- Undefinition of |
*A*|, defined |*A*|≤|*B*|, |*A*|≥|*B*|, |*A*|=|*B*| - Listed properties to check
- Defined countability
- Considered some infinite sets
*p**o**w*(ℕ) is an uncountable set- board image

**Undefinition:** Although we denote |*A*| to mean the number of elements of *A* for finite sets, we will **not** use this definition in this lecture (**or the homework!**), because it does not apply to infinite sets. We will not give a definition of |*A*| at all, in fact.

**Definitions:** - We define the expression |*A*|≥|*B*| to indicate that there exists a surjection from *A* to *B*. Equivalently, using the MCS notation, *A* *s**u**r**j* *B*. - We define the expression |*A*|≤|*B*| to indicate that there exists an injection from *A* to *B*. Equivalently, using the MCS notation, *A* *i**n**j* *B*. - We define the expression |*A*|=|*B*| to indicate that there exists a bijection from *A* to *B*. Equivalently, using the MCS notation, *A* *b**i**j* *B*.

**Mnemonic note:** I find it convenient to remember that all the functions go from the set on the left hand side (*A*) to the set on the right hand side (*B*). To remember whether injection or surjection goes with bigger or smaller, I usually draw a picture.

We have redefined ≤, ≥, and = for cardinalities (even though we haven't defined cardinalities!). In order for this to be sensible, we should check several things:

- |
*A*|≤|*A*|, |*A*|≥|*A*|, and |*A*|=|*A*| (reflexivity) - if |
*A*|≤|*B*| and |*B*|≤|*C*| then |*A*|≤|*C*| (transitivity) - if |
*A*|≤|*B*| and |*A*|≥|*B*| then |*A*|=|*B*| (antisymmetry) - |
*A*|≤|*B*| if and only if |*B*|≥|*A*| - if we consider the other definition of cardinality for finite sets, these definitions match

We checked one of them; we'll put another on the homework.

**Claim**: If |*A*|≤|*B*| and |*B*|≤|*C*| then |*A*|≤|*C*|.

**Proof**: Suppose |*A*|≤|*B*| and |*B*|≤|*C*|. Then there exist injections *f* : *A* → *B* and *g* : *B* → *C*. Consider the function *g* ∘ *f* : *A* → *C*. I claim that *g* ∘ *f* is injective (and thus |*A*|≤|*C*|). To see this, we must show that for all *a*_{1}, *a*_{2} ∈ *A*, if *g* ∘ *f*(*a*_{1})=*g* ∘ *f*(*a*_{2}) then *a*_{1} = *a*_{2}, so let's choose an arbitrary *a*_{1} and *a*_{2} and assume *g* ∘ *f*(*a*_{1})=*g* ∘ *f*(*a*_{2}).

Now, if we define *b*_{1} = *f*(*a*_{1}) and *b*_{2} = *f*(*a*_{2}) then *g*(*b*_{1})=*g*(*f*(*a*_{1})) = (*g* ∘ *f*)(*a*_{1})=(*g* ∘ *f*)(*a*_{2})=*g*(*f*(*a*_{2})) = *g*(*b*_{2})

Since *g* is injective, this means *b*_{1} = *b*_{2}, and thus *f*(*a*_{1})=*f*(*a*_{2}). Since *f* is also injective, we see that *a*_{1} = *a*_{2}, which is what we were trying to prove. Thus *g* ∘ *f* is injective, so |*A*|≤|*C*|, as desired.

**Definition:** A set *X* is **countable** if |ℕ|≥|*X*| (equivalently, by unchecked fact above, if |*X*|≤|ℕ|).

Informally, ∞ ≥ ∞ + 1:

**Claim:** |ℕ|≥|ℕ ∪ { − 1}|. **Proof:** let *f* : ℕ → ℕ ∪ { − 1} be defined by *f*(*n*): := *n* − 1. I claim *f* is a surjection. To see this, choose an arbitrary *y* in ℕ ∪ { − 1}. Then *y* ≥ −1, so *x* = *y* + 1 ∈ ℕ. But *f*(*x*)=*y* + 1 − 1 = *y*. Thus we have shown that for every *y* in the codomain of *f*, there exists an *x* in the domain of *f* with *f*(*x*)=*y*. In other words, *f* is surjective, and thus |ℕ|≥|ℕ ∪ { − 1}| as claimed.

**Note:** *f* is actually a bijection, and this is not hard to prove. Thus |ℕ| is actually equal to |ℕ ∪ { − 1}|. But our proof only shows surjectivity.

Informally, ∞ ≥ ∞ × ∞:

**Claim:** |ℕ × ℕ| is countable (i.e. |ℕ|≥|ℕ × ℕ|). **Proof:** We define *f* : ℕ → ℕ × ℕ by the following procedure. Start by writing all of the elements of ℕ × ℕ in a table:

(0,0) | (0,1) | (0,2) | ⋯ |

(1,0) | (1,1) | (1,2) | ⋯ |

⋮ | ⋱ |

Now, to find *f*(*n*), traverse up-and-to-the-right diagonals, starting from the top left and working your way down. *f*(*n*) is the *n*th pair traversed. Here are the first few values of *f*:

n |
f(n) |
---|---|

0 | (0,0) |

1 | (1,0) |

2 | (0,1) |

3 | (2,0) |

4 | (1,1) |

5 | (0,2) |

6 | (3,0) |

7 | (2,1) |

This is a surjection, because every pair in the table eventually gets traversed. Thus |ℕ|≥|ℕ × ℕ| as claimed.

**Claim:** *p**o**w*(ℕ) is uncountable. **Proof:** By contradiction. Assume that *p**o**w*(ℕ) is countable. Then there is a surjection *f* : ℕ → *p**o**w*(ℕ). We will derive a contradiction by finding a set *S*_{D} that is not in the image of *f*.

We start with an example: suppose the first few values of *f* are given as follows:

n |
f(n) |
---|---|

0 | ∅ |

1 | ℕ |

2 | {2, 3, 4} |

3 | all evens |

We can expand out each row by indicating, for each *j*, whether *j* ∈ *f*(*i*):

i↓ j→ | 0 | 1 | 2 | 3 | 4 | 5 | … | j |
---|---|---|---|---|---|---|---|---|

0 | no | no | no | no | no | no | ||

1 | yes | yes | yes | yes | yes | yes | ||

2 | no | no | yes | yes | yes | no | ||

3 | yes | no | yes | no | yes | no | ||

⋮ | ⋱ | |||||||

i | j ∈ f(i)? |

Now we create the diabolical set by changing everything on the diagonal:

j→ | 0 | 1 | 2 | 3 | 4 | 5 | … | j |
---|---|---|---|---|---|---|---|---|

S_{d} |
yes | no | no | yes | … | j ∉ f(j) |

We see that *S*_{D} ≠ *f*(0) because 0 ∈ *S*_{D} but 0 ∉ *f*(0). *S*_{D} cannot be *f*(1) because 1 ∈ *f*(1) but 1 ∉ *S*_{D}. In general, *S*_{D} ≠ *f*(*i*) because if *i* ∈ *f*(*i*) then *i* ∉ *S*_{D}, and if *i* ∉ *f*(*i*) then *i* ∈ *S*_{D}.

Now the table we drew in our example is only one example, so it is not a proof. However, regardless of what *f* is, we can construct *S*_{D} = {*i*|*i* ∉ *f*(*i*)}, and the argument that *S*_{D} ≠ *f*(*i*) for any *i* is also general.

However, since *S*_{D} is not *f*(*i*) for any *i*, we see that *f* cannot be surjective! This is a contradiction, and thus *p**o**w*(ℕ) is not countable.