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The question below is related to the fundamental idea that every basis in a finite-dimensional vector space is finite.

Suppose that $V$ is a vector space with countable basis $\mathcal{B}$, i.e., $|\mathcal{B}| = |\mathbb{N}|$. If $\hat{\mathcal{B} }$ is another basis, must it be countable as well or is it possible that it is uncountable? If so, is the proof (non-)trivial?

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marked as duplicate by Community Feb 26 '18 at 4:27

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Interesting question. The statement is true, and I suggest a proof - not necessarily it is the shortest one.

Write $\mathcal{B}=\{v_1,v_2,\ldots\}$ and for each $m\in\mathbb{N}$, let $\mathcal{B}_m=\{v_1,v_2,\ldots,v_m\}$ be the first $m$ vectors of your basis. Denote $V_m=\text{Span}\,\mathcal{B}_m$ the vector subspace generated by $\mathcal{B}_m$. Since $\mathcal{B}$ is a basis for $V$, necessarily $$ V=\cup_{m\in\mathbb{N}}V_m.\tag{1} $$

Let $\hat{\mathcal{B}}$ be another basis. Then for every $w\in\hat{\mathcal{B}}$, we necessarily have $w\in V_m$ (by $(1)$), for some $m\in\mathbb{N}$. Moreover, each $V_m$ can contain at most $m$ elements of $\hat{\mathcal{B}}$ (since $V_m$ has dimension $m$, and every finite subset of $\hat{\mathcal{B}}$ is linearly independent, by definition). So, defining $$ \hat{\mathcal{B}}_m=\hat{\mathcal{B}}\cap V_m, $$ we have $\hat{\mathcal{B}}_m$ finite and $$ \hat{\mathcal{B}}=\cup_{m\in\mathbb{N}}\hat{\mathcal{B}}_m, $$ proving that $\hat{\mathcal{B}}$ is countable.

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