Do any proofs/properties rely on the distinction between some uncountable size and a larger uncountable size, in order for the proof/property to hold?

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 Chlebík 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.


Sure. A large class of examples comes from the partition calculus. A simple result of the kind I have in mind is the following: Any infinite graph contains either a copy of the complete graph on countably many vertices or of the independent graph on countably many vertices. However, if we want to find an uncountable complete or independent graph, it is not enough that we begin with an uncountable graph. Instead, we need one of size strictly larger than the continuum.

For an encyclopedic reference on the partition calculus, including the result mentioned above, see

MR0795592 (87g:04002). Erdős, Paul; Hajnal, András; Máté, Attila; Rado, Richard. Combinatorial set theory: partition relations for cardinals. Studies in Logic and the Foundations of Mathematics, 106. North-Holland Publishing Co., Amsterdam, 1984. 347 pp. ISBN: 0-444-86157-2.


It is not a trivial theorem, but $(\ell^\infty)^*$ is the $\rm ba$ space, whose cardinality is $2^{2^{\aleph_0}}$. The reason is that we can identify this space with finitely additive measures, and every ultrafilter on $\Bbb N$ induces such measure, and by a fairly straightforward argument, there are $2^{2^{\aleph_0}}$ such ultrafilters. The upper bound can be obtained by noting that the algebraic dual, which is strictly larger, has cardinality $2^{2^{\aleph_0}}$, since the dimension of $\ell^\infty$, as a linear space, is $2^{\aleph_0}$.

 

Now, since $\ell^1$ is a separable Banach space, its cardinality is only $2^{\aleph_0}$. This gives a "quick" proof as to why $(\ell^\infty)^*\ncong\ell^1$.

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