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Let $F$ be a no finite field. How can I show the equality $$ \left|\prod_{j = 1}^{\infty} F\right| = |F|? $$ The inequality $|F| \leq \left|\prod_{j = 1}^{\infty} F\right|$ is clear, but how can I show that $\left|\prod_{j = 1}^{\infty} F\right| \leq |F|$? Must I differenciate the two cases $|F| = |\mathbb{R}|$ and $|F| = |\mathbb{Q}|$? Is there no possibility to get $\left|\prod_{j = 1}^{\infty} F\right| = |\mathbb{R}|$ and $|F| = |\mathbb{Q}|$?

joseabp91
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    Well it's actually not true, for $F=\mathbb{Q}$ for instance. More generally, pick a cardinal $\lambda$ such that $\lambda^{\aleph_0} \neq \lambda$ (those can be very large) and then there is a field of size $\lambda$ that contradicts your statement – Maxime Ramzi Jul 10 '18 at 18:08
  • that's not true, for example, taking $F=\mathbb{Q}$ you get $|\prod(\mathbb{Q})|=|\mathbb{R}|$ – Gaston Burrull Jul 10 '18 at 18:09
  • Then do you mean that my statement is false? – joseabp91 Jul 10 '18 at 18:13
  • @Max - one needs to add the existence of fields of any given infinite cardinality $\lambda$, for example the field of rational functions in $\lambda$ indeterminates with $\mathbb{Q}$-coefficients. – Nicky Hekster Aug 14 '18 at 11:34
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    @NickyHekster : yes that's what I meant by "then there is a field of size $\lambda$ that contradicts your statement" - though I agree it was perhaps not clear enough (another proof can use the Löwenheim-Skolem theorem for instance) – Maxime Ramzi Aug 14 '18 at 12:05

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