{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

View Full Document Right Arrow Icon
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.
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}