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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 $n$.

Michael Hardy
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Chiu-Tzu-Hsuan
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  • I mean how to prove the dimension of $W_i$ is $n$ – Chiu-Tzu-Hsuan Dec 12 '17 at 06:19
  • Is for some $i$ not for all? – Chiu-Tzu-Hsuan Dec 12 '17 at 06:24
  • A general case has been discussed here: https://math.stackexchange.com/q/10760 –  Dec 12 '17 at 06:38
  • Also, $k$ does not need to be given, the statement is true for (countably) infinitely many subspaces. –  Dec 12 '17 at 06:50
  • @stressed-out : no, actually it suffices that the field be infinite for it to work – Maxime Ramzi Dec 12 '17 at 06:54
  • @Max But not countably infinite; $\mathbb Q^2$ can be written as union of countably many 1-dimensional subspaces as it is itself countable. –  Dec 12 '17 at 06:57
  • @user8734617 : I thought we were talking about a finite $k$, in which case it holds for any infinite field – Maxime Ramzi Dec 12 '17 at 06:58
  • @Max Looking at the question’s title, I am not sure... –  Dec 12 '17 at 07:01
  • A linear subspace $W_i$ of $\Bbb R^n$ is closed and if $W_i\ne\Bbb R^n$ then $W_i$ has empty interior. If there are are finitely many $W_i$ or a countably infinite set of $W_i$ and none of them is $\Bbb R^n$ then their union is co-dense because $\Bbb R^n$ ( & any complete metric space) is a Baire space. – DanielWainfleet Dec 12 '17 at 07:32

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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 $W_l=\mathbb{R}^n$.

stressed out
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