{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# sol2 - SOLUTIONS FOR HOMEWORK 2 1.3.8 For k N let Ak =...

This preview shows pages 1–2. Sign up to view the full content.

SOLUTIONS FOR HOMEWORK 2 1.3.8. For k N let A k = { 1 , 2 ,...,k } . Then | A k | = k , yet k =1 A k = N . 1.3.12. Let F m be the family of subsets of N with precisely m elements. Let N m = N × ... × N ( m times) be the m -fold Cartesian product of N with itself, that is, the set of all ordered m -tuples whose elements are positive integers. Use induction on m to prove that N m is countable. The case of m = 1 is clear. The case of m = 2 was proved in class: there exists a bijection from N to N × N . This forms the base of induction. Now, suppose N m is countable, and prove the same for N m +1 . By the induction hypothesis, there exists an injection f m : N m N . For a = ( a 1 ,...,a m ,a m +1 ) N m +1 , let a = ( a 1 ,...,a m ). Define f m +1 by setting f m +1 ( a ) = f 2 ( f m ( a ,a m +1 )). You can check that f m +1 is injective, hence N m is countable. Next we define a map g : F m N m , taking a set of m elements to the m -tuple, listing them in the increasing order. This is an injection, hence so is f m +1 g . Therefore, F m is countable.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}