MATH1373
... T= { R , , all La} prove that T is a topology on R. Q13: Let R be the usual top. And let p be a point not in R. Let R R p. Define a collection T of subsets of R to be of two types: Type1: U , where U is open subset of R. Type 2: R B , where B is closed and bounded subset of R. Show that ...
... T= { R , , all La} prove that T is a topology on R. Q13: Let R be the usual top. And let p be a point not in R. Let R R p. Define a collection T of subsets of R to be of two types: Type1: U , where U is open subset of R. Type 2: R B , where B is closed and bounded subset of R. Show that ...
Topics in uniform continuity
... space is thin, but there exist complete thin spaces that are not UA. The main result of this section is a separation property of the complete thin spaces. The last section is dedicated to “additivity”, which turns out to be quite a non-trivial question in the case of uniform continuity. More precise ...
... space is thin, but there exist complete thin spaces that are not UA. The main result of this section is a separation property of the complete thin spaces. The last section is dedicated to “additivity”, which turns out to be quite a non-trivial question in the case of uniform continuity. More precise ...
Paths in hyperspaces
... Clearly, S is open in V because Uk− is open in CL(X). On the other hand, we may observe that S = V ∩[cl Uk ]− : indeed, every element of V is a subset of U and hence it cannot contain any point of (cl Uk ) \ Uk (because the sets Ui are pairwise disjoint). Therefore, S is also closed in V. ...
... Clearly, S is open in V because Uk− is open in CL(X). On the other hand, we may observe that S = V ∩[cl Uk ]− : indeed, every element of V is a subset of U and hence it cannot contain any point of (cl Uk ) \ Uk (because the sets Ui are pairwise disjoint). Therefore, S is also closed in V. ...
DEFINITIONS AND EXAMPLES FROM POINT SET TOPOLOGY A
... compactness this cover has a finite subcover {A0i }ni=1 . After removing A if necessary this gives a finite cover of C which is a subset of {Ai }i∈I . (10) A continuous image of a compact set is compact. Let f : X → Y be a continuous map between topological spaces and suppose that K ⊆ X is compact. ...
... compactness this cover has a finite subcover {A0i }ni=1 . After removing A if necessary this gives a finite cover of C which is a subset of {Ai }i∈I . (10) A continuous image of a compact set is compact. Let f : X → Y be a continuous map between topological spaces and suppose that K ⊆ X is compact. ...
The way-below relation of function spaces over semantic domains
... In the Compendium [2], one finds two characterizations of the way-below relation in these function spaces: Firstly, in I.1.21.1, in the special case where X is a compact Hausdorff space and L the extended real line, secondly for the general case in II.4.20. While in the special case they are correct ...
... In the Compendium [2], one finds two characterizations of the way-below relation in these function spaces: Firstly, in I.1.21.1, in the special case where X is a compact Hausdorff space and L the extended real line, secondly for the general case in II.4.20. While in the special case they are correct ...
Partial Metric Spaces
... considering. Let X = S ω = {x : ω → S}, the set of all infinite sequences in a set S, and let dS : X × X → IR be defined by: dS (x, y) = inf{2−k :xi = yi for each i < k}. It can be shown that (S ω , dS ) is a metric space. But computer scientists must compute the infinite sequence x, that is, write ...
... considering. Let X = S ω = {x : ω → S}, the set of all infinite sequences in a set S, and let dS : X × X → IR be defined by: dS (x, y) = inf{2−k :xi = yi for each i < k}. It can be shown that (S ω , dS ) is a metric space. But computer scientists must compute the infinite sequence x, that is, write ...