Abstract:
Let a finite non-empty X is equipped with discrete topology. We prove that S \subseteq X^\omega is of second category if and only if for each f:\omega -> \bigcup_{n \in \omega} X^n there exists a sequence {a_n}_{n \in \omega} belonging to S such that for infinitely many i \in \omega the infinite sequence {a_{i+n}}_{n \in \omega} extends the finite sequence f(i).

Abstract:
Let K be a field and F denote the prime field in K. Let \tilde{K} denote the set of all r \in K for which there exists a finite set A(r) with {r} \subseteq A(r) \subseteq K such that each mapping f:A(r) \to K that satisfies: if 1 \in A(r) then f(1)=1, if a,b \in A(r) and a+b \in A(r) then f(a+b)=f(a)+f(b), if a,b \in A(r) and a \cdot b \in A(r) then f(a \cdot b)=f(a) \cdot f(b), satisfies also f(r)=r. Obviously, each field endomorphism of K is the identity on \tilde{K}. We prove: \tilde{K} is a countable subfield of K, if char(K) \neq 0 then \tilde{K}=F, \tilde{C}=Q, if each element of K is algebraic over F=Q then \tilde{K}={x \in K: x is fixed for all automorphisms of K}, \tilde{R} is equal to the field of real algebraic numbers, \tilde{Q_p}={x \in Q_p: x is algebraic over Q}.

Abstract:
Let varphi_n:C^n times C^n->C, varphi_n((x_1,...,x_n),(y_1,...,y_n))=sum_{i=1}^n (x_i-y_i)^2. We say that f:C^n->C^n preserves distance d>=0, if for each X,Y in C^n varphi_n(X,Y)=d^2 implies varphi_n(f(X),f(Y))=d^2. We prove: if n>=2 and a continuous f:C^n->C^n preserves unit distance, then f has a form I circ (rho,...,rho), where I:C^n->C^n is an affine mapping with orthogonal linear part and rho:C->C is the identity or the complex conjugation. For n >=3 and bijective f the theorem follows from Theorem 2 in [8].

Abstract:
Let K be a field and \tilde{K} denote the set of all r \in K for which there exists a finite set A(r) with {r} \subseteq A(r) \subseteq K such that each mapping f:A(r) \to K that satisfies: if 1 \in A(r) then f(1)=1, if a,b \in A(r) and a+b \in A(r) then f(a+b)=f(a)+f(b), if a,b \in A(r) and a \cdot b \in A(r) then f(a \cdot b)=f(a) \cdot f(b), satisfies also f(r)=r. We prove: \tilde{K} is a subfield of K, \tilde{K}={x \in K: {x} is existentially first-order definable in the language of rings without parameters}, if some subfield of K is algebraically closed then \tilde{K} is the prime field in K, some elements of \tilde{K} are transcendental over Q (over R, over Q_p) for a large class of fields K that are finitely generated over Q (that extend R, that extend Q_p), if K is a Pythagorean subfield of R, t is transcendental over K, and r \in K is recursively approximable, then {r} is \emptyset-definable in (K(t),+,\cdot,0,1), if a real number r is recursively approximable then {r} is existentially \emptyset-definable in (R,+,\cdot,0,1,U) for some unary predicate U which is implicitly \emptyset-definable in (R,+,\cdot,0,1).

Abstract:
Let G: C^n \times C^n -> C, G((x_1,...,x_n),(y_1,...,y_n))=(x_1-y_1)^2+...+ (x_n-y_n)^2. We say that f: R^n -> C^n preserves distance d>0 if for each x,y \in R^n G(x,y)=d^2 implies G(f(x),f(y))=d^2. Let A(n) denote the set of all positive numbers d such that any map f: R^n -> C^n that preserves unit distance preserves also distance d. Let D(n) denote the set of all positive numbers d with the property: if x,y \in R^n and |x-y|=d then there exists a finite set S(x,y) with {x,y} \subseteq S(x,y) \subseteq R^n such that any map f:S(x,y)->C^n that preserves unit distance preserves also the distance between x and y. We prove: (1) A(n) \subseteq {d>0: d^2 \in Q}, (2) for n>=2 D(n) is a dense subset of (0,\infty). Item (2) implies that each continuous mapping f from R^n to C^n (n>=2) preserving unit distance preserves all distances.

