2
$\begingroup$

$\newcommand{\card}{\operatorname{card}}$

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?)

  • 1
    Do you want the inequality in your first display to go in the other direction?2012-07-14
  • 0
    @DavidMitra: Do you mean in (*)? If so then, no, I want it "$\geq$", since if it was $\leq$ then we are done.2012-07-14
  • 1
    @Sarah; And you want $a$ and $b$ as written? Then the inequality is false.2012-07-14
  • 0
    Cardinalities are not really real numbers, they are ordinals. The infinimum is actually minimum, and the results may not be real numbers at all. Using the notation of $\inf$ is confusing and may be ill-defined (if someone wants to argue that $\min$ requires the axiom of choice, let me clarify that $\inf$ requires it just as well).2012-07-14
  • 0
    You cannot apply the parenthetical comment at the end; that would let you deduce that $\inf(A\cap[x,x+a])\geq \inf(B\cap [x,x+a]$. But here you are comparing the sets of cardinals, not the sets of reals.2012-07-14
  • 0
    @Asaf: In this case, since $A$ is countable, the infimum is either $\aleph_0$ or a natural number, no?2012-07-14
  • 0
    @Arturo: In the countable case yes. It can be considered in a broad context where $A$ is any set of reals (although the question becomes vastly more difficult then). Last time I checked, though $\aleph_0$ was not a real number. :-)2012-07-14
  • 0
    @Asaf: Oh, I agree; but even in the context of sets of ordinals, it's true that if $\varnothing\neq A\subseteq B$ then $\inf(A)\geq\inf(B)$ (in fact, that's true in any lattice in which both infima exist, e.g., a complete lattice). The error here lies in that the inclusion that was established is between sets *other than* those whose infima we are trying to compare (your criticism, on the other hand, hits the nail right on the head for the second answer, since the characteriziation of infima in terms of $\epsilon$s is not valid in more general contexts).2012-07-14
  • 0
    @Arturo: I becomes apparent that I have to sleep. I'll get back to you on that...2012-07-14

2 Answers 2