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Exercise 2.2: Let and let . Verify that is a topology on . (Hence, is a topological space.)

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baruch_shahi[]

That and are in is obvious.

Now, if we intersect with anything else in , we get back. Consider the following pairwise intersections

Since intersection is a commutative operation on sets, we have accounted for all possible pairwise intersections of elements of . Notice that every intersection is an element of . Thus, is closed with respect to finite intersection.

By inspection, an arbitrary union of elements in can never yield anything other than the elements of . More rigorously, any union containing must be equal to . Any union containing but not must be equal to . Any union containing just and is equal to . Obviously any union only containing is just . Hence, is closed with respect to arbitrary union.

Thus, is a topology on .

baruch_shahi 09:21, March 22, 2011 (UTC -5)

Thanks for working through the details! --Steven.clontz 19:32, March 22, 2011 (UTC)

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