Proper subsets of R

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Proper subsets of R

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lgx57

#1 h Mar 30, 2025, 8:19 AM

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Let $S_1,S_2 \cdots S_n$ are proper subsets of $\mathbb{R}$ and they are closed for addition and subtraction. Try to prove that:

$\displaystyle\bigcup_{i=1}^n S_i \ne \mathbb{R}$

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#2 h Mar 30, 2025, 1:14 PM

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Suppose we have $G=\cup_1^n x_kH_k$ where each $H_k$ is a subgroup of $G$ . Then it is a result by von Neumann that at least one of the subgroups has finite index in $G$ . This solves the problem since $R$ has no proper subgroups of finite index.
Indeed, suppose $A\subset R$ is a subgroup of finite index $n>1$ . Then $nx\in A$ for all $x\in R$ . Hence $t=n(t/n)\in A$ for all $t$ and $A=R$ , contradiction.
Now we give an outline of the proof of von Neumann's result. Write $r$ for the number of distinct sets among the $H_i$ , then we perform induction wrt $r$ . The lemma is clear for $r=1$ . Now suppose that it is true for $r-1$ and that we have $G=\cup_1^n x_kH_k$ and $r>1$ .
We may assume $H_1\not=H_n$ and wlog $H_1,\cdots, H_m\not=H_n$ and $H_{m+1},\cdots, H_n=H_n$ .
If $G=\cup_{m+1}^n x_iH_i$ then $[G:H_n]<\infty$ . Otherwise choose $h\not\in \cup_{m+1}^n x_iH_i$ .
Then $hH_n$ is disjoint with $x_iH_i$ for $i>m$ hence $hH_n\subset \sum_1^m x_kH_k$ .
Hence for $i>m$ we have $x_i H_i=x_iH_n=x_ih^{-1} hH_n\subset \cup_1^m (x_ih^{-1}x_k)H_k$ .
It means that we also have decomposition of $G$ with strictly less than $r$ distinct $H_i$ and we can apply induction.

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