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How to prove the set of real numbers under addition; i.e., $(\mathbb{R}, +)$, is not the direct sum of two of its proper subgroups?

Ken
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Via
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  • @ Seirios I know a finite dimensional vs over infinite field can't be written as a union of finitely many of its proper subspaces,but R over Q is uncountable dimension! – Via Sep 10 '14 at 15:32
  • Where is the problem? I don't understand your argument. – Seirios Sep 10 '14 at 15:36
  • Related: http://math.stackexchange.com/q/302514/, http://math.stackexchange.com/questions/232568/isomorphism-between-mathbb-r2-and-mathbb-r?lq=1 – Jonas Meyer Jun 04 '15 at 06:08

1 Answers1

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For me, your statement is false: $\mathbb{R}$ is a $\mathbb{Q}$-vector space of infinite dimension, so two proper subspaces $A,B$ may be found so that $\mathbb{R}= A \oplus B$. In particular, it gives a decomposition of $\mathbb{R}$ as an additive group.

Seirios
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  • the problem is about Z submodules of R,not Q submodules of R.Because subgroups of R may not be subspace of R (as vs over Q) – Via Sep 10 '14 at 16:20
  • also you have to prove that it can not be written as a direct sum of its proper subgroups – Via Sep 10 '14 at 16:42
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    @Via clearly any $\mathbb{Q}$-submodule will also be a $\mathbb{Z}$-submodule. It is clear from this answer that Seiros understood what you asked, but what you were asked to prove is not correct. – Tobias Kildetoft Sep 10 '14 at 17:08
  • (well, assuming AC that is). – Tobias Kildetoft Sep 10 '14 at 17:09
  • @ Tobias KildetoftCan you give me example of two non trivial proper subgroups of R whose direct sum is R – Via Sep 10 '14 at 17:25
  • @Via Take a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Take the subspace generated by one vector and the subspace generated by the rest of the vectors. – Tobias Kildetoft Sep 10 '14 at 17:28
  • Yea I understand all of u are right..thank u – Via Sep 10 '14 at 18:19