Let $H$ be $H_i$ 's are vector subspaces of dimension $n$. If $\bigcup\limits_{i=1}^r H_i$ contains $H$, then $H = H_i$ for some $i$.

Question

Can you provide a solution for the case where there are finitely many $H_i$? When i could take only two values, this is trivial. What about bigger values?

Answer 1

Hint: Use induction and also the fact that the complent of a vector space always containes another (moved) vector space.

P.S: This is rather trivial if the underlying field is $\mathbb{R}$ or $\mathbb{C}$ as every proper subspace has measure $0$ .

Answer 2

Lemma:

Let $X$ be a topological space. If $U \subseteq X$ is a nonempty open set and $F \subseteq X$ is a closed set with empty interior, then $U \setminus F$ is a nonempty open set.

Indeed, if $U \setminus F = \emptyset$ then $U \subseteq F$ so $\emptyset\ne U \subseteq \operatorname{Int} F$.

Another lemma:

Let $X$ be a normed space and $M$ its proper subspace. Then $M$ has empty interior.

Indeed, if $B(x, r) \subseteq M$ then for any $y \in X$ we have that $x + \frac{r\|y-x\|}{2}(y - x) \in M$, so $$y = \frac{2}{r\|y-x\|}\left(\left(x + \frac{r\|y-x\|}{2}(y - x)\right) - x\right) \in M$$

Therefore $M = X$.

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Equip your vector space $H$ with a norm and assume $H_i$ are proper subspaces of $H$. Now the subspaces $H_i$ are closed with empty interior.

However, $H$ is open so $H \setminus H_1$ is a nonempty open set. Therefore, $(H \setminus H_1) \setminus H_2 = H \setminus (H_1 \cup H_2)$ is also a nonempty open set.

Continuing inductively, we obtain that $\emptyset = H \setminus H = H \setminus \bigcup_{i=1}^r H_i$ is also a nonempty open set.

This is a contradiction so there exists $H_i$ which is equal to $H$.