Number of subspaces not less than cardinality of the field

Asked Feb 16 '12 at 16:17

Active Feb 16 '12 at 16:31

Viewed 81 times

Possible Duplicate:
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?

edited Apr 13 '17 at 12:20

Community

asked Feb 16 '12 at 16:17

Goodarz Mehr

2,309

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 $1$. – Alex Becker Feb 16 '12 at 16:21
1

Here are two related threads: http://math.stackexchange.com/q/10760 and http://math.stackexchange.com/q/60698/ – t.b. Feb 16 '12 at 16:23
3

Pete L. Clark just published a note about this in the Monthly. It's available in PDF format from his website. – Arturo Magidin Feb 16 '12 at 17:03
This MO thread is relevant as well. – Mar 09 '12 at 12:11

Number of subspaces not less than cardinality of the field

0 Answers0