Prove that if a set U is countable and a set V is countable, the set U union V i

Question

Prove that if a set U is countable and a set V is countable, the set U union V is countable. To simplify your proof, assume U and V have no elements in common. b. What's wrong with the following proof? (Irrational numbers are any real number that cannot be written as a fraction, for example the squareroot of 2). The set of irrational numbers I is not countable. To show this, we use diagonalization. Assume for sake of contradiction I was countable. There would exist a bijection f from N to I and so we could list the elements of I as I_0, I_1, I_2, ... Now define the number x, where the ith decimal digit of x is 3 if the ith digit of I_i is 2, and the ith decimal digit of x is 2 otherwise. x is an irrational number, but x is not included anywhere in that list of elements. Thus, our original bijection didn't include all of the irrational numbers, we reach a contradiction, and the set cannot be countable. c. Prove correctly that the set of irrational numbers is not countable.

Explanation / Answer

a)
If U and V are countable, then (UuV) is also countable since there is one-to-end onto map
f:U->N. And also it has one to one mapping between all natiral numbers from both sets.

b)
In the first statement, we assumed that irrational numbers are countable.
But in the second statement that again discussed that x is rational number and it is not countable (given no bijections --> countable sets has bijection mapping)

c)
Irrational numbers have differnt patterns but rational numbers have fully sequence digits.
So they are countable. If a irrational numbers are countable, then union of both numbers
also must be countable. But it gives contradiction. So Irrational numbers are not countable.

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