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Preliminaries
Published in Hugo D. Junghenn, Principles of Analysis, 2018
A collection U $ \boldsymbol{{ \fancyscript {U}}} $ of open subsets of a topological space X such that ⋃U=X $ \bigcup \boldsymbol{{ \fancyscript {U}}}= X $ is called an open cover of X . A subcollection of U $ \boldsymbol{{ \fancyscript {U}}} $ that is a cover of X is called a subcover. A space X is said to be compact if each open cover of X has a finite subcover. A subset Y of X is compact if it is compact in the relative topology, that is, if for each collection U $ \boldsymbol{{ \fancyscript {U}}} $ of open sets in X there exists a finite subcollection U0 $ \boldsymbol{{ \fancyscript {U}}}_0 $ such that Y⊆⋃U0 $ Y \subseteq \bigcup \boldsymbol{{ \fancyscript {U}}}_0 $ . A subset Y of X is relatively compact if its closure is compact.
Application of Measure of Noncompactness on Infinite System of Functional Integro-differential Equations with Integral Initial Conditions
Published in S. A. Mohiuddine, Bipan Hazarika, Sequence Space Theory with Applications, 2023
Let M and S be subsets of a metric space (X,d). If ϵ>0, then the set S is called ϵ−net of M if for any x∈M there exists s∈S, such that d(x,s)<ϵ. If S is finite, then the ϵ−netS of M is called finite ϵ−net. The set M is said to be totally bounded if it has a finite ϵ−net for every ϵ>0. A subset M of a metric space X is said to be compact if every sequence (xn) in M has a convergent subsequence and the limit of that subsequence is in M. The set M is called relatively compact if the closure M¯ of M is a compact set. If a set M is relatively compact, then M is totally bounded. If the metric space (X,d) is complete, then the set M is relatively compact if and only if it is totally bounded.
Finite-time attractivity for semilinear functional differential inclusions
Published in Applicable Analysis, 2022
Let (A) hold. If is integrably bounded, then is equicontinuous in , where the operator is given by (4). In addition, if is a semicompact sequence then is relatively compact in . Moreover, if weakly in then strongly in .
Solvability of Langevin equations with two Hadamard fractional derivatives via Mittag–Leffler functions
Published in Applicable Analysis, 2022
Mohamed I. Abbas, Maria Alessandra Ragusa
Thus we get as which proves that the operator is equicontinuous operator. Hence, by Arzelà-Ascoli theorem, we conclude that is relatively compact on . Therefore, the Schauder's fixed point theorem shows that the operator has a fixed point which corresponds to the solution of (1). This completes the proof.
Analysis of age wise fractional order problems for the Covid-19 under non-singular kernel of Mittag-Leffler law
Published in Computer Methods in Biomechanics and Biomedical Engineering, 2023
Bibi Fatima, Mati ur Rahman, Saad Althobaiti, Ali Althobaiti, Muhammad Arfan
As can be seen, as In view of the Arzelà-Ascoli theorem, is relatively compact, and hence F is completely continuous. Thus, the suggested problem (2) has one solution. □