Abstract:
Let K be a ring and let A be a subset of K. We say that a map f:A \to K is arithmetic if it satisfies the following conditions: if 1 \in A then f(1)=1, if a,b \in A and a+b \in A then f(a+b)=f(a)+f(b), if a,b \in A and a \cdot b \in A then f(a \cdot b)=f(a) \cdot f(b). We call an element r \in K arithmetically fixed if there is a finite set A \subseteq K (an arithmetic neighbourhood of r inside K) with r \in A such that each arithmetic map f:A \to K fixes r, i.e. f(r)=r. We prove: for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is a neighbourhood of r inside R and is not a neighbourhood of r inside Z[\sqrt{-1}]; for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is not a neighbourhood of r inside Q; if K=Q(\sqrt{5}) or K=Q(\sqrt{33}), then for infinitely many rational numbers r for some arithmetic neighbourhood of r inside Q this neighbourhood is not a neighbourhood of r inside K; for each n \in (Z \cap [3,\infty)) \setminus {2^2,2^3,2^4,...} there exists a finite set J(n) \subseteq Q such that J(n) is an arithmetic neighbourhood of n inside R and J(n) is not an arithmetic neighbourhood of n inside C.

Abstract:
We conjecture that if a system S \subseteq {x_i=1, x_i+x_j=x_k, x_i \cdot x_j=x_k: i,j,k \in {1,...,n}} has only finitely many solutions in integers x_1,...,x_n, then each such solution (x_1,...,x_n) satisfies |x_1|,...,|x_n| \leq 2^{2^{n-1}}. By the conjecture, if a Diophantine equation has only finitely many solutions in integers (non-negative integers, rationals), then their heights are bounded from above by a computable function of the degree and the coefficients of the equation. The conjecture implies that the set of Diophantine equations which have infinitely many solutions in integers (non-negative integers) is recursively enumerable. The conjecture stated for an arbitrary computable bound instead of 2^{2^{n-1}} remains in contradiction to Matiyasevich's conjecture that each recursively enumerable set M \subseteq {\mathbb N}^n has a finite-fold Diophantine representation.

Abstract:
Let E_n={x_i=1, x_i+x_j=x_k, x_i*x_j=x_k: i,j,k \in {1,...,n}}. We prove: (1) there is an algorithm that for every computable function f:N-->N returns a positive integer m(f), for which a second algorithm accepts on the input f and any integer n>=m(f), and returns a system S \subseteq E_n such that S is consistent over the integers and each integer tuple (x_1,...,x_n) that solves S satisfies x_1=f(n), (2) there is an algorithm that for every computable function f:N-->N returns a positive integer w(f), for which a second algorithm accepts on the input f and any integer n>=w(f), and returns a system S \subseteq E_n such that S is consistent over N and each tuple (x_1,...,x_n) of non-negative integers that solves S satisfies x_1=f(n).

Abstract:
For K \subseteq C, let B_n(K)={(x_1,...,x_n) \in K^n: for each y_1,...,y_n \in K the conjunction (\forall i \in {1,...,n} (x_i=1 => y_i=1)) AND (\forall i,j,k \in {1,...,n} (x_i+x_j=x_k => y_i+y_j=y_k)) AND (\forall i,j,k \in {1,...,n} (x_i*x_j=x_k => y_i*y_j=y_k)) implies that x_1=y_1}. We claim that there is an algorithm that for every computable function f:N->N returns a positive integer m(f), for which a second algorithm accepts on the input f and any integer n>=m(f), and returns a tuple (x_1,...,x_n) \in B_n(Z) with x_1=f(n). We compute an integer tuple (x_1,...,x_{20}) for which the statement (x_1,...,x_{20}) \in B_{20}(Z) is equivalent to an open Diophantine problem. We prove that if the set B_n(Z) (B_n(N), B_n(N \setminus {0})) is not computable for some n, then there exists a Diophantine equation whose solvability in integers (non-negative integers, positive integers) is logically undecidable.

Abstract:
Let E_n={x_i=1, x_i+x_j=x_k, x_i \cdot x_j=x_k: i,j,k \in {1,...,n}}. If Matiyasevich's conjecture on single-fold Diophantine representations is true, then for every computable function f:N->N there is a positive integer m(f) such that for each integer n>=m(f) there exists a system U \subseteq E_n which has exactly f(n) solutions in non-negative integers x_1,...,x_n.