# Hamel Bases: Cardinality? [duplicate]

Every vector space admits a Hamel basis by AC.
That is there are maximally linear independent sets.
But how to prove their cardinalities necessarily agree?

..I couldn't really find any reference.

## marked as duplicate by Najib Idrissi, Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 15 '16 at 13:56

• The easy answer is: this is just like the finite dimensional case, only with transfinite recursion to construct the bijection between the two bases, rather than just an induction up to $n$ or whatever. – Asaf Karagila Jun 15 '16 at 13:59
If there is a finite Hamel base, the vector space is finite dimensional and we can assume to be known that any basis has the same number of elements. Suppose that $\mathcal B$ is an infinite Hamel basis for the $F$-vector space $V$. Here $F$ is the field of the scalars and I assume in this answer that $F$ has the cardinality of $\mathbb R$. Any vector is a finite linear combination of element of $\mathcal B$ in a unique way. Note that this imply that $V$ has the cardinality of $F$, because there is an injection. $$V\to \bigcup_{n=1}^\infty\left(F\cup F^2\ldots\cup F^n\right)$$ So, there are two possibilities: $\mathcal B$ is infinite but countable or $\mathcal B$ has the continuum cardinality of $F$. For the case of topological VS The first possibility is ruled out here; No infinite-dimensional $F$-space has a countable Hamel basis.