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Exercise 5.2.4
The set of all closed subsets of reals has cardinality .
Answers
Proof. Let denote all the closed subsets of and the open sets. We form a mapping defined by
for . Clearly by definition is open for every since is closed.
Now consider any where . Then there is an such that or vice versa. Without loss of generality we can assume the former. Then since it follows that . But also since (but ) we have that . Thus
so that is injective.
Now consider any and let .
so that is also surjective. Hence we have that
by Theorem 2.6b. □