(Sorry if my terminology is a bit imprecise; I'm trying to describe a rather vague notion in my head about when something "relies on" the distinction between finite/countable/uncountable, but I am finding it difficult to put into words precisely.)
There are proofs or properties that only hold when something is finite, but breaks down when it becomes infinite. For example, the finite intersection of open sets is open, but not arbitrary infinite intersections. As another example, $ a_n>0 \ \forall n \in \mathbb{N} $ does not imply $ \lim_{n \rightarrow \infty}{a_n}>0 $.
Similarly, there are also proofs or properties that only hold when something is countable. For example, normal induction can only be used if the variable takes values from a countable set.
I was wondering, are there any common proofs or properties that rely on a similar distinction between some uncountable size and another larger uncountable size? For example, property X is true iff some set S has size $\leq \aleph_n$ where $n>0$ ?
(Not sure what to tag; please edit if necessary 🙂 )
Best Answer
There are several relatively straightforward existence proofs in analysis that make use of $c < 2^c.$
1. "Most" Lebesgue measure zero sets are not Borel sets, because there are $2^c$ many Lebesgue measure zero sets (consider all subsets of a measure zero Cantor set) and there are only $c$ many Borel sets.
2. "Most" Riemann integrable functions are not Borel measurable, because the characteristic function of any subset of a measure zero Cantor set is Riemann integrable and there are only $c$ many Borel measurable functions.
3. "Most" complete Borel measures on $\mathbb R$ are not $\sigma$-finite. In fact, there are $2^c$ many complete Borel measures on $\mathbb R$ and only $c$ many $\sigma$-finite Borel measures (complete or not complete) on ${\mathbb R}.$ To see the first claim, let $B$ be a Borel set of cardinality $c$ (e.g. $B$ could be a Cantor set or the interval $[0,1]).$ For each $A \subseteq B,$ define ${\mu}_A(E) = \infty$ if $A \cap E \neq \emptyset$ and ${\mu}_A(E) = 0$ if $A \cap E = \emptyset.$ To see the second claim, note that every finite Borel measure on $\mathbb R$ is the Lebesgue-Stieltjes measure of some monotone function, and there are only $c$ many monotone functions (several ways to prove this). Now observe that every $\sigma$-finite Borel measure on $\mathbb R$ can be associated with a sequence of finite Borel measures on ${\mathbb R}.$ (Recall that there are only $c$ many sequences whose terms all come from a given set of cardinality $c.)$
4. "Most" convex subsets of ${\mathbb R}^2$ are not Borel sets, since removing any subset of the boundary of the unit disk results in a convex set and there are only $c$ many Borel sets. Note how badly this fails for ${\mathbb R}.$
5. "Most" functions $f:{\mathbb R} \rightarrow {\mathbb R}$ that are symmetrically continuous at each point (i.e. for each $x \in \mathbb R$ we have $\lim_\limits{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0)$ are not continuous, or even Borel measurable. Miroslav ChlebÃk proved in this 1991 Proc. AMS paper that there are $2^c$ symmetrically continuous functions, and there are only $c$ many continuous functions (indeed, only $c$ many Borel measurable functions).
6. "Most" subsets of the boundary of the unit disk are not a divergence set for any power series with complex coefficients and radius of convergence $1,$ since there are $2^c$ many subsets of the boundary of the unit disk and only $c$ many power series with complex coefficients. For more details about the possible divergence sets of a power series with complex coefficients, see this answer. Note how different this is for power series with real coefficients, in which there are only $2^2 = 4$ possible subsets of the boundary of an interval (there are only $4$ subsets of a $2$-element set) and it is not difficult to see that any of these subsets can be a divergence set.