Lecture 11: Countability Infjnity is weeeeeeird 1 / 20 What Is - PowerPoint PPT Presentation

Same Infjnities How can we prove this? Need a bijection! Don’t need all functions bijective! Only need one. Adding one elt to infjnite set doesn’t seem to change size... what if we added more? 5 / 20 Claim : | N | = | Z + | 1 Claim : f ( x ) = x + 1 is bijection N → Z + Why? Has inverse f − 1 ( y ) = y − 1 But what about f ( x ) = x ? Not onto! 1 Here | S | means the cardinality or “size” of S

Same Infjnities How can we prove this? Need a bijection! Don’t need all functions bijective! Only need one. Adding one elt to infjnite set doesn’t seem to change size...what if we added more? 5 / 20 Claim : | N | = | Z + | 1 Claim : f ( x ) = x + 1 is bijection N → Z + Why? Has inverse f − 1 ( y ) = y − 1 But what about f ( x ) = x ? Not onto! 1 Here | S | means the cardinality or “size” of S

Inverse is f 1 y More Infjnities x 1 0 y 2 y 0 y 1 2 y x is even 2 x x is odd 2 Take f x ... 2 4 2 3 1 2 1 1 0 0 ? to How can we map from 6 / 20 Claim : | N | = | Z |

Inverse is f 1 y More Infjnities 2 0 y 2 y 0 y 1 2 y x is even 2 x x is odd x 1 Take f x ... 2 4 2 3 1 2 1 1 0 0 6 / 20 Claim : | N | = | Z | How can we map from N to Z ?

Inverse is f 1 y More Infjnities 2 0 y 2 y 0 y 1 2 y x is even 2 x x is odd x 1 Take f x ... 2 4 2 3 1 2 1 1 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0

Inverse is f 1 y More Infjnities x is odd 0 y 2 y 0 y 1 2 y x is even 2 x 2 x 1 Take f x ... 2 4 2 3 1 2 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1

Inverse is f 1 y More Infjnities x is odd 0 y 2 y 0 y 1 2 y x is even 2 x 2 x 1 Take f x ... 2 4 2 3 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1 2 → − 1

Inverse is f 1 y More Infjnities x 0 y 2 y 0 y 1 2 y x is even 2 x is odd 2 x 1 Take f x ... 2 4 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1 2 → − 1 3 → 2

Inverse is f 1 y More Infjnities x 0 y 2 y 0 y 1 2 y x is even 2 x is odd 2 x 1 Take f x ... 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1 2 → − 1 3 → 2 4 → − 2

Inverse is f 1 y More Infjnities x 0 y 2 y 0 y 1 2 y x is even 2 x is odd 2 x 1 Take f x ... 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1 2 → − 1 3 → 2 4 → − 2

More Infjnities ... x is even 2 x is odd 2 6 / 20 Claim : | N | = | Z | How can we map from N to Z ? 0 → 0 1 → 1 2 → − 1 3 → 2 4 → − 2 { x + 1 Take f ( x ) = − x { 2 y − 1 y > 0 Inverse is f − 1 ( y ) = − 2 y y ≤ 0

Bijection Alternatives 2 3 not valid for 3 2 1 Ex: 0 1 2 Careful — need fjnite position for any element! 3 1 2 Explicitly stating a bijection can be a pain... 0 1 , can enumerate as Ex: For Need to eventually hit every element List “1st” elt of S , then “2nd”, then “3rd”, etc. , can give enumeration of S : To prove S What alternatives do we have? 7 / 20

Bijection Alternatives 2 3 not valid for 3 2 1 Ex: 0 1 2 Careful — need fjnite position for any element! 3 1 2 Explicitly stating a bijection can be a pain... 0 1 , can enumerate as Ex: For Need to eventually hit every element List “1st” elt of S , then “2nd”, then “3rd”, etc. , can give enumeration of S : To prove S What alternatives do we have? 7 / 20

