Sec.%208%20Cardinality%20II

Sec.%208%20Cardinality%20II - Math 350 Homework Solutions...

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

View Full Document Right Arrow Icon
Math 350 Advanced Calculus I Fall 2009 Homework Solutions Section 8 CARDINALITY Text Problems 8.10 Prove: For all sets S , if S is denumerable, then S is equinumerous with a proper subset of itself. Proof: Suppose S is denumerable. We need to show there exists a proper subset T of S such that S T . So we need to ±rst come up with a subset T and then show that S is equinumerous with this subset T . Since S is denumerable, then S is equinumerous with N so we can list the elements in S S = { s 1 , s 2 , s 3 , . . . } , where all elements in the listing are distinct. Let T be the set of all elements of S with an even subscripts in this listing T = { s 2 , s 4 , s 6 , . . . } . Then T is a proper subset of S . We now need to show that S is equinumerous with T . To show this, we need to show there exists a bijection g : S T . De±ne g : S T by g ( s n ) = s 2 n for each s n S (or for each n N ). Injective. To show that g is injective, we need to show ( s n S ) ( s m S ) [ g ( s m ) = g ( s n ) s m = s n ] . Suppose s m and s n are arbitrary elements in S . Suppose g ( s m ) = g ( s n ) . We need to show s m = s n . Since g ( s m ) = s 2 m and g ( s n ) = s 2 n , then g ( s m ) = g ( s n ) implies s 2 m = s 2 n . Since the elements in the listing for S are distinct, then this implies 2 m = 2 n so m = n . But then s m = s n , as required. Therefore
Background image of page 1

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

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

Page1 / 3

Sec.%208%20Cardinality%20II - Math 350 Homework Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online