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Let $V$ be a finite dimensional vector space over $R$, then is it possible to write $V$ as union of finitely many proper subspaces?

I am not sure but suppose if we have $B$ as an ordered basis for our vector space $V$, then we may break $B$ into $B_1$ and $B_2$ such that $B1\cap B2=\phi$ and $B1\cup B2=B$

So, isn't it true that $V=L(B1) \cup L(B2)$?

Where $L(B1)$ is linear span of $B1$

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    $\begingroup$ It's true with $+$ instead of $\cup$ on the right hand side. Think about the cross in 2d. $\endgroup$ – Berci Sep 25 '19 at 22:27
  • $\begingroup$ It is possible if and only if it is a space of dimension $\geq 2$ over a FINITE field. If the dimension is 1 or 0, then any subspace is improper. $\endgroup$ – N. S. Sep 25 '19 at 23:54
  • $\begingroup$ This question has been asked, and answered, several times on this website. $\endgroup$ – Gerry Myerson Sep 26 '19 at 0:39
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The Lebesgue measure we have $mB = 0$ for any proper subspace. Hence any countable or finite union of proper subspaces $\cup_n B_n$ has measure zero. Since $m\mathbb{R}^n = \infty$ we see that it cannot be written as a countable or finite union of proper subspaces.

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There is a general argument: consider an infinite field $K$ and $V$ any $K$-vector space.

Then $V$ is not covered by finitely many proper subspaces $W_i$.

The proof is by induction on the number $p$ of subspaces. If $p=1$, there is nothing to prove. If the statement holds till rank $p-1$, and $V$ is covered by the proper subspaces $W_1, \ldots W_p$, there is some $x \in W_1$ that does not belong to any $W_k$, $k\geq 2$ (else the $W_k$, $k \geq 2$, cover $V$, impossible by induction hypothesis).

Of course, there is some $y \in W_2$ but not in $W_1$ (else, the induction hypothesis applies to $W_1, W_3,\ldots, W_p$). Consider $\Delta=\{z_t=(1-t)x+ty,\,t \in K\}$: it has infinitely many points so meets some $W_t$ at least twice. Since $W_t$ is a proper subspace, $\Delta \subset W_t$. So $x \in W_t$ so $t=1$. But also $y \in \Delta$ so $y \in W_t$, so $t \neq 1$. We have a contradiction.

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No.

Take for instance $\Bbb{R}^2$

The only proper subspaces of this space,are the lines that pass through the origin and $\{(0,0)\}$

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