Bijection Alternatives 2 3 not valid for 3 2 1 Ex: 0 1 2 Careful — need fjnite position for any element! 3 1 2 Explicitly stating a bijection can be a pain... 0 1 , can enumerate as Ex: For Need to eventually hit every element List “1st” elt of S , then “2nd”, then “3rd”, etc. What alternatives do we have? 7 / 20 To prove | S | = | N | , can give enumeration of S :

Bijection Alternatives Explicitly stating a bijection can be a pain... What alternatives do we have? List “1st” elt of S , then “2nd”, then “3rd”, etc. Need to eventually hit every element Careful — need fjnite position for any element! Ex: 0 1 2 1 2 3 not valid for 7 / 20 To prove | S | = | N | , can give enumeration of S : Ex: For Z , can enumerate as 0 , 1 , − 1 , 2 , − 2 , 3 , − 3 , ...

Bijection Alternatives Explicitly stating a bijection can be a pain... What alternatives do we have? List “1st” elt of S , then “2nd”, then “3rd”, etc. Need to eventually hit every element Careful — need fjnite position for any element! 7 / 20 To prove | S | = | N | , can give enumeration of S : Ex: For Z , can enumerate as 0 , 1 , − 1 , 2 , − 2 , 3 , − 3 , ... Ex: 0 , 1 , 2 , ..., − 1 , − 2 , − 3 , ... not valid for Z

Enumeration Example Theorem : 0 1 Could give bijection, but lots of words Instead, enumerate: 0 1 00 01 10 11 000 001 010 Any string with fjnite length hit eventually! 8 / 20 Defjnition : { 0 , 1 } ∗ is set of fjnite bit strings

Enumeration Example Could give bijection, but lots of words Instead, enumerate: 0 1 00 01 10 11 000 001 010 Any string with fjnite length hit eventually! 8 / 20 Defjnition : { 0 , 1 } ∗ is set of fjnite bit strings Theorem : |{ 0 , 1 } ∗ | = | N |

Enumeration Example Could give bijection, but lots of words Instead, enumerate: 0 1 00 01 10 11 000 001 010 Any string with fjnite length hit eventually! 8 / 20 Defjnition : { 0 , 1 } ∗ is set of fjnite bit strings Theorem : |{ 0 , 1 } ∗ | = | N |

Enumeration Example Could give bijection, but lots of words Instead, enumerate: Any string with fjnite length hit eventually! 8 / 20 Defjnition : { 0 , 1 } ∗ is set of fjnite bit strings Theorem : |{ 0 , 1 } ∗ | = | N | ϵ, 0 , 1 , 00 , 01 , 10 , 11 , 000 , 001 , 010 , ...

Enumeration Example Could give bijection, but lots of words Instead, enumerate: Any string with fjnite length hit eventually! 8 / 20 Defjnition : { 0 , 1 } ∗ is set of fjnite bit strings Theorem : |{ 0 , 1 } ∗ | = | N | ϵ, 0 , 1 , 00 , 01 , 10 , 11 , 000 , 001 , 010 , ...

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N |

Have Some More Enumeration Should be surprising — seems like many more pairs! Proof by picture: 9 / 20 Theorem : | Z × Z | = | N | Gives an enumeration of Z × Z !

b in lowest terms, f q 1 4 , f 0 66 Be Rational! Can we conclude anything from this? But notice: is one-to-one 1 , 2 4 , ...) 1 No! Not onto (eg 1 0 , Is f a bijection? 2 3 , etc. 2 1 , f 0 25 f 2 a b a If q as follows: Can create function f 10 / 20 How does | Q | compare to | Z × Z | ?

