Mathematics Asked by Samuel Muldoon on December 11, 2020
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 Lebesque 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 Lebesque 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 $XS < TS$
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 $XS$
Let $XS = {1/n: n in mathbb{N}}$
$XS = {1, hspace{0.1 in} 0.5,hspace{0.1 in} 0.3overline3, hspace{0.1 in} 0.25,hspace{0.1 in} 1/5,hspace{0.1 in} 1/6,hspace{0.1 in} 1/7,hspace{0.1 in} …}$
The $0.1$-intervalization of $XS$ is:
$(0.9, 1.1) cup (0.4, 0.6) cup (0.2overline3, 0.4overline3) cup (0.15, 0.35) cup …$
Let $XS = (-1, 1) = {x in mathbb{R}: -1< x < 1}$
The $w$-intervalization of $XS$ is:
$(w-1, w+1)$
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.
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 +∞.
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 $YS$
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 $S$ ⋨ $T$.
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 $|S| leq |T|$
$S ≼ T$ 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:
$mathbb{R} – S < mathbb{R} – T$ if and only if $T < S$
As per my comments, let me start by rephrasing the definition more simply (and with different notation to avoid clash):
For $Xsubseteq mathbb{R}$ and $cinmathbb{R}$, let $$Xlangle crangle={a+b: ain X, vert bvert<c},$$ and let $mu$ denote Lebesgue measure. Then we set $Xtriangleleft Y$ iff there is some $r$ such that for every $cin (0,r)$ we have $$mu(Xlangle crangle)<mu(Ylangle crangle).$$
This unfortunately has a very serious problem, even if we restrict attention to very nice sets - namely, that we can have $Xlangle crangle=Ylangle crangle$ for every $c>0$ even if $X$ and $Y$ are very different. For example, note that if $c>0$ then $$(mathbb{Q}cap [0,1])langle crangle=[0,1]langle crangle=(-c,1+c),$$ 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 with the same closure: if $cl(A)=cl(B)$ then for all $c>0$ we have $Alangle crangle=Blangle crangle$.
We can avoid this issue if we look only at closed sets, but that throws away a lot of interesting sets and I suspect you won't find that satisfying.
Answered by Noah Schweber on December 11, 2020
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