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$\newcommand{\c}{\text{card}}$For an infinite set $S$ with cardinality $\kappa$, is it fair to think of any subset $A \subset S$ with $\c (A) < \kappa$ as playing the role of a "measure $0$" subset of $S$, and is there a more "proper" way to do this than simply noting the difference in cardinalities?

I know this is vague so here's my motivation.

First, consider the set of all real-valued functions $\mathscr F := \{f : \mathbb R \to \mathbb R\}$ which has cardinality $2^\mathfrak c$. Then it can be shown that the continuous functions, say $\mathcal C$, are a subset of $\mathscr F$ with cardinality $\mathfrak c$ which is strictly less than $2^\mathfrak c$. Intuitively it makes sense that "almost all" functions are not continuous, but without actually having a measure on $\mathscr F$ is there a better way to get at this notion of "almost all" that's more insightful than what I've already done, namely $\mathfrak c < 2^\mathfrak c$? A similar example is how the cardinality of the Borel $\sigma$-algebra is $\mathfrak c$ while the Lebesgue $\sigma$-algebra has cardinality $2^\mathfrak c$ so "almost all" measurable sets are not Borel.

Another example: any countable subset of $\mathbb R$ has Lebesgue measure $0$. Also any subspace of $\mathbb R^n$ has Lebesgue measure $0$, so even though the cardinalities are the same in that case it makes me wonder if there's a way to think of lower cardinality sets almost as subspaces, which then would have measure $0$ in the containing space.

Ultimately my goal is to know if there's a more insightful way to think about a statement like

if $A \subset B$ with $\c (A) < \c (B)$ then "almost all" elements of $B$ are in $B\backslash A$

than $\c (A) < \c (B)$ which we already have. Perhaps we could define a "null set" in this context to be any set $A \subset S$ such that $\c (S \backslash A) = \c (S)$?

If this is too vague to answer I will try to add more motivating examples to better show what I'm trying to get at, but my whole difficulty here is I don't know how to properly describe what I mean, and maybe the notion of cardinality already captures it and I'm just not sufficiently appreciating that. Thank you.

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Well. Even in the case of the Lebesgue measure, it is consistent with $\sf ZFC$ that $\aleph_0<\aleph_1<\frak c$, but there are sets of size $\aleph_1$ which are non-measurable.

So just smaller cardinality than your space does not necessarily mean "null". Even in the case of the Lebesgue measure.

The commonly used analog for "null" in the uncountable case is "non-stationary". We have several reasons to view the closed and unbounded sets as "almost everywhere", so their complement is non-stationary. If $X\subseteq\kappa$, and $|X|<\operatorname{cf}(\kappa)$, then $X$ is certainly non-stationary and thus can be seen as null.

The problem is that you want to think about some sort of structure, and non-stationary sets are given by a fixed well-ordering (of uncountable cofinality). This makes the problem a bit awkward, since now things may depend on how you enumerate things.

But, at least for the examples you suggested with $\frak c$ and $2^\frak c$, this problem doesn't really pose a real problem. By König's theorem $\operatorname{cf}(\mathfrak c)<\operatorname{cf}(2^\mathfrak c)$. So any subset of $\Bbb{R^R}$ whose cardinality is $\frak c$ must be non-stationary in any enumeration of $\Bbb{R^R}$, so it is always null in this aspect.

Ultimately the problem boils down to structure. Even if you assume the Continuum Hypothesis, you can enumerate $[0,1]$ such that the Cantor set is enumerated on a closed and unbounded set, and its remainder on a non-stationary set. So moving from "everyday sets" to "cardinals" depends quite a lot on your choice of enumeration.

Another possible solution is to consider the product measure on $2^X$ for a suitable $X$, and ask if your collection is measurable, and if so, if it is null. This might also not work very well, e.g. the Lebesgue measurable sets are not even measurable. Indeed, moving from "countably-generated" to "arbitrary" is one of the hardest thing you can do in this mathematical world.

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  • $\begingroup$ thank you very much for this answer, I think stationarity is exactly the kind of notion i was hoping to come across! $\endgroup$ – alfalfa May 11 '18 at 14:31

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