(α,β)-SEMI OPEN SETS AND SOME NEW GENERALIZED
... 4. (α,β)-generalized semi closed sets and (α, β) − semi T1/2 spaces We recall that if A ⊆ X and α,β:P (X)→P (X) are associated operators to a topology τ on X, then the (α, β) − sCl(A) is (α, β)-semi closed set. In consequence, we can introduce the notions of (α,β)-semi T1/2 spaces in a natural way, ...
... 4. (α,β)-generalized semi closed sets and (α, β) − semi T1/2 spaces We recall that if A ⊆ X and α,β:P (X)→P (X) are associated operators to a topology τ on X, then the (α, β) − sCl(A) is (α, β)-semi closed set. In consequence, we can introduce the notions of (α,β)-semi T1/2 spaces in a natural way, ...
Let (X, τ) be a topological space, a base B is a
... Let (X, τ ) be a topological space, a base B is a subset of τ such that each element of τ could be written as an union of element from B, every base i) covers X and ii) has the property that the intersection of any two base sets contains another base set (and conversely every subset of 2X which sati ...
... Let (X, τ ) be a topological space, a base B is a subset of τ such that each element of τ could be written as an union of element from B, every base i) covers X and ii) has the property that the intersection of any two base sets contains another base set (and conversely every subset of 2X which sati ...
On the topological Hochschild homology of bu. I.
... dimension 2p − 1 (see [5]). Note that this cannot be true for τ for the trivial reason that HH∗ Z is zero in all positive dimensions. In the cases R = S 0 and R = Σ∞ (ΩX)+ mentioned above one can give explicit descriptions of T HH(R): T HH(S 0) = S 0 and ...
... dimension 2p − 1 (see [5]). Note that this cannot be true for τ for the trivial reason that HH∗ Z is zero in all positive dimensions. In the cases R = S 0 and R = Σ∞ (ΩX)+ mentioned above one can give explicit descriptions of T HH(R): T HH(S 0) = S 0 and ...
Math 55a: Honors Advanced Calculus and Linear Algebra Metric
... Proof : Let {Vα } be an open cover of Y. Then {f −1 (Vα )} is a cover of K, which is open because f is continuous. Thus it has a finite subcover {f −1 (Vαi )}. Then {Vαi } is a finite subcover of {Vα }. Sequential compactness. [See Rudin, pages 51–52.] Compactness can be formulated in several ways w ...
... Proof : Let {Vα } be an open cover of Y. Then {f −1 (Vα )} is a cover of K, which is open because f is continuous. Thus it has a finite subcover {f −1 (Vαi )}. Then {Vαi } is a finite subcover of {Vα }. Sequential compactness. [See Rudin, pages 51–52.] Compactness can be formulated in several ways w ...
Norm continuity of weakly continuous mappings into Banach spaces
... distinct classes of Banach spaces. On the other hand we prove in Section 3, Corollary 3, that the classes T and L coincide. We also show (see Proposition 3) that E = l∞ and E = l∞ /c0 do not belong to T . In both cases we explicitly describe how to construct a weakly continuous mapping h : Z → E def ...
... distinct classes of Banach spaces. On the other hand we prove in Section 3, Corollary 3, that the classes T and L coincide. We also show (see Proposition 3) that E = l∞ and E = l∞ /c0 do not belong to T . In both cases we explicitly describe how to construct a weakly continuous mapping h : Z → E def ...
