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Let $X$ be a (real or complex) infinite dimensional vector space. (Not Normed or Banach one).

Is every Hamel Basis for $X$ necessarily uncountable ?

Neil hawking
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    The $\mathbb{R}$-vector space of sequences of real numbers in which almost all terms are zero is generated by the sequences that are all zero except for one value. – conditionalMethod Nov 25 '19 at 18:54
  • @conditionalMethod Post your comment as an answer. In fact I think it's the answer, up to isomorphism. – Ethan Bolker Nov 25 '19 at 19:00
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    not necessarily, by example the polynomial ring, that it is also a vector space over $\Bbb R $ or $\Bbb C $, have an infinite countable basis. However if the space is complete (if it is Banach) then the basis is necessarily uncountable –  Nov 25 '19 at 19:30
  • Generally, no. But yes, in the case of a Banach space. – Michael Nov 15 '22 at 03:33

1 Answers1

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Take any infinite dimensional vector space. Take a set $S$ of countably infinitely many linearly independent vectors from that vector space. Be $V$ the subspace spanned by $S$ (using finite linear combinations). Then the vectors in $S$ form a Hamel basis of $V$.

Thus $V$ has a countably infinite Hamel basis.

Note however, that in a vector space that has an uncountable Hamel basis, all other Hamel bases are also uncountable, as all Hamel bases of the same vector space have the same cardinality.

celtschk
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