Proposition I.2

Question

If $V$ is a vector space and $(W_\alpha)_{\alpha in A}$ is any collection of subspaces, then $W= \bigcap_{\alpha \in A}$ is a subspace.

Toğrul Topal · Accepted Answer

By Proposition 1 it suffices to show that $W$ is closed under linear combinations. Let $c \in \mathbf{K}$ and let $v,w \in W$ be arbitrary. Since all of the $W_\alpha$ are subspaces themselves, and since $\forall \alpha in A: v, w \in W_\alpha$, we must also have $\forall \alpha \in A: cv + w \in W_\alpha \implies cv + w \in W$.

Proposition I.2

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Proposition I.2

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