arXiv subjects
Publications and source records attributed to Apoloniusz Tyszka.
This paper has been withdrawn by the author because Conjecture 1 is false. Please see arXiv:0901.2093 for a justification that Conjecture 1 is false. The other main results are also available from the above URL.
We discuss two conjectures. (I) For each x_1,...,x_n \in R (C) there exist y_1,...,y_n \in R (C) such that \forall i \in {1,...,n} |y_i| \leq 2^{2^{n-2}} \forall i \in {1,...,n} (x_i=1 \Rightarrow y_i=1) \forall i,j,k \in {1,...,n} (x_i+x_j=x_k \Rightarrow y_i+y_j=y_k) \forall i,j,k \in {1,...,n} (x_i \cdot x_j=x_k \Rightarrow y_i \cdot y_j=y_k) (II) Let G be an additive subgroup of C. Then for each x_1,...,x_n \in G there exist y_1,...,y_n \in G \cap Q such that \forall i \in {1,...,n} |y_i| \leq 2^{n-1} \forall i \in {1,...,n} (x_i=1 \Rightarrow y_i=1) \forall i,j,k \in {1,...,n} (x_i+x_j=x_k \Rightarrow y_i+y_j=y_k)
We discuss conjectures related to the following two conjectures: (1) for each complex numbers x_1,...,x_n there exist rationals y_1,...,y_n \in [-2^{n-1},2^{n-1}] such that \forall i \in {1,...,n} (x_i=1 \Rightarrow y_i=1) \forall i,j,k \in {1,...,n} (x_i+x_j=x_k \Rightarrow y_i+y_j=y_k) (2) for each complex (real) numbers x_1,...,x_n there exist complex (real) numbers y_1,...,y_n such that \forall i \in {1,...,n} |y_i| \leq 2^{2^{n-2}} \forall i \in {1,...,n} (x_i=1 \Rightarrow y_i=1) \forall i,j,k \in {1,...,n} (x_i+x_j=x_k \Rightarrow y_i+y_j=y_k) \forall i,j,k \in {1,...,n} (x_i \cdot x_j=x_k \Rightarrow y_i \cdot y_j=y_k)
H. Steinhaus asked a question whether inside each acute triangle there is a point from which perpendiculars to the sides divide the triangle into three parts with equal areas. We present two methods of solving Steinhaus' problem.
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.
We consider partitions of n-dimensional boxes in R^n, n>1, into a finite number of boxes with pairwise disjoint interiors. We study sets X \subseteq (0,\infty) with the Property (W_n): for every n-dimensional box P and every partition of P, if each constituent box has one side with the length belonging to X, then the length of one side of P belongs to X. We prove that the set X \subseteq (0,\infty) has Property (W_n) if and only if X is closed with respect to the operations: x+y and x+y+z-2min(x,y,z).
Let G be a group and let M be an object of the topos G-Set. We prove that an object X of the category G-Set is isomorphic to some subobject of one of the objects P(M), P(P(M)), P(P(P(M))),... if and only if card X < sup{card P(M), card P(P(M)), card P(P(P(M))),...} and {g \in G: \forall m \in M gm=m} \subseteq {g \in G: \forall x \in X gx=x}.
Let F be a subfield of a commutative field extending R. Let phi_n:F^n \times F^n ->F, phi_n((x_1,...,x_n),(y_1,...,y_n))=(x_1-y_1)^2+...+(x_n-y_n)^2. We say that f:R^n->F^n preserves distance d>=0 if for each x,y \in R^n |x-y|=d implies phi_n(f(x),f(y))=d^2. Let A_n(F) denote the set of all positive numbers d such that any map f:R^n->F^n that preserves unit distance preserves also distance d. Let D_n(F) 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)->F^n that preserves unit distance preserves also the distance between x and y. Obviously, {1} \subseteq D_n(F) \subseteq A_n(F). We prove: A_n(C) \subseteq {d>0: d^2 \in Q} \subseteq D_2(F). Let K be a subfield of a commutative field Gamma extending C. Let psi_2: Gamma^2 \times Gamma^2->Gamma, psi_2((x_1,x_2),(y_1,y_2))=(x_1-y_1)^2+(x_2-y_2)^2. We say that f:C^2->K^2 preserves unit distance if for each X,Y \in C^2 psi_2(X,Y)=1 implies psi_2(f(X),f(Y))=1. We prove: if X,Y \in C^2, psi_2(X,Y) \in Q and X \neq Y, then there exists a finite set S(X,Y) with {X,Y} \subseteq S(X,Y) \subseteq C^2 such that any map f:S(X,Y)->K^2 that preserves unit distance satisfies psi_2(X,Y)=psi_2(f(X),f(Y)) and f(X) \neq f(Y).
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].
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).
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}.
