An Illustrated Introduction to Topology and Homotopy Solutions Manual for Part 1 Topology by Kalajdzievski Sasho;Krepski Derek;Kalajdzievski Damjan;

Author:Kalajdzievski, Sasho;Krepski, Derek;Kalajdzievski, Damjan;
Language: eng
Format: epub
Publisher: CRC Press LLC
Published: 2018-01-15T00:00:00+00:00

5.2 Infinite Products of Spaces

Solutions of the odd-numbered exercises

1. Show that each projection is open.

Solution. Consider first the standard basis set , where all Ui -s are open and where in all but finitely many cases Ui = Xi. Then , and this is open. Since every open set is union of basis sets, and since for every mapping, the claim in this exercise follows.

3. Show that it is not true that if each of the spaces Xi is separable (first countable, second countable), then the space must be separable (first countable, second countable, respectively).

Solution. ℝ is separable, first countable and second countable, and it is clear that is neither first countable nor second countable.

However, and somewhat surprisingly, this space is separable. (Exercise: prove it!] In order for not to be separable it is sufficient and necessary that the index set I be of cardinality larger than . Here is a proof that is not separable if .

Denote , and assume it is separable; we prove that in that case . Under our assumption there is a countable dense set D. Denote by pi the projection X → ℝ onto the i-th component of the product space, and let U be any (non-empty) open interval . Since D is dense, each subset of D, i ∈ I, is not empty. Now we show that i ↦ Di is a one-to-one mapping , where is the power set of D. It would follow from the definition of the relation < among cardinal numbers that , which is what we wanted to show.

Choose any i, j ∈ I, i ≠ j, and let V be any non-empty open subset of ℝ that is disjoint from U. Then is open and non-empty, and so, since D is dense, there is a point x in . Now, this x is not in , since U and V are disjoint. Hence , and , implying that Di ≠ Di. This proves that i ↦ Di is one-to-one.

5. Show that if I is infinite, and if for every i ∈ I, Xi is homeomorphic to a fixed space X, then for any fixed j ∈ I, the spaces and are homeomorphic.

Solution. There is a bijection f: I → I \ {j}; to find it isolate a countable subset C containing j, find a bijection C → C \ {j} and fix the elements in I \ C. Now define as follows:, where for x ∈ X, xk is the corresponding element in Xk. It is now straightforward to show that g is a homeomorphism.

7. Show that if I is infinite then the set , where each Ui is a non-empty proper open subset of Xi, is NOT open in .

Solution. Of course. None of the standard basis elements is contained in so it could not be a union of such.

9. (a) Let X = ℕ be equipped with any topology. Show that X∞ is separable.

(b) Show that if Xi, i = 1,2,…,n,…, are spaces such that

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