Solutions to Chapter 1: Topology, Hocking and Young

Exercise 1-4: Prove that the collection of all open half planes is a subbasis for the Euclidean topology of the plane.

Proof 1-4:

Note: In the Euclidean plane, we can take as a basis the collection of all interiors of squares.

Note also that a subcollection \mathcal{B} of all open sets of a topological space S is a SUBBASIS of S provided that the collection of all finite intersections of elements of $\mathcal{B}$ is a basis for S.

Clearly, from all open half planes (x > x_{i} and y > y_{i}), we can create a collection of all interior of squares.

Hence, the collection of all finite intersections of all open-half planes satisfies:

Axiom \mathcal{O_{2}}: the intersection of a finite intersection of open half planes is an open set. (interior of a square).

Axiom \mathcal{O_{1}}; (also). The union of any number of finite intersections of all open half planes is also open set (interior of a square).

Axiom O_{3}: (clearly). \phi and S are open.

QED.

Exercise 1-5:

Let S be any infinite set. Show that requiring every infinite subset of S to be open imposes the discrete topology on S.

Proof 1-5:

Case: S is countable. We neglect the subcase that a selected subcase is finite. (I am using Prof. Rudin’s definition of countable). The other subcase is that there exists a subset X_{1} \subset S, where X_{1} is countable. Let X_{1} be open. Hence, S-X_{1} is closed. But S-X_{1} is also countable. Hence, S-X_{1} is also open. Hence, there is no limit point. Hence, the topology is discrete, that is, there is no limit point.

Case: S is uncountable. Consider again a proper subset X_{1} \subset S; hence, X_{1} is open by imposition of hypothesis. Hence, S-X_{1} is closed. But, S-X_{1} \neq \phi and not finite also. Hence, S-X_{1} is infinite. Hence, S-X_{1} is open. Hence, there are no limit points. Hence, it is a discrete topology in this case also.

QED.

Your comments/observations are welcome !

Regards,

Nalin Pithwa.

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