Is this a superior measure of the size of sets, to cardinality?

Is this a superior measure of the size of sets, to cardinality?

If we use points on the real line as an example of the number of elements in a set, we know that any segment has uncountably many points and any set of discrete points is countable.

Surely therefore, we can create a measure of the number of points in any subset of $\mathbb{R}$ as follows:

Sum the length of any segments into one measure; a real number (or uncountable infinity) $r$.

Sum the number of discrete points (including any endpoints of closed segments) into another measure; a natural number or countable infinity $n$.

Then we will have a measure $\lvert X\rvert_m=\{n,r\}$ with $n$ a natural number or countable infinity, and $r$ a real number or uncountable infinity.

This measure has the advantage over cardinality that it can order the sizes of some sets which are in-between some uncountably infinite sets and others. For example, by this measure the open and closed line segments from $0$ to $1$ have measures:

$\lvert[0,1]\rvert_m=\{2,\infty\}$, and

$\lvert(0,1)\rvert_m=\{0,\infty\}$

respectively. Whereas by cardinality, the larger set has the same measure as the smaller.

Do measures of this nature exist, and can this be translated to set theory? It would seem that this measure, to some degree, sidesteps the problem of the continuum hypothesis, which limits the power of cardinality.

• The nice thing about cardinality is that it shows us that, for instance the two intervals you're looking at, are actually the same size (they contain the same number of elements)! This is the case if you can create a bijection from one set to the other. – Bobson Dugnutt Apr 19 '17 at 10:20
• @Lovsovs I agree but it's also true there are more whole numbers than there are even numbers, because the even numbers are a proper subset of the whole numbers and when you remove them all, you have something left over. – user334732 Apr 19 '17 at 10:34
• @Lovsovs it's like if on one table you have infinitely many apples and on the other you have infinitely many apples and infinitely many oranges. The table with apples and oranges on has more on it, because you can take infinitely many apples away from both tables, and only one will have infinitely many oranges left over. – user334732 Apr 19 '17 at 10:38
• That is actually not true (which is baffling, for sure!), when you define the size of a set in terms of bijections (which seems to be the best way). Even weirder than just the even/whole numbers having the same size, is that the number of rational numbers is the same as the number of whole numbers (even though there is an infinite amount of rational numbers in between each whole number). Moreover $\mathbb{R}^n$ has the same cardinality as $\mathbb{R}.$ ... Infinities are weird. :) – Bobson Dugnutt Apr 19 '17 at 10:46
• @Lovsovs you're still using cardinalities. You haven't grasped the fact that a proper superset is larger than its proper subset, even though by the measure of cardinality it might be the same. Even though they can be put into bijection with each other, one contains all that the other set does, and more. – user334732 Apr 19 '17 at 10:49

Sum the number of discrete points (including any endpoints of closed segments) into another measure; a natural number or countable infinity $n$.
If by "discrete point" you mean "point not contained in an interval" (note that usually "discrete point" would generally be taken to mean "isolated point") then there can be uncountably many discrete points! E.g. if we take $X$ to be the set of irrationals, then every point in $X$ is discrete in this sense, but there are continuum-many irrationals. So you need to allow $n$ to be an arbitrary cardinality $\le$ the continuum.
And this means you don't sidestep CH at all. Note, in particular, that if $X$ is an uncountable set of reals of size less than continuum then every element of $X$ is discrete in your sense, since each nontrivial interval has cardinality continuum. (This is related to your previous questions about intervals and CH; I suspect that ultimately you're conflating "point not contained in any interval in $X$" with "isolated point of $X$".) At the end of the day, for "interesting" sets of reals (that is, sets relevant to CH) counting the discrete points is exactly the same as counting the cardinality.
Finally, note that even if we expand your definition to allow $n$ to appropriately count the number of discrete points, your definition is actually less expressive than cardinality in some circumstances. E.g. consider a fat Cantor set of measure $1$ versus an interval of length $2$. Your size notion would give the first set $(0, 2^{\aleph_0})$ and the second set $(2, 0)$, so it wouldn't see that the interval is bigger than the fat Cantor set.
• Thanks; interesting. I think you have transposed $n,r$ compared with me in your last para? Anyhow... agreed $n$ can be uncountable. If $r$ takes precedence over $n$ in size, the Cantor set made up of isolated points would sit in $n$ and therefore is smaller in measure than the interval which is measured in $r$, so unless I'm mistaken I don't think it is less expressive? I'm not grasping what you mean by "every element of X is discrete in your sense". Only elements not in a segment would be discrete. – user334732 Apr 19 '17 at 19:59
• @RobertFrost Whoops, I left out the "of size $<$ continuum" there; fixed. – Noah Schweber Apr 19 '17 at 20:19
• Re: taking precedence, ok, what if the interval has length ${1\over 2}$? Then you've declared it bigger than the fat Cantor set, but that clashes wildly with measure, which is mathematically very important. Just making more distinctions isn't inherently valuable - they have to be mathematically meaningful. – Noah Schweber Apr 20 '17 at 4:16