How to prove if $\Bbb R^n=W_1\cup W_2\cup \cdots\cup $ where each $W_k$ is a subspace, then $\Bbb R^n=W_i$ holds for some $i$ I know if $W_1,\ldots,W_k$ are subspaces of a vector space, their sum is the span of their union. ie. $W_1+W_2+\cdots+W_k=W_1\cup W_2\cup \cdots \cup W_k$.
But how to prove $\Bbb R^n=W_i$ holds for some $i$? Is there anything to do with direct sum? Because if $\Bbb R^n$ is the direct sum of $W_1+W_2+\cdots+W_k\iff$ the sum of the dimensions of the $W_i$ is $n$.
 A: Use induction on $k$. Suppose that it's true for $k=2$:
Then $\mathbb{R}^n = W_1 \cup W_2$. Let $w_1 \in W_1$ be any vector and take $w_2 \in \mathbb{R}^n-W_1 \subseteq W_2$. Then $w_1 + \lambda w_2$ is not in $W_1$ for any $\lambda$ and therefore, $w_1 + \lambda w_2 = w_2' \in W_2$. This proves that $w_1 = w_2'-\lambda w_2 \in W_2$ and therefore, $W_1 \subseteq W_2$. i.e. $W_2 = \mathbb{R}^n$
Now suppose that $k=m$ is true, for $k=m+1$:
Same story. Then $\mathbb{R}^n = W_1 \cup \bigcup_{k=2}^{m+1} W_k$. Let $w_1 \in W_1$ be any vector and take $w_0 \in \mathbb{R}^n-W_1 \subseteq \bigcup_{k=2}^{m+1} W_k$. Then $w_1 + \lambda w_0$ is not in $W_1$ for any $\lambda$ and therefore, $w_1 + \lambda w_0 \in \bigcup_{k=2}^{m+1} W_k$. Since the number of proper sets in $\bigcup_{k=2}^{m+1} W_2$ is finite but the number of elements in $\mathbb{R}$ is infinite, at least one of them, let's say $W_j$ must contain $w_1 + \lambda w_0 = w_j \in W_j$ for some $\lambda$. This proves that $w_1 = w_j-\lambda w_0 \in W_j \subset \bigcup_{k=2}^{m+1} W_k$ and therefore, $W_1 \subseteq \bigcup_{k=2}^{m+1} W_k$. i.e. $\bigcup_{k=2}^{m+1} W_k = \mathbb{R}^n$. But $\bigcup_{k=2}^{m+1} W_k$ is the union of $m$ proper subspaces equal to $\mathbb{R}^n$ and by the induction hypothesis, we find that some $l \in \{2,\cdots,m+1\}$ exists such that $W_l=\mathbb{R}^n$.
