A universally measurable homomorphism between Polish groups is automatically continuous

You need to know: Groups and homomopshisms between them, topological spaces and homeomopshisms between them, continuous function on a topological space, probability measure on a topological space, metric space, complete metric space, dense subset of a metric space.

Background: A topological group G is a topological space that is also a group such that the group operations of product $G\times G \to G: (x,y) \to xy$ and taking inverses $G \to G: x \to x^{-1}$ are continuous. A topological space is called Polish space if it is homeomorphic to a complete metric space that has a countable dense subset. A Polish group is a topological group G that is also a Polish space. A Borel probability measure on a topological space is a probability measure that is defined on all open sets. A subset of a Polish group G is called universally measurable if it is measurable with respect to
every Borel probability measure on G. A homomorphism $\phi: G\to H$ between Polish groups G and H is called universally measurable if $\phi^{-1}(U)$ is a universally measurable set in G for every open set $U \subseteq H$ .

The Theorem: On 8th December 2018, Christian Rosendal submitted to the Forum of mathematics, Pi a paper in which he proved that every universally measurable homomorphism between Polish groups is automatically continuous.

Short context: The Theorem resolves a longstanding problem posed by Christensen in 1971. It can be viewed as a generalisation of old classical theorem stating that any Lebesgue measurable function $f:{\mathbb R}\to{\mathbb R}$ satisfying the functional equation $f(x+y)=f(x)+f(y), \, x,y \in {\mathbb R}$ must be continuous.

Links: The original paper is available here.

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Tarski’s circle squaring problem has a constructive solution

You need to know: Euclidean plane ${\mathbb R}^2$ , notation $A+v$ for the translation of set $A\subset {\mathbb R}^2$ by a vector $v\in {\mathbb R}^2$ , open subset of ${\mathbb R}^2$ , countable union, countable intersection, and set difference ( $B \setminus A = \{x\in B, \, x\not\in A\}$ ) of sets. You also need to know what is the Axiom of choice to fully understand the context.

Background: A subset $A\subset {\mathbb R}^2$ is a Borel set if it can be formed from open sets through the operations of countable union, countable intersection, and set difference. We call two sets $A,B \subset {\mathbb R}^2$ equidecomposable by translations if there are partitions $A = A_1 \cup \dots \cup A_m$ and $B = B_1 \cup \dots \cup B_m$ , such that $B_i = A_i + v_i$ , $i=1,\dots,m$ , for some vectors $v_1, \dots, v_m \in {\mathbb R}^2$ . If, moreover, all $A_i$ (and thus $B_i$ ) are Borel sets, we say that A and B are equidecomposable by translations with Borel parts.

The Theorem: On 17th December 2016, Andrew Marks and Spencer Unger submitted to arxiv a paper in which they proved that a circle and a square of the same area on the plane are equidecomposable by translations with Borel parts.

Short context: In 1990, Laczkovich, answering a 1925 question of Tarski, proved that circle and a square of the same area are equidecomposable by translations. In a paper submitted in 2015, Grabowski, Máthé, and Pikhurko proved that this is possible even using only Lebesgue measurable pieces (that is, those having a well-define area). However, both results use axiom of choice and the resulting pieces $A_i$ are impossible to construct explicitly. The Theorem states that the circle can be squared with only Borel pieces. The proof does not use the axiom of choice. If such a proof is possible, we say that a problem has a constructive solution.

Links: Free arxiv version of the original paper is here, journal version is here.

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Cardinal numbers p and t are equal

You need to know: Notation ${\mathbb N}$ for the set of natural numbers. Cardinality of a set: sets A and B have the same cardinality (we write $|A|=|B|$ ) if there exists a bijection from A to B. We write $|A|\leq |B|$ if there exists a bijection from A to a subset of B. It is known that $|A|\leq |B|$ or $|B|\leq |A|$ for every $A,B$ .

Background: Let $A\subseteq^*B$ mean that set $\{x: x\in A, x\not\in B\}$ is finite. Let ${\mathbb N}^{\mathbb N}$ be the family of all infinite sequences of natural numbers, and let $D\subset {\mathbb N}^{\mathbb N}$ . We say that D has a pseudo-intersection if there is an infinite $A \subseteq {\mathbb N}$ such that $A\subseteq^*B$ for all $B \in D$ . We say that D has the strong finite intersection property (s.f.i.p. in short) if every nonempty finite subfamily of D has infinite intersection. D is called well ordered by $\subseteq^*$ if $A\subseteq^*B$ or $B\subseteq^*A$ for all $A,B \in D$ . D is called a tower if it is well ordered by $\subseteq^*$ and has no pseudo-intersection. Let $\textbf{p}$ be the smallest cardinality of $D\subset {\mathbb N}^{\mathbb N}$ which has the s.f.i.p. but has no pseudo-intersection. Let $\textbf{t}$ be the smallest cardinality of $D\subset {\mathbb N}^{\mathbb N}$ which is a tower.

The Theorem: On 27th August 2012, Maryanthe Malliaris and Saharon Shelah submitted to arxiv and the Journal of the AMS a paper in which they proved that $\textbf{p}=\textbf{t}$ .

Short context: Cantor proved in 1874 that the cardinality of set of integers (denoted $\aleph_0$ ) is strinctly less than the cardinality of set of real numbers (denoted $2^{\aleph_0}$ ). Clearly, both $\textbf{p}$ and $\textbf{t}$ are at least $\aleph_0$ and no more than $2^{\aleph_0}$ . It is easy to see that $\textbf{p}\leq\textbf{t}$ , since a tower has the s.f.i.p. An old open question asks whether equality holds. The Theorem asnwers this question affirmatively.

Links: Free arxiv version of the original paper is here, journal version is here.

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Z is definable in Q by a universal first-order formula in the language of rings

You need to know: Set ${\mathbb Z}$ of integers, set ${\mathbb Q}$ of rational numbers, standard notations $\forall$ (for all) and $\exists$ (there exists), polynomials.

Background: Let ${\mathbb Z}[t, x_1, \dots, x_n]$ denote the set of polynomials in $n+1$ variable $t, x_1, \dots, x_n$ with integer coefficients.

The Theorem: On 15th October 2010, Jochen Koenigsmann submitted to arxiv a paper in which he proved the existence of positive integer n, and a polynomial $g \in {\mathbb Z}[t, x_1, \dots, x_n]$ such that, for any $t\in{\mathbb Q}$ , we have $t\in{\mathbb Z}$ if and only if $\forall x_1, \dots, \forall x_n \in {\mathbb Q}: g(t,x_1,\dots,x_n) \neq 0$ .

Short context: Hilbert’s 10th problem was to find a general algorithm for deciding, given any n and any polynomial $f \in {\mathbb Z}[x_1, \dots, x_n]$ , whether or not f has a zero in ${\mathbb Z}^n$ . In 1970, Matiyasevich proved that there can be no such algorithm. Hilbert’s 10th problem over ${\mathbb Q}$ remains open. If we could define ${\mathbb Z}$ in ${\mathbb Q}$ as in the Theorem but with existential quantifiers $\exists$ instead of universal ones $\forall$ , this together with Matiyasevich theorem would give a (negative) answer to this problem. The Theorem may be considered as a major step in this direction.

Links: Free arxiv version of the original paper is here, journal version is here.

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