# Let V be a KK-Vectorspace, and let B_1,B_2,... be subsets of V such that B_1 supe B_2 supe ... and for all i in NN ∖{0} is ⟨B_i⟩=V. Prove or disprove that ⟨nn_(i in NN) B_i⟩=V.

Karsyn Beltran 2022-07-17 Answered
Let V be a $\mathbb{K}$-Vectorspace, and let ${B}_{1},{B}_{2},...$ be subsets of V such that ${B}_{1}\supseteq {B}_{2}\supseteq ...$ and for all $i\in \mathbb{N}\setminus \left\{0\right\}$ is $⟨{B}_{i}⟩=V$. Prove or disprove that $⟨{\cap }_{i\in \mathbb{N}}{B}_{i}⟩=V$. The hint from our professor was that this is not true. Let's choose ${B}_{k}$ such that it is a subset of all other ${B}_{i}$ and also the smallest set. So every vector of ${B}_{k}$ is also in every other ${B}_{i}$. Therefore, the intersection of all ${B}_{i}$ would be ${B}_{k}$ and we know that $⟨{B}_{k}⟩=V$ which would make the statement true. Where am I wrong? Any solution or hint would be highly appreciated.
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Nobody said that we have only finitely many ${B}_{i}$'s. Suppose that $V=\mathbb{K}$ and that
${B}_{n}=\left\{x\in \mathbb{K}\mid |x|<\frac{1}{n}\right\}.$
Then $\left(\mathrm{\forall }n\in \mathbb{N}\right):⟨{B}_{n}⟩=\mathbb{K}$. However, $\bigcap _{n\in \mathbb{N}}{B}_{n}=\left\{0\right\}$, and so
$⟨\bigcap _{n\in \mathbb{N}}{B}_{n}⟩=\left\{0\right\}.$
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