Jachin Shen

# homework

## Feasible Intersection Patterns

### Problem

Let us call a sequence $(a_1, a_2, \dots, a_n) \in \mathbb{N}_0 \quad feasible$ if there are sets $A_1, \dots, A_n$ such that all k-wise intersections have size $a_k$.

That is, $\lvert A_i \rvert = a_1$ for all $i$, $\lvert A_i \cap A_j \rvert = a_2$ for all $i \neq j$ and so on.

For example, $(5, 3, 1, 0)$ is not feasible, but $(6, 3, 1, 0)$ is.

### Hint

A problem may become easier if you:

1. make it more general
2. change it a little bit

### A-Table

#### Definition

Given $A_1, A_2, \dots, A_n$ , $I \subseteq {1, 2, \dots, n}$

define $A_I = \bigcap_{i \subseteq I} A_i$

#### Example

$A_1 = \lbrace 1, 2, 3, 4 ,7 \rbrace$ $A_2 = \lbrace 3, 4, 5, 6 \rbrace$ $A_3 = \lbrace 1, 2, 3, 5, 8, 9 \rbrace$

Then, $A_{ \lbrace 1, 2\rbrace } = A_1 \bigcap A_2 = \lbrace 3, 4 \rbrace$

#### Intersection table

I {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3}
$\lvert A_I \rvert$ 5 4 6 2 3 2 1

#### $(a_1, a_2, \dots, a_n)$ to A_Table

I {1} {2} $\dots$ {n} {1, 2} $\dots$ {n-1, n} $\dots$ {1, 2, $\dots$, n}
$\lvert A_I \rvert$ $a_1$ $a_1$ $\dots$ $a_1$ $a_2$ $\dots$ $a_2$ $\dots$ $a_n$

#### Is it easy to judge the table feasible?

Absolutely, No!!!

### B-Table

#### Definition of B-Table

Given $A_1, A_2, \dots, A_n$, $\varnothing \neq I \subseteq [n]$

define $B_I = (\bigcap_{i \in I} A_i) \setminus (\bigcup_{j \notin I} A_j)$

#### Example of B-Table

$A_1 = \lbrace 1, 2, 3, 4 ,7 \rbrace$ $A_2 = \lbrace 3, 4, 5, 6 \rbrace$ $A_3 = \lbrace 1, 2, 3, 5, 8, 9 \rbrace$

Then, $B_{ \lbrace 1 \rbrace } = A_1 \setminus (A_2 \bigcap A_3) = \lbrace 7 \rbrace$

#### Table

I {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3}
$\lvert A_I \rvert$ 5 4 6 2 3 2 1
$\lvert B_I \rvert$ 1 1 2 1 2 1 1

#### Given a B-Table, is it easy to judge it feasible

Yes. Just every element is Non-negative integer.

#### Given a B-Table, how to compute an A-Table

For a set P in A-Table, adding all the set Q that $P \subset Q$.

#### Given a A-Table, how to compute an B-Table

Suppose we know B-Table. Express A-Table as above. Then solve equations to get B-Table.

## Infinity sets

### Examples

$\mathbb{I, N_0, Q, R, N}$

### Claim

#### $\mathbb{N_0, N}$ have same size

$\mathbb{N}$ 1 2 3 4 5 $\dots$
$\mathbb{N_0}$ 0 1 2 3 4 $\dots$

There is a bijection function that:

$f: \mathbb{N} \to \mathbb{N_0} \quad x \mapsto x-1$

#### Definition of cardinality

Let $\mathbf{A, B}$ be sets if there is a bijection $\mathbf{A} \leftrightarrow \mathbf{B}$, then we can say $\mathbf{A}$ and $\mathbf{B}$ have the same cardinality (not use size because it sounds like science) and mark $\mathbf{A} \cong \mathbf{B}$

#### More examples

• $\mathbb{N_0} \cong \mathbb{Z}$
$\mathbb{Z}$ $\dots$ -4 -3 -2 -1 0 1 2 3 4 $\dots$
$\mathbb(N_0)$ $\dots$ 7 5 3 1 0 2 4 6 8 $\dots$
$f: \mathbb{Z} \to \mathbb{N_0} \quad x \mapsto \begin{cases} 2x, & \text{if x \le 0 } \\ -2x-1, & \text{if x < 0 } \\ \end{cases}$
• $\mathbb{Z} \times \mathbb{Z} \cong \mathbb{N}$

$f: \mathbb{N} \to \mathbb{Z} \times \mathbb{Z} \quad x \mapsto the \ xth \ point \ on \ the \ path$
• $\mathbb{N_0} \cong \mathbb{Q}$

$f: \mathbb{Q} \to \mathbb{Z} \times \mathbb{Z} \\ q = \frac{a}{b} \ where \ b \in \mathbb{N} \ a \in \mathbb{Z}, gcd(a, b) = 1$ $see \quad q \to (a, b) \\ \begin{cases} injection: q_1 \neq q_2 \Rightarrow f(q_1) \neq f(q_2) \\ surjection: \forall (a, b) \in \mathbb{Z} \times \mathbb{Z} \quad \exists q \in \mathbb(Q) \quad f(q) = (a, b) ? \quad No! \end{cases}$

So how?

• Another Definition

If there is an injection $f: \mathbf{A} \to \mathbf{B}$, then we can say $\mathbf{A} \subseteq \mathbf{B}$ and $\mathbf{A}$’s cardinality is at most $\mathbf{B}$’s cardinality.

• Therefore, $\mathbb{Q} \le \mathbb{Z} \times \mathbb{Z}$

• How about $\mathbb{Z} \times \mathbb{Z} \le \mathbb{Q}$? $\mathbb{Z} \times \mathbb{Z} \to \mathbb{N_0} \to \mathbb{Q} \quad if (x \mapsto x )$

• So, how about $\mathbb{N_0} \cong \mathbb{Q}$

• $\mathbb{Q} \subseteq \mathbb{Z} \times \mathbb{Z} \subseteq \mathbb{N_0}$
• Obviously, $\mathbb{N} \subseteq \mathbb{Q}$
• So, $\mathbb{N} \cong \mathbb{Q}$
• $\lbrace 0, 1 \rbrace^N \cong 2^N$

• $2^N$: power set, the set of all subsets. So every number $\in 2^N$
• $\lbrace 0, 1 \rbrace^N$: the set of all infinite but sequence $(a_1, a_2, \dots) \quad a_i \in \lbrace 0, 1 \rbrace$

$010101 \dots \mapsto \lbrace 2, 4, 6, \dots \rbrace$ $0110101 \dots \mapsto \lbrace 2, 3, 5, 7, \dots \rbrace$ $1111 \dots \mapsto \mathbb{N}$

• $\lbrace 0, 1 \rbrace^N \cong \mathbb{R}$
• $f: \mathbb{R} \mapsto \lbrace 0, 1 \rbrace^N$, injection?
• $g: [0, 1) \mapsto \lbrace 0, 1 \rbrace^N$:x write in binary as $(x)_2 = 0.a_1 a_2 \dots$
• Every $x \in [0, 1) \mapsto$binary, $x \mapsto (a_1, a_2, \dots, a_n) \in \lbrace 0, 1 \rbrace^N$
• $\frac{1}{2}: 0.1000\dots$
• $\frac{1}{3}: 0.010101\dots$
• $\frac{1}{4}: 0.0100\dots$
• Injection, but not surjection!