Let F be a subfield of a commutative field extending R. Let ϕ_2: F^2 \times F^2 \to F, ϕ_2((x_1,x_2),(y_1,y_2))=(x_1-y_1)^2+(x_2-y_2)^2. We say that f:R^2 \to F^2 preserves distance d \geq 0 if for each x,y \in R^2 |x-y|=d implies ϕ_2(f(x),f(y))=d^2. We prove that each unit-distance preserving mapping f:R^2 \to F^2 has a form I \circ (ρ,ρ), where ρ: R \to F is a field homomorphism and I: F^2 \to F^2 is an affine mapping with orthogonal linear part.
Let phi: C^2 times C^2 -> C, phi((x_1,x_2),(y_1,y_2))=(x_1-y_1)^2+(x_2-y_2)^2. We say that f:C^2->C^2 preserves unit distance, if for each X,Y in C^2 phi(X,Y)=1 implies phi(f(X),f(Y))=1. We prove that each unit-distance preserving mapping f:C^2->C^2 has a form J circ (gamma,gamma), where gamma:C->C is a field homomorphism and J:C^2->C^2 is an affine mapping with orthogonal linear part. We also prove a more general result for any commutative field K for which {x \in K: x^2+1=0} \neq \emptyset and char(K) \not\in {2,3,5}.
Let F be a commutative field of characteristic 0, G_n: F^n \times F^n -> F, G_n((x_1,...,x_n),(y_1,...,y_n))=(x_1-y_1)^2+...+(x_n-y_n)^2. We say that g:R^n->F^n preserves distance d>=0 if for each x,y \in R^n |x-y|=d implies G_n(g(x),g(y))=d^2. Let f:R^n->F^n preserve unit distance. We prove: (1) if n>=2, x,y \in R^n and x \neq y, then G_n(f(x),f(y)) \neq 0, (2) if A,B,C,D \in R^2, r \in Q and \overrightarrow{CD}=r\overrightarrow{AB}, then \overrightarrow{f(C)f(D)}=r\overrightarrow{f(A)f(B)}, (3) if A,B,C,D \in R^2 and \overrightarrow{AB} and \overrightarrow{CD} are linearly dependent, then \overrightarrow{f(A)f(B)} and \overrightarrow{f(C)f(D)} are linearly dependent, (4) if A,B,C,D \in R^2 and \overrightarrow{AB} is perpendicular to \overrightarrow{CD}, then \overrightarrow{f(A)f(B)} is perpendicular to \overrightarrow{f(C)f(D)}, (5) if A,B,C,D \in R^2 and |AB|=|CD|, then G_2(f(A),f(B))=G_2(f(C),f(D)). Let D_n(F) 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 g:S(x,y)->F^n that preserves unit distance preserves also the distance between x and y. Obviously, {1} \subseteq D_n(F) \subseteq A_n(F). We prove: (6) A_n(C) \subseteq {d>0: d^2 \in Q}, (7) {d>0: d^2 \in Q} \subseteq D_2(F).
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.
Let ϕ((x_1,x_2),(y_1,y_2))=(x_1-y_1)^2+(x_2-y_2)^2. We say that f:R^2 -> C^2 preserves distance d>=0 if for each x,y \in R^2 ϕ(x,y)=d^2 implies ϕ(f(x),f(y))=d^2. We prove that if x,y \in R^2 and |x-y|=(2\sqrt{2}/3)^k \cdot (\sqrt{3})^l (k,l are non-negative integers) then there exists a finite set {x,y} \subseteq S(x,y) \subseteq R^2 such that each unit-distance preserving mapping from S(x,y) to C^2 preserves the distance between x and y. It implies that each continuous map from R^2 to C^2 preserving unit distance preserves all distances.
Let ϕ((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 ϕ(x,y)=d^2 implies ϕ(f(x),f(y))=d^2. We prove that if x,y \in R^n (n>=3) and |x-y|=(\sqrt{2+2/n})^k \cdot (2/n)^l (k,l are non-negative integers) then there exists a finite set {x,y} \subseteq S(x,y) \subseteq R^n such that each unit-distance preserving mapping from S(x,y) to C^n preserves the distance between x and y. It implies that each continuous map from R^n to C^n (n>=3) preserving unit distance preserves all distances.
On every set A there is a rigid binary relation i.e. such a relation R \subseteq A \times A that there is no homomorphism (A,R) \rightarrow (A,R) except the identity (Vop{ě}nka et al. [1965]). We prove that for each infinite cardinal number κif card A \leq 2^κ, then there exists a relation R \subseteq A \times A with the following property: \forall (x \in A) \exists ({x} \subseteq A(x) \subseteq A, card A(x) \leq κ) \forall (f: A(x) \rightarrow A, f \neq id_A(x)) f is not a homomorphism of R. The above property implies that R is rigid. If a relation R \subseteq A \times A has the above property, then card A \leq 2^κ.