Let $V$ an infinite dimensional vector space.
How to show that the cardinality of $V$ is the same of a basis of $V$?
I saw this argument here link in the main answer (MathOverFlow).
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1$\begingroup$ If $V$ is nontrivial $K$-vector space then $|V|\ge|K|$. Especially if $K$ is uncountable and $V$ has countable basis (e.g. $\Bbb{R}^\infty$) then the cardinality of $V$ is not equal to the cardinality of the basis of $V$. $\endgroup$– Hanul JeonOct 27, 2013 at 4:41
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You need the additional assumption that $\dim_F V \ge |F|$ (where $F$ is the base field).
Let $V$ be a vector space over a field $F$ and let $B \subseteq V$ be a basis. Suppose $B$ is infinite. There is a surjection $(F \times B)^{<\omega} \to V$ given by $$(( a_i, v_i) : i < n) \mapsto \sum_{i<n} a_iv_i$$ So we have $$|V| \le |(F \times B)^{<\omega}| = \sum_{n<\omega} |F \times B|^n \le \aleph_0 \cdot |F| \cdot |B| = \max \{ \aleph_0, |F|, |B| \}$$ So if $B$ is infinite and $|B| \ge |F|$ then we get $|V| \le |B| = \dim_F V$. (And obviously $|V| \ge \dim_F V$ so we have equality.)
answered Oct 27, 2013 at 4:59
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$\begingroup$ I guess I got it, but to be sure that I correctly understood: what you mean by $(F\times B)^{<\omega}$? $\endgroup$– BinaiOct 27, 2013 at 5:05
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1$\begingroup$ Sorry I should have clarified. If $X$ is a set then $X^{<\omega}$ denotes the set of finite sequences of elements of $X$. That is $$X^{<\omega} = \bigcup_{n=0}^{\infty} X^n$$ $\endgroup$ Oct 27, 2013 at 5:06