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The answer is yes. Theorem. Every function $f:\newcommand\R{\mathbb{R}}\R\to\R$ is the sum of two surjective functions, $f=g+h$. Indeed, we can find such $g$ and $h$ that are surjective on every nontr …

This is not a full answer, but let me just point out that the statement is relatively consistent both with CH and also with $\neg$CH. CH implies the statement, since we can take $S$ to be the (convers …

The answer is yes. Every function on the reals is the sum of two injective functions, and this can be done in a highly effective manner, constructing the two functions $g,h$ from $f$ without any need …

We can make those limits be any two positive numbers that add to $1$, or we can make them nonconvergent. The reason is that in any interval, we can construct $A$ and $B$ on that interval by using the …

The answer is yes. We construct $f$ by transfinite recursion using a well ordering of the reals. (So this may not be explicit enough for you.) In fact, we can make the function $f$ bijective, with the …

Let me mount the kind of cardinality argument to which you allude. You had asked for a proof that the $\sigma$-algebra of Lebesgue measurable sets is not countably generated. But in fact, a much stron …

The assertion that $2^{\aleph_0}=2^{\aleph_1}$ is known as Luzin's hypothesis, and was presented by Luzin as an alternative to Cantor's continuum hypothesis. This is now known to be independent of ZFC …

Such an approach will violate the Cantor-Hume principle, which asserts that "the number of elements" of a set $A$ should be invariant under equinumorsity. That is, if $A$ and $B$ can be placed into on …

One might hope to handle proper classes as objects by working in one of the standard second-order set theories. For example, there is Gödel–Bernays set theory GBC, which has classes as objects, and in …

The answer is: zero. The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with …

This can fail, even if you assume that the lower cones of the order are measurable with respect to the algebra $\Sigma$. For example, consider the real numbers $X=\mathbb{R}$ and let $\Sigma$ be the u …

The answer is no, by a construction using the axiom of choice. We shall build a counterexample set $A$ by a transfinite recursive process of length continuum. At each stage, we shall promise that cer …

This is a great question! But unfortunately, the answer is no, the Lebesgue measure on the unit interval is not finitely $\vee$-additive. Theorem. There are two disjoint regular open sets $L$ and $R …

Background. The field of Borel equivalence relation theory provides a robust, unifying theory that organizes most of the classification problems of classical mathematics into a hierarchy, allowing us …

Countable choice is sufficient to prove that there is a non-Borel set, since under countable choice, the collection of sets of reals with a Borel code (which is a well-founded countable tree labeled w …