# Monday, November 7, 2011 Introduction to set theory

Maintainer: admin

And now onto the last serious topic, one even more basic than Peano arithmetic. The axiom system we'll be using for set theory is ZFC (Zermelo-Fraenkel set theory with the axiom of choice).

## 1Set Theory¶

### 1.1Notation¶

The $\sum$ signature for this system is $\{ \in \}$ which is a binary relation symbol and that's it.

We often will write $x \not \in X$ instead of $\lnot(x \in X)$

$x \subseteq y$ abbreviates $\forall z (z \in x \to z \in y)$. $\subseteq$ is read as "is a subset of". $\subset$, on the other hand, means "is a proper subset of" and does not include the possibility that $x$ and $y$ are the same set. We won't be using

### 1.2Russell's paradox¶

Consider the set $R = \{x: x \notin X \}$ is $R \in R$? This is called Russell's paradox.

If yes then $R \notin R$ but then... contradiction.

The usual way out of this paradox: $R$ is not a set.

### 1.3Axioms of Set Theory¶

Most axioms of set theory are constructive principles, for making sets.

#### 1.3.1Extensionality¶

$\forall x \forall y (\forall z (z \in x \iff z \in y) \to x = y)$ (a set is completely defined by what things it has as elements)

We can also write extensionality as $\forall x \forall y (x = y \leftrightarrow (x \subset y \wedge y \subset x))$.

Note that extensionality tells us something about the nature of sets, but doesn't actually give us any. The next axiom does, however.

#### 1.3.2Empty set axiom¶

$\exists x \forall y (y \notin x)$ (there is a set that has no elements)

This is called the empty set, which can be defined as follows: $\varnothing = \{y : y \neq y) = \{\}$. By extensionality, there is only one such set.

#### 1.3.3Comprehension (separation) scheme¶

The sweeping construction principle.

Let $\Phi(x, \bar{y})$ be the $\Sigma$-formula $\forall y (y \notin x)$. We then have the following axiom:

$$\forall \bar{y} [ \forall z (\exists w \forall x (x \in w \iff (x \in z \land \Phi(x, \bar{y})))]$$

In other words, there must be a set $z$ that is "bigger" than $w$1. But we need $z$ firstwhat?.

N.B. for all $x, y, z$:

• $x \subseteq x$
• $x \subseteq y \land y \subseteq z \to x = y$
• $x \subseteq y \land y \subseteq z \to x \subseteq z$
• $\varnothing \subseteq x$
• $x \subseteq \varnothing \iff x = \varnothing$

#### 1.3.4Pairing¶

$\forall x \forall y \exists z (x \in z \land y \in z)$ (note: $x$ and $y$ can be equal)

Accepting this, we can now show the existence of a set $Z$ such that $\varnothing \in Z$.

In fact, an instance of comprehension now gives us $\{ \varnothing \}$ - the set composed of only the empty set - which we can think of as the number "1" (with the empty set being the number "0"). $\Phi(x)$ in this case is $\forall u (u \in x)$.what does this mean?

Now that we have two things to play with, let's make more! But first, let's look at singletons:

#### 1.3.5Singletons¶

A singleton x ($\{ x \}$) is the set with only x as an element.

A "doubleton $x, y$" is $\{x,y\}$ where $x \neq y$.

We can now form some more sets: $\{\{ \varnothing \}\}$ from before, and $\{\varnothing, \{\varnothing\}\}$ ("2", as it has two elements). Or, $\{ 1 \}$ and $\{0, 1\}$.

Now, we would like to get $\{0, 1, 2\} = 3$ but we can't yet make a set with three elements, as pairing doesn't guarantee us anything beyond "2". We can form $\{0, 1\}$ and $\{2\}$, but because we don't have any sort of "union" axiom yet, we can't really do anything with that.

#### 1.3.6Intersections¶

We can, however, form an axiom for intersections. We define $z \cap y$ as the set $w = \{ x \in z: x \in y\}$, where $\Phi(x, y)$ is $x \in y$.

We can also form $z \backslash y$: $\{x \in z : x \notin y$.

#### 1.3.7Unions¶

(Not the kind that likes to go on strike, the other kind)

$\forall x \exists y (\forall z \forall w (z \in w \land w \in x \to z \in y))$ (so $y$ is the union of all the $x$'s - $\bigcup x$

We can now form finite and infinite sets. If the sets in $x$ are $z_1, z_2, z_3, \ldots$ then $\bigcup x = z_1 \cup z_2 \cup \ldots$. $A \cup B = \bigcup \{A, B \}$; $3 = \cup \{ \{0,1\}, \{2\}\}$. We can also form 4, 5, 6 etc.

1. As Russell's paradox demonstrates, some care needs to be taken about accepting the "set" $\{, x \Phi (x)\}$. This isn't really relevant though.