Is this a reasonable measure for comparing the sizes of two sets of real numbers?

Question

Introduction

We want to define a measure for determining whether one subset of $\mathbb{R}$ is bigger than the other. I am have defined such a measure, but I am not sure if it is a very useful one.

I do not think we want to use cardinality.
For instance, I want the real-interval (0.1, 0.2) to be smaller than the interval (100, 200) However, |(0.1, 0.2)| = |(100, 200)|

I do not think think that the "Labesgue Measure" is an appropriate alternative to cardinality.

Notably, I think that the Lebesgue measure of a set of "isolated points" is zero.
We say that a set $S$ is "isolated" if and only if for all $x∈S$, there exists $ε >0$ such that $x$ is the only element of $S$ in the interval $(x−ε,x+ε)$.

The Lebesgue measure would indicate that the two sets of isolated points have the same size. However, I can envision isolated sets which have different sizes from an intuitive standpoint. For example, if $S=\{1/n:n∈\mathbb{N}\}$ and $T=\{1/(2^{n}):n∈\mathbb{N}\}$then the Lebesgue measure of $S$ is equal to the Lebesgue measure of $T$, which equals zero.

I am kind of hoping for a measure where if $XS$ is a proper subset of $TS$, then $XS < TS$

Begin Section on "the $r$-intervalization"

Begin Definition of "the $r$-intervalization"

Let $XS$ be a set of real numbers.
Let $r$ be a real number.
Let $YS$ be the set of all real numbers such that:
For all $y$ in $YS$, there exists $x \in XS$ such that $|x-y|<r$
Then we say that $YS$ is the the $r$-intervalization of $XS$

Begin Examples of "the $r$-intervalization"

Begin Example One of "the $r$-intervalization"

Let $XS = \{1/n: n \in \mathbb{N}\}$
$XS = \{1, \hspace{0.1 in} 0.5,\hspace{0.1 in} 0.3\overline3, \hspace{0.1 in} 0.25,\hspace{0.1 in} 1/5,\hspace{0.1 in} 1/6,\hspace{0.1 in} 1/7,\hspace{0.1 in} ...\}$ The $0.1$-intervalization of $XS$ is:
$(0.9, 1.1) \cup (0.4, 0.6) \cup (0.2\overline3, 0.4\overline3) \cup (0.15, 0.35) \cup ...$

Begin Example Two of "the $r$-intervalization"

Let $XS = (-1, 1) = \{x \in \mathbb{R}: -1< x < 1\}$
The $w$-intervalization of $XS$ is:
$(w-1, w+1)$

"well-behaved" sets of real numbers

Definition of "well-behaved"

Let $XS$ be a set of real numbers.
Suppose that there exists $r \in \mathbb{R}$ such that $r > 0$ and for all $\delta \in \mathbb{R}$, if $0 < \delta < r$, then the lebesgue measure of the $\delta$-intervalization of $XS$ is well-defined.

Additional Explanation of "well-behaved"

What does it mean for "the Lebesgue measure" of the $\delta$-intervalization of $XS$ to be well-defined?

Let $IS$ be the $\delta$-intervalization of $XS$.
The set $IS$ can be viewed as the union of some set of intervals. If two intervals overlap, they simply merge to become one big interval. "the Lebesgue measure" of is the sum of the lengths of the intervals. For example, the length of the interval $(3.5, 16)$ is $12.5$.

Consider the $\delta$-intervalization of $\mathbb{N}$. It looks like $(1-\delta, 1+\delta) \cup (2-\delta, 2+\delta) \cup (3-\delta, 3+\delta) \cup (4-\delta, 4+\delta) \cup ...$. If you tried to sum the lengths of intervals, you would be in trouble. It's basically $2*\delta + 2*\delta + 2*\delta + ...$ ad infinitum.

The "the lebesgue measure" is well-defined if the sum of interval lengths is not +∞.

Definition of binary "$<$" operator on sets of real numbers###

Let $XS$ and $YS$ be well-behaved sets of real numbers.

