Apr 17 '20 at 15:17
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    If $|S|\geq|K|\geq\aleph_0$, then the vector space has cardinality $2^{|S|}>|K|$, so by basic cardinal arithmetic arguments, the dimension is $2^{|S|}$. I imagine that this holds even if we only require that $S$ is infinite. – Asaf Karagila Apr 17 '20 at 19:49
  • @AsafKaragila I'm not quite seeing the justification for your first or second claim, though my cardinal arithmetic is a bit rusty. – WillG Apr 17 '20 at 21:22
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    @Will: If $|V|>|K|$, then $\dim V=|K|$. So if $2^{|S|}>|K|$, we're there already. See also https://math.stackexchange.com/a/194287/622. – Asaf Karagila Apr 17 '20 at 21:24
  • See https://mathoverflow.net/questions/49551/dimension-of-infinite-product-of-vector-spaces – Eric Wofsey Apr 22 '20 at 21:10
  • @EricWofsey If I'm reading that post correctly, then $\dim K^S=|K^S|$, is that right? I'd still love to see a proof or proof sketch of this theorem (I don't have access to the book referenced in the other post). – WillG Apr 23 '20 at 15:25
  • That's correct (assuming $S$ is infinite). There is a complete proof in Pierre-Yves Gaillard's answer. – Eric Wofsey Apr 23 '20 at 16:13
  • @EricWofsey I guess I'm confused by the notation. Is his $\Pi_{i\in I}V_i$ the same as my $K^S$, where $K$ is thought of as a $K$–vector space and the "product" over $S$ in $\Pi_{s\in S}$ is the same as "exponentiating" to the power $S$? Also is $\Pi_{i \in I}$ the same as $\oplus_{i\in I}$? As you can tell I'm pretty new to this subject. – WillG Apr 23 '20 at 16:24
  • If $V_i=K$ for all $i$ then $\prod_{i\in I}K$ is the same as $K^S$. $\bigoplus_{i\in I}K$ is different, it is the elements of $K^S$ with only finitely many nonzero coordinates. – Eric Wofsey Apr 23 '20 at 16:30
  • @EricWofsey Got it, thanks. – WillG Apr 23 '20 at 16:39

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