1 4 , f 0 66 Be Rational! f 2 2 1 , f 0 25 2 3 , etc. Is f a bijection? No! Not onto (eg 1 0 , 1 1 , 2 4 , ...) But notice: is one-to-one Can we conclude anything from this? 10 / 20 How does | Q | compare to | Z × Z | ? Can create function f : Q → Z × Z as follows: If q = a b in lowest terms, f ( q ) = ( a , b )

Is f a bijection? Be Rational! No! Not onto (eg 1 0 , 1 1 , 2 4 , ...) But notice: is one-to-one Can we conclude anything from this? 10 / 20 How does | Q | compare to | Z × Z | ? Can create function f : Q → Z × Z as follows: If q = a b in lowest terms, f ( q ) = ( a , b ) f ( 2 ) = ( 2 , 1 ) , f ( 0 . 25 ) = ( 1 , 4 ) , f ( 0 . 66 ) = ( 2 , 3 ) , etc.

Is f a bijection? Be Rational! No! Not onto (eg 1 0 , 1 1 , 2 4 , ...) But notice: is one-to-one Can we conclude anything from this? 10 / 20 How does | Q | compare to | Z × Z | ? Can create function f : Q → Z × Z as follows: If q = a b in lowest terms, f ( q ) = ( a , b ) f ( 2 ) = ( 2 , 1 ) , f ( 0 . 25 ) = ( 1 , 4 ) , f ( 0 . 66 ) = ( 2 , 3 ) , etc.

Is f a bijection? Be Rational! But notice: is one-to-one Can we conclude anything from this? 10 / 20 How does | Q | compare to | Z × Z | ? Can create function f : Q → Z × Z as follows: If q = a b in lowest terms, f ( q ) = ( a , b ) f ( 2 ) = ( 2 , 1 ) , f ( 0 . 25 ) = ( 1 , 4 ) , f ( 0 . 66 ) = ( 2 , 3 ) , etc. No! Not onto (eg ( 1 , 0 ) , ( − 1 , − 1 ) , ( 2 , 4 ) , ...)

Is f a bijection? Be Rational! But notice: is one-to-one Can we conclude anything from this? 10 / 20 How does | Q | compare to | Z × Z | ? Can create function f : Q → Z × Z as follows: If q = a b in lowest terms, f ( q ) = ( a , b ) f ( 2 ) = ( 2 , 1 ) , f ( 0 . 25 ) = ( 1 , 4 ) , f ( 0 . 66 ) = ( 2 , 3 ) , etc. No! Not onto (eg ( 1 , 0 ) , ( − 1 , − 1 ) , ( 2 , 4 ) , ...)

What Is An Outjection? B and B A ! A means B So surjection B A B ifg have surject B Note: Have inject A B ! A , CSB says A If A Cantor-Schröder-Bernstein Theorem B A injection f B if Can say A What does this mean to us? Proof in Bonus Lecture tomorrow! 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection

What Is An Outjection? B and B A ! A means B So surjection B A B ifg have surject B Note: Have inject A B ! A , CSB says A If A Cantor-Schröder-Bernstein Theorem B A injection f B if Can say A What does this mean to us? Proof in Bonus Lecture tomorrow! 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection

What Is An Outjection? B and B A ! A means B So surjection B A B ifg have surject B Note: Have inject A B ! A , CSB says A If A Cantor-Schröder-Bernstein Theorem B A injection f B if Can say A What does this mean to us? Proof in Bonus Lecture tomorrow! 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection

What Is An Outjection? Cantor-Schröder-Bernstein Theorem Proof in Bonus Lecture tomorrow! What does this mean to us? Note: Have inject A B ifg have surject B A So surjection B A means B A ! 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection Can say | A | ≤ | B | if ∃ injection f : A → B If | A | ≤ | B | and | B | ≤ | A | , CSB says | A | = | B | !

What Is An Outjection? Cantor-Schröder-Bernstein Theorem Proof in Bonus Lecture tomorrow! What does this mean to us? So surjection B A means B A ! 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection Can say | A | ≤ | B | if ∃ injection f : A → B If | A | ≤ | B | and | B | ≤ | A | , CSB says | A | = | B | ! Note: Have inject A → B ifg have surject B → A

What Is An Outjection? Cantor-Schröder-Bernstein Theorem Proof in Bonus Lecture tomorrow! What does this mean to us? 11 / 20 If ∃ injections f : A → B and g : B → A , ∃ bijection Can say | A | ≤ | B | if ∃ injection f : A → B If | A | ≤ | B | and | B | ≤ | A | , CSB says | A | = | B | ! Note: Have inject A → B ifg have surject B → A So surjection B → A means | B | ≥ | A | !