We write $XS < YS$ if and only if there exists $r \in \mathbb{R}$ such that $r > 0$ and for all $\delta \in \mathbb{R}$, if $0 < \delta < r$ then the Lebesgue measure of $XS$ is strictly less than he Lebesgue measure of $YS$

Question

What I would like to know is if our definition of the "$<$" operator for comparing sets, has various desirable properties. If not, can we tweak the definition slightly to force the "$<$" operator to be more amenable? Some properties we would like to see are as follows:

For all $S$, $T$ subsets of $[0, 1]$ we will say ...

• If $S ⊊ T$, then $S$ ⋨ $T$.

  • If $S$ is a proper subset of $T$, then set $S$ is strictly smaller than set $T$

• For most $S$, $T$ subsets of $[0, 1]$, we want $S ≼ T$ and $T ≼ S$ to not both true. However, there are some weird cases for which we allow $S ≼ T$ and $T ≼ S$. For example, suppose that for all $x$ in $[0, 1]$, $x \in S$ if and only if $1-x \in T$. In that case, both $S ≼ T$ and $T ≼ S$.

• If $S$ and $T$ are finite, then $S ⪵ T$ if and only if $|S| \leq |T|$

  • EXAMPLE:
    • $S = \{0.25, 0.75\}$
    • $T = \{0.6, 0.7, 0.8, 0.9 \}$
    • set $S$ is strictly smaller than set $T$

• $S ≼ T$ if and only if $S- S ∩T$ ≼ $T- S∩T$

• If $S ⊆ T$ then $S ≼ T$

• If $S ≼ T$ then $|S| \leq |T|$. $|S|$ denotes cardinality. $|S| \leq |T|$ if and only if there exists an injective function from $S$ into $T$

• If you can shift all elements in one set up or down the number line, to make one a subset of the other, then that tells us something about the size. $S ⊄ T$, but there exists $a \in \mathbb{R}$ such that $S ⊊T′$ where $T′$ = $\{t – a: t \in T \}$ then $S$ ⋨ $T$

• $S ≼ T$ if all of the following statements are true:

  • $S$ is the union of a set of intervals and + $T$ is the union of a set of intervals
    • For set $S$ there exists a set of intervals, σ, such that:
      • every interval in σhas cardinality at least 2, and σ.
        • The interval $[5, 5] = \{r \in \mathbb{R}: 5 \leq r \leq 5\} = \{5\}$ is not allowed.
      • $S$ is the union of intervals in σ
  • $Σ(S) \leq Σ(T)$
  • $Σ(S)$ is defined to be the sum of the lengths of the intervals in σ, where σ is any set of intervals whose set-union is equal to set $S$

• $\mathbb{R} - S < \mathbb{R} - T$ if and only if $T < S$

Answer 1

This doesn't behave very well; in particular, it can't distinguish between sets with the same closure. Note that this isn't fixed by restricting attention to bounded sets (consider e.g. $[0,1]$ vs. $\mathbb{Q}\cap [0,1]$). We can avoid this issue by restricting attention to closed sets, but I suspect you won't find this satisfying.

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As per my comments, let me start by rephrasing the definition more simply (and with different notation to avoid clash):

For $X\subseteq \mathbb{R}$ and $c\in\mathbb{R}$, let $$X\langle c\rangle=\{a+b: a\in X, \vert b\vert<c\},$$ and let $\mu$ denote Lebesgue measure. Then we set $X\triangleleft Y$ iff there is some $r$ such that for every $c\in (0,r)$ we have $$\mu(X\langle c\rangle)<\mu(Y\langle c\rangle).$$

This unfortunately has a very serious problem, even if we restrict attention to very nice sets - namely, that we can have $X\langle c\rangle=Y\langle c\rangle$ for every $c>0$ even if $X$ and $Y$ are very different. For example, note that if $c>0$ then $$(\mathbb{Q}\cap [0,1])\langle c\rangle=[0,1]\langle c\rangle=(-c,1+c),$$ so we don't have $\mathbb{Q}\cap [0,1]\triangleleft[0,1]$ in your sense. More generally, this notion fails to distinguish between two sets (bounded or not) with the same closure: if $cl(A)=cl(B)$ then for all $c>0$ we have $A\langle c\rangle=B\langle c\rangle$.