Results 151 to 160 of about 1,323,506 (183)
Some of the next articles are maybe not open access.
MLQ, 2002
Every countable Scott set is the standard system of a model of PA, and for every countable Scott \(S\) set there is a completion \(T\) of PA such that \(S\) is the family of sets represented in \(T\). Every countable recursively saturated model \(M\) of PA is SSy\((M)\)-saturated and SSy\((M)\) (the standard system of \(M\)) is the unique Scott set \(S\
openaire +2 more sources
Every countable Scott set is the standard system of a model of PA, and for every countable Scott \(S\) set there is a completion \(T\) of PA such that \(S\) is the family of sets represented in \(T\). Every countable recursively saturated model \(M\) of PA is SSy\((M)\)-saturated and SSy\((M)\) (the standard system of \(M\)) is the unique Scott set \(S\
openaire +2 more sources
Approximation of Distributions by Bounded Sets
Acta Applicandae Mathematicae, 2007Let \((S,d)\) be a separable metric space, mostly supposed to be locally compact. Let \(\mathcal{A}\) be a collection of subsets of \(S\). \(\mathcal{A}\) is called \(K\)-bounded if the diameters of \(A\in \mathcal{A}\) do not exceed \(K\). Hence \(\mathcal{A}^K_k\) denotes the collection of \(k\)-unions \(\bigcup_1^k A_j\) of sets \(\mathcal{A}_j ...
Käärik, Meelis, Pärna, Kalev
openaire +1 more source
On Closed, Totally Bounded Sets
Canadian Mathematical Bulletin, 1966C. Goffman asserts that "… in a metric space X a set S is compact if and only if it is closed and totally bounded." [1] and "… every totally bounded sequence in a metric space has convergent subsequence." [2].The statements (incidentally, equivalent to each other) are both wrong, as the following counter-example shows.
openaire +2 more sources
Archive for Mathematical Logic, 2017
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Bernard A. Anderson +2 more
openaire +1 more source
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Bernard A. Anderson +2 more
openaire +1 more source
Nonautonomous Bounded Remainder Sets
Russian Mathematics, 2018zbMATH Open Web Interface contents unavailable due to conflicting licenses.
openaire +2 more sources
When Is a Family of Sets a Family of Bounded Sets?
The American Mathematical Monthly, 2000Let (X, r) be a metrizable topological space. A metric p for X is called admissible provided it is compatible with the topology r for X. Given any admissible metric p for the topology, the equivalent metric d defined by d(x, y) = min{1, p(x, y)} makes each subset of X d-bounded. Now let v be a family of subsets of X.
openaire +2 more sources
On the Circumradius of a Bounded Set
Journal of the London Mathematical Society, 1952openaire +2 more sources
Ars Comb.
Let \(G\) be a finite group and \(x,y\) not necessarily distinct elements in \(G\). The square of the set \(\{x,y\}\) is the set \(\{x^2,xy,yx,y^2\}\). Let \[ P_i(G)=|\{(x,y)\in G^2:|\{x,y\}^2|=i\}|/|G|^2 \] for \(1\leq i\leq 4\). The values of the \(P_i\)'s depend on the proportion of pairs that commute, the proportion of pairs that have equal squares,
Slilaty, Daniel, Vanderkam, Jeffrey
openaire +1 more source
Let \(G\) be a finite group and \(x,y\) not necessarily distinct elements in \(G\). The square of the set \(\{x,y\}\) is the set \(\{x^2,xy,yx,y^2\}\). Let \[ P_i(G)=|\{(x,y)\in G^2:|\{x,y\}^2|=i\}|/|G|^2 \] for \(1\leq i\leq 4\). The values of the \(P_i\)'s depend on the proportion of pairs that commute, the proportion of pairs that have equal squares,
Slilaty, Daniel, Vanderkam, Jeffrey
openaire +1 more source
The fusion process of interval opinions based on the dynamic bounded confidence
Information Fusion, 2016Haiming Liang +2 more
exaly