Karsyn Beltran

2022-07-17

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 $i\in \mathbb{N}\setminus \{0\}$ is $\u27e8{B}_{i}\u27e9=V$. Prove or disprove that $\u27e8{\cap}_{i\in \mathbb{N}}{B}_{i}\u27e9=V$. 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 $\u27e8{B}_{k}\u27e9=V$ which would make the statement true. Where am I wrong? Any solution or hint would be highly appreciated.

akademiks1989rz

Beginner2022-07-18Added 16 answers

Nobody said that we have only finitely many ${B}_{i}$'s. Suppose that $V=\mathbb{K}$ and that

${B}_{n}=\{x\in \mathbb{K}\mid |x|<\frac{1}{n}\}.$

Then $(\mathrm{\forall}n\in \mathbb{N}):\u27e8{B}_{n}\u27e9=\mathbb{K}$. However, $\bigcap _{n\in \mathbb{N}}{B}_{n}=\{0\}$, and so

$\u27e8\bigcap _{n\in \mathbb{N}}{B}_{n}\u27e9=\{0\}.$

${B}_{n}=\{x\in \mathbb{K}\mid |x|<\frac{1}{n}\}.$

Then $(\mathrm{\forall}n\in \mathbb{N}):\u27e8{B}_{n}\u27e9=\mathbb{K}$. However, $\bigcap _{n\in \mathbb{N}}{B}_{n}=\{0\}$, and so

$\u27e8\bigcap _{n\in \mathbb{N}}{B}_{n}\u27e9=\{0\}.$

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