I am familiar with the case where the basis is not infinite (it becomes the case that it CANNOT be represented as such). However, many sources suggest that this is not the case for an infinite basis. How could one go about proving this?
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Yes. See here and reference therein. An infinite dimensional vector space can be covered by $\aleph_0$-many proper subspaces. – Arturo Magidin Jan 23 '21 at 05:25
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Suppose that $V$ is infinite dimensional and fix a basis $B$. Then, by assumption, we have $|B| \geq \aleph_0$, and thus there exists a countable subset $S \subset B$. For each $s \in S$, consider $V_s = \langle B \setminus \{s\} \rangle$. Then $V = \bigcup_{s \in S}V_s$.
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