Hence, Notice, have injection by “inclusion” So Thus ! 12 / 20 Back To Q Previously: found injection Q → Z × Z

Notice, have injection by “inclusion” So Thus ! 12 / 20 Back To Q Previously: found injection Q → Z × Z Hence, | Q | ≤ | Z × Z | = | N |

! Thus 12 / 20 Back To Q Previously: found injection Q → Z × Z Hence, | Q | ≤ | Z × Z | = | N | Notice, have injection N → Q by “inclusion” So | N | ≤ | Q |

12 / 20 Back To Q Previously: found injection Q → Z × Z Hence, | Q | ≤ | Z × Z | = | N | Notice, have injection N → Q by “inclusion” So | N | ≤ | Q | Thus | Q | = | N | !

Brake Time for a 4-minute break! Today’s Discussion Question : https://tinyurl.com/70-discussion-q 13 / 20

Brake Time for a 4-minute break! Today’s Discussion Question : https://tinyurl.com/70-discussion-q 13 / 20

Countability So far, all sets we’ve seen are countable! Natural question: are all sets countable? Turns out, no! 14 / 20 Say a set S is countable if | S | ≤ | N |

Countability So far, all sets we’ve seen are countable! Natural question: are all sets countable? Turns out, no! 14 / 20 Say a set S is countable if | S | ≤ | N |

Countability So far, all sets we’ve seen are countable! Natural question: are all sets countable? Turns out, no! 14 / 20 Say a set S is countable if | S | ≤ | N |

Countability So far, all sets we’ve seen are countable! Natural question: are all sets countable? Turns out, no! 14 / 20 Say a set S is countable if | S | ≤ | N |

Not With That Attitude You Cant-or 1 1 1 0 1 ... o n for all n ! s 1101... Consider s . . . . . . 0 1 0 0 0 ... 3 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 o n n 0 1 onto fn o Suppose for contra Proof : 0 1 Theorem : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings

Not With That Attitude You Cant-or 3 o n for all n ! s 1101... Consider s . . . . . . 0 1 0 0 0 ... 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 o n n 0 1 onto fn o Suppose for contra Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N |

Not With That Attitude You Cant-or 0 1 0 0 0 ... o n for all n ! s 1101... Consider s . . . . . . 3 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n )

Not With That Attitude You Cant-or 0 1 0 0 0 ... o n for all n ! s 101... . . . . . . 3 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 1

Not With That Attitude You Cant-or 0 1 0 0 0 ... o n for all n ! s 01... . . . . . . 3 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 11

Not With That Attitude You Cant-or 0 1 0 0 0 ... o n for all n ! s 1... . . . . . . 3 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 110

Not With That Attitude You Cant-or 0 1 0 0 0 ... o n for all n ! s ... . . . . . . 3 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 1101

Not With That Attitude You Cant-or 3 o n for all n ! s . . . . . . 0 1 0 0 0 ... 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 1101...

Not With That Attitude You Cant-or 2 . . . . . . 0 1 0 0 0 ... 3 1 1 1 0 1 ... 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 1101... s ̸ = o ( n ) for all n !

Not With That Attitude You Cant-or 3 Method known as Cantor Diagonalization . . . . . . 0 1 0 0 0 ... 1 1 1 0 1 ... 2 1 0 1 0 1 ... 1 0 0 0 0 0 ... 0 n Proof : 15 / 20 Def : Let { 0 , 1 } ∞ be set of infjnite length bit strings Theorem : |{ 0 , 1 } ∞ | > | N | Suppose for contra ∃ onto fn o : N → { 0 , 1 } ∞ o ( n ) Consider s = 1101... s ̸ = o ( n ) for all n !