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Let $A$ be a nonempty countable set of real numbers, and $0< a\leq b$. Is the following true:

$ \tag{*} \inf_{x\in \mathbb R} \card(A\cap [x,x+a]) \geq \inf_{x\in \mathbb R} \card(A\cap [x,x+b]) $

where $\card$ means number of elements in the set.

As far as I know, since $a\leq b$ then $A\cap [x,x+a] \; \subseteq \; A\cap [x,x+b] $

so $ \card(A\cap [x,x+a]) \; \leq\; \card(A\cap [x,x+b]) $ for all $x\in \mathbb R$. How I can complete the proof (if the result is correct)?

(I know that if we have two sets $B\subseteq C$ then $\inf B \geq\inf C$. But how I can use this in terms of cardinality of sets?)

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    @Arturo: I becomes apparent that I have to sleep. I'll get back to you on that...2012-07-14

2 Answers 2

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The inequality displayed in ($*$) is incorrect.

Consider $A=\mathbb{Z}$, $a=\frac{1}{2}$, $b=\frac{3}{2}$. An interval of length $\frac{1}{2}$ contains at most one integer, so for every $x\in\mathbb{R}$, $A\cap[x,x+a]$ is either empty, or has cardinality $1$. In particular, since there are $x$ for which the intersection is empty, then it follows that $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+a]) = 0.$ On the other hand, an interval of length $\frac{3}{2}$ contains always at least one integer, and sometimes two; so $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+b]) = 1.$ Therefore $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+a]) = 0 \not\geq 1 = \inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+b]).$

On the other hand, if you meant whether $\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap[x,x+a])\leq\inf_{x\in\mathbb{R}}\mathrm{card}(A\cap [x,x+b]),$ then that inequality does hold.

Since $A$ is countable, the infimum on the left is either infinite, or a natural number.

If it is infinite, then $A\cap[x,x+a]$ is infinite for every $x$, and therefore $A\cap[ x,x+b]$ is infinite for every $b$, so both infima are infinite, hence equal.

Now suppose that the infimum on the left is finite, equal to $n$. For every $x$, $A\cap [x,x+a]\subseteq A\cap [x,x+b]$, so $\mathrm{card}(A\cap [x,x+a])\leq\mathrm{card}(A\cap [x,x+b])$. Since the infimum is less than or equal to every element of the set, we have $n\leq \mathrm{card}(A\cap [x,x+a])\leq\mathrm{card}(A\cap [x,x+b]),$ hence $n$ is a lower bound for the set $\{\mathrm{card}(A\cap[x,x+b])\mid b\in\mathbb{R}\},$ and hence we have $n\leq \inf_{x\in \mathbb{R}}\mathrm{card}(A\cap [x,x+b]).$ Thus, the desired inequality holds.

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    @AsafKaragila, not funny!!2012-07-14
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$\newcommand{\card}{\operatorname{card}}$The correct is $ \inf_{x \in \mathbb{R}} \card(A\cap [x,x+a]) \le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]).$ By definition of $\inf$ there exists $x_n \in \mathbb{R}$ such that $\card(A\cap [x_n,x_n+b]) \le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]) + 1/n.$ Hence, $ \inf_{x \in \mathbb{R}} \card(A\cap [x,x+a]) \le \card(A\cap [x_n,x_n+a]) \le \card(A \cap [x_n,x_n+b]) $ $\le \inf_{x \in \mathbb{R}}\card(A\cap [x,x+b]) +1/n.$ Do $n \rightarrow \infty.$

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    The mistake is that cardinalities are not *really* real numbers, they are ordinals. The infinimum is actually minimum, and the results may not be real numbers at all.2012-07-14