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... Proof. Suppose X is first countable, and A ⊆ X has the property that, if C is any compact set in X, the set A ∩ C is closed in C. We want to show tht A is closed in X. Since X is first countable, this is equivalent to showing that any sequence (xi ) in A converging to x implies that x ∈ A. Let C = { ...
... Proof. Suppose X is first countable, and A ⊆ X has the property that, if C is any compact set in X, the set A ∩ C is closed in C. We want to show tht A is closed in X. Since X is first countable, this is equivalent to showing that any sequence (xi ) in A converging to x implies that x ∈ A. Let C = { ...
Some results on linearly Lindelöf spaces
... Therefore, the subspace C 0 is a space of countable spread. By Lemma 3.3, the subspace C 0 is a Lindelöf space. Consequently, in view of the proof of Theorem 3.1, we can get B 0 ⊂ B such that B 0 is discrete in X 0 and |B 0 | = |B|. Then we let C = {x ∈ X : B 0 is not discrete at x in X}. Obviously ...
... Therefore, the subspace C 0 is a space of countable spread. By Lemma 3.3, the subspace C 0 is a Lindelöf space. Consequently, in view of the proof of Theorem 3.1, we can get B 0 ⊂ B such that B 0 is discrete in X 0 and |B 0 | = |B|. Then we let C = {x ∈ X : B 0 is not discrete at x in X}. Obviously ...
Free Topological Groups - Universidad Complutense de Madrid
... In the Abelian case, the meaning of the word “free” is, of course, slightly different. Again, one can show that if x1 , . . . , xn are pairwise distinct elements of X and k1 , . . . , kn are arbitrary integers, then the equality k1 x1 + k2 x2 + · · · + kn xn = 0A(X ) implies that k1 = k2 = · · · = ...
... In the Abelian case, the meaning of the word “free” is, of course, slightly different. Again, one can show that if x1 , . . . , xn are pairwise distinct elements of X and k1 , . . . , kn are arbitrary integers, then the equality k1 x1 + k2 x2 + · · · + kn xn = 0A(X ) implies that k1 = k2 = · · · = ...
Smooth fibrations
... Exercise 6.19. Let G be an abelian group, and k a field. Suppose that for each nonzero λ ∈ k there exists an automorphism φλ : G −→ G, such that φλ ◦ φλ0 = φλλ0 , and φλ+λ0 (g) = φλ (g) + φλ0 (g). Show that G is a vector space over k. Show that all vector spaces can be obtained this way. π ...
... Exercise 6.19. Let G be an abelian group, and k a field. Suppose that for each nonzero λ ∈ k there exists an automorphism φλ : G −→ G, such that φλ ◦ φλ0 = φλλ0 , and φλ+λ0 (g) = φλ (g) + φλ0 (g). Show that G is a vector space over k. Show that all vector spaces can be obtained this way. π ...
M132Fall07_Exam1_Sol..
... Fix an index k and show that the factor space Xk is Hausdorff. Let x, y be points in Xk with x 6= y. Since the factor spaces are all nonempty [this is where we need that hypothesis], in each space Xn (n 6= k) pick a point tn . Let x̂ ∈ Πn Xn be the point that is x in coordinate k and tn in each othe ...
... Fix an index k and show that the factor space Xk is Hausdorff. Let x, y be points in Xk with x 6= y. Since the factor spaces are all nonempty [this is where we need that hypothesis], in each space Xn (n 6= k) pick a point tn . Let x̂ ∈ Πn Xn be the point that is x in coordinate k and tn in each othe ...
AN ABSTRACT ALGEBRAIC-TOPOLOGICAL APPROACH TO THE
... For our dual spaces we clearly define algebraic operations pointwise, for instance, h1 , h2 ∈ X d , ∀x ∈ X : (h1 h2 )(x) := (h1 (x))(h2 (x)) or h ∈ X d∗ , ∀x ∈ X : h(x∗ ) := h(x). Concerning the algebraic structure of our spaces we get: (1) For the Banach space X, X is a IK-vector space and X 0 is a ...
... For our dual spaces we clearly define algebraic operations pointwise, for instance, h1 , h2 ∈ X d , ∀x ∈ X : (h1 h2 )(x) := (h1 (x))(h2 (x)) or h ∈ X d∗ , ∀x ∈ X : h(x∗ ) := h(x). Concerning the algebraic structure of our spaces we get: (1) For the Banach space X, X is a IK-vector space and X 0 is a ...
Introduction to derived algebraic geometry
... want to go that far afield, so we’ll just squeak by by only defining homotopies (and hence homotopy classes of maps). ...
... want to go that far afield, so we’ll just squeak by by only defining homotopies (and hence homotopy classes of maps). ...
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... a simplicial complex and has some nice formal properties that make it ideal for studying topology. Simplicial sets are useful because they are algebraic objects and they make it possible to do topology indirectly, using only algebra. In this paper we illustrate the use of simplicial sets in algebrai ...
... a simplicial complex and has some nice formal properties that make it ideal for studying topology. Simplicial sets are useful because they are algebraic objects and they make it possible to do topology indirectly, using only algebra. In this paper we illustrate the use of simplicial sets in algebrai ...