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A vector space over $R$ is not a countable union of proper subspaces

Today we had a Linear algebra exam. (Iran PPCE $2012$, you can find the problems on AoPS). problem $2$ of the exam was this:

Prove that if a vector space is the union of some of its proper subspaces, then the number of these subspaces cannot be less than the number of elements of the field of that vector space.

I'm really eager to see its solution. any Ideas?

  • 2
    Every proper subspace has codimension at least $1$, so the union must have at least as many subspaces as there are spaces of dimension $1$.2012-02-16
  • 1
    Here are two related threads: http://math.stackexchange.com/q/10760 and http://math.stackexchange.com/q/60698/2012-02-16
  • 3
    [Pete L. Clark](http://math.stackexchange.com/users/299/pete-l-clark) just published a note about this in the Monthly. It's available in PDF format from [his website](http://math.uga.edu/~pete/coveringnumbersv2.pdf).2012-02-16
  • 0
    [This MO thread is relevant as well.](http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces)2012-03-09

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