Topology I
... checked that this is an equivalence relation. The equivalence classes (which form a partition of U) must be open intervals (because U is open). Since each open interval can be coded by one of the rational numbers that it contains and since there are countably many rational numbers, there are countab ...
... checked that this is an equivalence relation. The equivalence classes (which form a partition of U) must be open intervals (because U is open). Since each open interval can be coded by one of the rational numbers that it contains and since there are countably many rational numbers, there are countab ...
Lecture 1
... ⇒ Let (X, τ ) be a topological space, x ∈ X and F(x) the filter of neighbourhoods of x. Then (N1) trivially holds by definition of neighbourhood of x. To show (N2), let us take A ∈ F(x). Since A is a neighbourhood of x, there exists B ∈ τ s.t. x ∈ B ⊆ A. Then clearly B ∈ F(x). Moreover, since for an ...
... ⇒ Let (X, τ ) be a topological space, x ∈ X and F(x) the filter of neighbourhoods of x. Then (N1) trivially holds by definition of neighbourhood of x. To show (N2), let us take A ∈ F(x). Since A is a neighbourhood of x, there exists B ∈ τ s.t. x ∈ B ⊆ A. Then clearly B ∈ F(x). Moreover, since for an ...
4 Open sets and closed sets
... exists aSball B(x) about x lying in A. We have A = x∈A B(x). Indeed, the union x∈A B(x) is a subset of A because every ball B(x) is a subset of A, and the union contains every point x ∈ A because x ∈ B(x). Definition 4.7. The interior of a set A is the union of all open sets contained in A, that is, ...
... exists aSball B(x) about x lying in A. We have A = x∈A B(x). Indeed, the union x∈A B(x) is a subset of A because every ball B(x) is a subset of A, and the union contains every point x ∈ A because x ∈ B(x). Definition 4.7. The interior of a set A is the union of all open sets contained in A, that is, ...
b*-Continuous Functions in Topological Spaces
... Proof: (i)Assume that f : X → Y is b∗ continu- ous. Let M be open in Y . Then M c is closed in Y . Since Y is b∗ -continuous f −1 (M c ) is b∗ -closed in X . But f −1 (M c ) = X − f −1 (G). Thus X − f −1 (G) is b∗ -closed in X and so f −1 (M ) is b∗ -open in X . Therefore (i) ⇒ (ii). conversely assu ...
... Proof: (i)Assume that f : X → Y is b∗ continu- ous. Let M be open in Y . Then M c is closed in Y . Since Y is b∗ -continuous f −1 (M c ) is b∗ -closed in X . But f −1 (M c ) = X − f −1 (G). Thus X − f −1 (G) is b∗ -closed in X and so f −1 (M ) is b∗ -open in X . Therefore (i) ⇒ (ii). conversely assu ...
Separation Properties - University of Wyoming
... U = f −1 ( 0, 13 ) and V = f −1 ( 23 , 1 ) are disjoint open sets separating K from L. Conversely, suppose X is normal, and let K, L ⊆ X be disjoint closed sets. Let U1 = X ........ L. By Lemma 2.2 we choose an open set U0 with K ⊆ U0 ⊆ U0 ⊆ U1 = X ........ L. Similarly, we find an open set U 21 wit ...
... U = f −1 ( 0, 13 ) and V = f −1 ( 23 , 1 ) are disjoint open sets separating K from L. Conversely, suppose X is normal, and let K, L ⊆ X be disjoint closed sets. Let U1 = X ........ L. By Lemma 2.2 we choose an open set U0 with K ⊆ U0 ⊆ U0 ⊆ U1 = X ........ L. Similarly, we find an open set U 21 wit ...
Separation axioms
... These are disjoint neighborhoods of x and y respectively showing that (R, Tll ) is T 2. To verify separation axiom T5 , let A, B ⊆ X be two separated sets. Then X − B̄ is an open set and so we can, for each a ∈ A ⊂ X − B̄, find an xa ∈ X such that [a, xa i ⊂ X − B̄ (since the half-open intervals are ...
... These are disjoint neighborhoods of x and y respectively showing that (R, Tll ) is T 2. To verify separation axiom T5 , let A, B ⊆ X be two separated sets. Then X − B̄ is an open set and so we can, for each a ∈ A ⊂ X − B̄, find an xa ∈ X such that [a, xa i ⊂ X − B̄ (since the half-open intervals are ...
PRESERVATION OF COMPLETENESS BY SOME CONTINUOUS
... Proof of Theorem 3.1. Let Cn form a sequence of scattered (or open, or locally finite open, or S uniform open, respectively) covers of X. Put Pn = { F : F ⊂ C1 ∧· · ·∧Cn finite}. Then each Pn is obviously closed with respect to unions of finite subfamilies, and it is easy to check that the condition ...
... Proof of Theorem 3.1. Let Cn form a sequence of scattered (or open, or locally finite open, or S uniform open, respectively) covers of X. Put Pn = { F : F ⊂ C1 ∧· · ·∧Cn finite}. Then each Pn is obviously closed with respect to unions of finite subfamilies, and it is easy to check that the condition ...
