A is set of irrational number. x &#x2208;<!-- ∈ --> A , It must exist inteval that contai

Roland Waters

Roland Waters

Answered question

2022-06-27

A is set of irrational number. x A, It must exist inteval that contain y R .
Prove that the set of irrational number is neither closed or open subset on Set of real number

Answer & Explanation

Jerome Page

Jerome Page

Beginner2022-06-28Added 16 answers

Definition. Let A R . We say that A is not open if there exists x A such that for every open interval I that contains x, we have I A.
Let Q c denotes the set of irrational numbers. Choose p Q c . Then any open interval I that contains p, we have I Q c (this is because there are infinitely many rational numbers in I). Thus, Q c is not open. In the same manner, the set Q , which is the set of rationals, is not open, which means that the complement of Q c is not open. This proves that Q c is not closed.

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