Exercise 5.2.4

The set of all closed subsets of reals has cardinality 2 0 .

Answers

Proof. Let C denote all the closed subsets of 𝑹 and O the open sets. We form a mapping f : C O defined by

f ( A ) = 𝑹 A

for A C . Clearly by definition f ( A ) is open for every A C since A is closed.

Now consider any A , B C where A B . Then there is an a A such that a B or vice versa. Without loss of generality we can assume the former. Then since a A it follows that a 𝑹 A . But also since a B (but a 𝑹 ) we have that a 𝑹 B . Thus

f ( A ) = 𝑹 A 𝑹 B = f ( B )

so that f is injective.

Now consider any B O and let A = 𝑹 B .

f ( A ) = 𝑹 A = 𝑹 ( 𝑹 B ) = B

so that f is also surjective. Hence we have that

| C | = | O | = 2 0

by Theorem 2.6b. □

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2024-07-15 11:42
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