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Let $\nu\geq4$ be any composite integer, and let $d\in\left\{ 2,\ldots,\nu-1\right\}$ be any non-trivial divisor of $\nu$. Since $d\mid\nu$, note that any sequence $\left\{ \mathfrak{y}_{n}\right\} _{n\geq1}\subseteq\mathbb{Z}_{\nu}$ which converges in $\nu$-adic absolute value to a limit $\mathfrak{y}$ necessarily converges to $\mathfrak{y}$ with respect to the $d$-adic absolute value, as well. Since $\mathbb{Z}_{\nu}$ (resp. $\mathbb{Z}_{d}$) is simply the closure/completion of $\mathbb{Z}$ with respect to the $\nu$-adic (resp. $d$-adic) topology, this implies that the inclusion $\iota:\mathbb{Z}_{\nu}\hookrightarrow\mathbb{Z}_{d}$ is an embedding of $\mathbb{Z}_{\nu}$ in $\mathbb{Z}_{d}$.

Now, let $\mu_{\nu}$ (resp. $\mu_{d}$) be the Haar measure on $\mathbb{Z}_{\nu}$ (resp. $\mathbb{Z}_{d}$), normalized so that $\mu_{\nu}\left(\mathbb{Z}_{\nu}\right)=1$ (resp. $\mu_{d}\left(\mathbb{Z}_{d}\right)=1$). How does the inclusion map $\iota$ affect the measures of sets? That is, given a measurable set $V\subseteq\mathbb{Z}_{\nu}$, and letting $\iota\left(V\right)$ denote the copy of $V$ embedded in $\mathbb{Z}_{d}$, how does $\mu_{\nu}\left(V\right)$ compare to $\mu_{d}\left(\iota\left(V\right)\right)$?

My intuition is that $V$ should be $\mu_{d}$-measurable, and that $\mu_{d}\left(\iota\left(V\right)\right)$ should be $\geq\mu_{\nu}\left(V\right)$ (i.e., $\mathbb{Z}_{\nu}$ is a “small” subset of $\mathbb{Z}_{d}$). I'm wondering if this is correct, and if it can be taken further; for instance, is $\mu_{d}\left(\iota\left(V\right)\right)=0$ for all $V\subseteq\mathbb{Z}_{\nu}$? An answer and/or a reference to an answer would be much appreciated.

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The contrapositive statement of Steinhaus' Theorem says that for a locally compact abelian group $G$ (written additively), a haar-measurable subset $A\subseteq G$ will have zero measure in $G$ if $A-A=\left\{ a-b:a,b\in A\right\}$ does not contain an open neighborhood of the identity element of $G$.

So, letting $G=\mathbb{Z}_{d}$ and letting $A=\mathbb{Z}_{\nu}$, define for each positive integer $N$ the $d$-adic number:

$$\mathfrak{c}_{N}=\sum_{n=N}^{\infty}d^{n}\in\mathbb{Z}_{d}$$

Note that $\mathfrak{c}_{N}\rightarrow0$ $d$-adically as $N\rightarrow\infty$. If $\nu$ is not of the form $d^{m}$ for some integer $m\geq1$, it follows that the $\mathfrak{c}_{N}$s are a sequence of elements in $\mathbb{Z}_{d}\backslash\mathbb{Z}_{\nu}$ which converge to $0$ in $\mathbb{Z}_{d}$. Since $0$ is therefore an accumulation point of the complement of $\mathbb{Z}_{\nu}$ in $\mathbb{Z}_{d}$, the set $$\mathbb{Z}_{\nu}-\mathbb{Z}_{\nu}=\left\{ \mathfrak{a}-\mathfrak{b}:\mathfrak{a},\mathfrak{b}\in\mathbb{Z}_{\nu}\right\}$$ does not contain a $d$-adically open neighborhood of $0$. Thus, by Steinhaus' Theorem, $\mathbb{Z}_{\nu}$ must have zero measure in $\mathbb{Z}_{d}$ whenever $\nu$ is a positive integer multiple of $d$ which is not a power of $d$.

Q.E.D.

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