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A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are such that $(1,\alpha,\dots,\alpha_m)$ is linearly independent over $\mathbb{Q}$. For any $\varepsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \varepsilon$ for all $1 \leq i \leq m$.

Edit: it was originally asked assuming only that all $\alpha_i$ are irrational. As observed in the comments, for $m\ge 2$ and $\varepsilon<\frac12$, $\alpha_2=-\alpha_1$ yields an obvious counterexample.

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    $\begingroup$ Yes. This is because the infinite cyclic subgroup generated by $(\alpha_1,\dots,\alpha_m)$ in $(\mathbf{R}/\mathbf{Z})^m$ cannot be closed (since the latter is compact), and hence it accumulates at zero. $\endgroup$
    – YCor
    Oct 1, 2019 at 8:32
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    $\begingroup$ No. Check out $\alpha_1=1-\alpha_2=\sqrt3$. If you want the $k\alpha_i$ to be just close to integers (from either side) --- then yes, e.g., by the reasons @YCor mentions. $\endgroup$ Oct 1, 2019 at 8:36
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    $\begingroup$ The answer is yes under the assumption that $1,\alpha_1,\dots,\alpha_m$ are linearly independent, by Kronecker's theorem $\endgroup$
    – Wojowu
    Oct 1, 2019 at 8:42
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    $\begingroup$ @IlyaBogdanov Ah, I see, I tend to work in the quotient (and hence interpreted as "$d(\alpha_i,\mathbf{Z})\le\varepsilon$"). Indeed taking the representative in $[0,1[$ yields a different conclusion. Given such a trivial counterexample I guess my interpretation was the OP's intent, so hopefully OP will clarify. $\endgroup$
    – YCor
    Oct 1, 2019 at 9:20
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    $\begingroup$ It's necessary. I've done so, adding the context which makes the comments understandable. $\endgroup$
    – YCor
    Oct 5, 2019 at 8:41

1 Answer 1

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(reflecting the comments)

The answer is yes, by Kronecker's theorem. Namely, the assumption that $(1,\alpha_1,\dots,\alpha_m)$ is linearly independent over $\mathbf{Q}$ means that $(\alpha_1,\dots,\alpha_m)$ generates a dense sub(semi)group of the torus $(\mathbf{R}/\mathbf{Z})^m$, and hence this subsemigroup meets $[0,\varepsilon]^m$ (infinitely many times).

In the original setting (assuming only $\alpha_i$ irrational for some $i$), it is still true that it meets $[-\varepsilon,\varepsilon]^m$ infinitely many times: indeed this assumption ensures that the closure of the sub(semi)group is non-discrete and accumulates at zero. (And using that every nonempty closed subsemigroup of a compact group is a subgroup.) But for $m\ge 2$, taking $\alpha_2=-\alpha_1$ irrational shows that it doesn't work with $[0,\varepsilon]^m$ instead.

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