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We know that $(\mathbb R\setminus \{0\},.)$ is decomposible as it can be expressed as internal direct product of $\mathbb R^{+}$ and $\{-1,1\}$. I have also found somewhere that $(\mathbb R,+)$ is decomposible. Can anyone explain how to show it? Any help will be appreciated.

Anupam
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1 Answers1

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We can see $\mathbb{R}$ as a vector space over $\mathbb{Q}$ (so using rationals as scalars). This has a base (using the axiom of choice, there is no explicit description for such a base!) $B = \{x_i : i \in I \}$, that we can split in to linearly independent subsets, say $B_1 = \{x_i: i \in I_1 \}$ and $B_2 = \{x_i: i \in I_2\}$. Then $(\mathbb{R}, +)$ is a direct sum of the span of $B_1$ and the span of $B_2$.

Henno Brandsma
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  • What are $I_1$ and $I_2$? – Anupam Jan 03 '16 at 15:46
  • Any decomposition of $I$, the index set for the base – Henno Brandsma Jan 03 '16 at 16:01
  • Maybe it is more clear if you mention that $B_1$ and $B_2$ should be disjoint. I wonder if there is any solution that would avoid the axiom of choice.. – spin Jan 03 '16 at 18:04
  • @spin That's part of decomposition (for me). AC might be unavoidable, I'm not sure. – Henno Brandsma Jan 03 '16 at 18:16
  • @HennoBrandsma Some amount of choice is necessary; ZF alone is not sufficient to prove that such a basis exists. (Both the answers here mention the fact; I can't seem to find a good reference in literature for it though, but everyone seems to know it) – Milo Brandt Jan 03 '16 at 18:26
  • @MiloBrandt Does the decomposition alone imply the existence of a basis? I don't think so. Hence my doubt. I know for sure that we cannot prove the existence of a base for the reals over the rationals without some choice (it's in the book Consequences of the Axiom of Choice). But the decomposition of the reals in 2 summands is probably weaker still. – Henno Brandsma Jan 03 '16 at 18:29
  • @HennoBrandsma I think choice still is needed. To sketch a proof: Choose some $x\in B_1$ greater than $0$. Now, consider the sets $S_n=(B_1\cap [nx,(n+2)x)) + (B_2\cap (-nx,(1-n)x])$. We have that $[x,2x)\subseteq \bigcup S_n \subseteq [0,2x)$ so the measure of the union of finite and positive. However, I think one can translate $S_n$ to coincide with $S_m$ except for an arbitrarily small difference. Thus, they have the same measure - but this means $S_n$ is not measurable - but in ZF, it is consistent that no such sets exists. – Milo Brandt Jan 03 '16 at 18:45