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Assume that $V$ is an infinite dimensional vector space. I know that if $V$ is a vector space over a field F, then $|V|=\max\{\text{dim}V,|F|\}$. So if we take $V=\mathbb{R}$ and $F=\mathbb{Q}$ then $|V|>|F|$ and $|V|=\text{dim}V$ (Cardinality of a basis of an infinite-dimensional vector space). Is there any example for the case $|V|>\text{dim}V$ and $|V|=|F|$?

Hopmaths
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uio666
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1 Answers1

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Let $ \ \big( \ell^{\infty} , \Vert \cdot \Vert \big) \ $ be the $\mathbb{R}$-vector space of all bounded real sequences with the norm $ \ \displaystyle \Vert (x_n) \Vert = \sup_{n \in \mathbb{N}} | x_n | \ $ and $$P = \big\{ (x_n) \in \ell^{\infty} : (\exists N \in \mathbb{N})(\forall n \in \mathbb{N})(n \geq N \to x_n = 0) \big\} $$ be the subspace of all eventually zero sequences. It is straightforward to show that $ \ \{ e_n \}_{n \in \mathbb{N}} \ $ is a countable (Hamel) basis for $P$, where $ \ e_n = ( \delta_{nk} )_{k \in \mathbb{N}} \ $ and $ \ \delta_{nk} \ $ is the Kronecker delta.

In advance note that $ \ \mathbb{R} \ni \alpha \mapsto \alpha \cdot e_1 \in P \ $ is an injection. Then clearly $$\dim_{\mathbb{R}}(P) = \big| \{ e_n \}_{n \in \mathbb{N}} \big| = \aleph_0 < | \mathbb{R} | \leq |P| \ \ \ . $$ So, because $ \ |P| = \max \big\{ \! \dim_{\mathbb{R}}(P) , | \mathbb{R} | \big\} \ $ we are left with $ \ |P| = | \mathbb{R} | \ $.

Gustavo
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