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Given a vector space of finite dimension, can we always find an injective map to the natural numbers?


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$\Bbb R$ is a one-dimensional vector space over $\Bbb R$. – Brian M. Scott Feb 4 '13 at 1:53
@BrianM.Scott, is this true in ZF+AD? – alancalvitti Feb 4 '13 at 2:59
@alancalvitti: The reals are always uncountable. – Brian M. Scott Feb 4 '13 at 3:05
@BrianM.Scott, what about in ZF? 1. Vector spaces may have no bases. 2. Vector spaces may have two bases with different cardinalities. (Herrlich AC) – alancalvitti Feb 4 '13 at 3:07
Furthermore what Brian wrote is always true. Every field is a one dimensional vector space over itself. – Asaf Karagila Feb 4 '13 at 9:14
up vote 12 down vote accepted

Depends over what field. If the field is finite or countable, e.g. $\mathbb Q$, then yes. If the field is uncountable, e.g. $\mathbb R$, then no.

The reason is that $|\mathbb F^n|=|\mathbb F|^n$, and if $|\mathbb F|\leq\aleph_0$ then $|\mathbb F|^n\leq\aleph_0$.

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You rock, thank you. – Ziggy Feb 6 '13 at 1:04

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